Inter 2nd Year – Page 18 – AP Board Solutions

AP Inter 2nd Year Physics Notes

September 3, 2022

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Physics Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Physics Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Physics Study Material and AP Inter 2nd Year Physics Important Questions for exam preparation.

AP Intermediate 2nd Year Physics Notes

TS AP Inter 2nd Year Physics Weightage Blue Print

These TS AP Intermediate 2nd Year Physics Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Physics Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

AP Inter 2nd Year Physics Important Questions Chapter Wise Pdf 2022-2023 | Sr Inter Physics Important Questions

September 3, 2022

TS AP Intermediate 2nd Year Physics Important Questions Chapter Wise 2022: Here we have created a list of Telangana & Andhra Pradesh BIEAP TS AP Intermediate Sr Inter 2nd Year Physics Important Questions Chapter Wise with Answers 2022-2023 Pdf Download just for you. Those who are preparing for Inter exams should practice the Intermediate 2nd Year Physics Important Questions Pdf and do so will clear their doubts instantly. These Senior Inter Physics Important Questions with Answers enhance your conceptual knowledge and prepares you to solve different types of questions in the exam.

Students must practice these AP Inter 2nd Year Physics Important Questions Pdf to boost their preparation for the exam paper. These AP Intermediate 2nd Year Physics Important Questions with Answers are prepared as per the latest exam pattern. Each of these chapters contains a set of solved questions and additional questions for practice.

Students can also read AP Inter 2nd Year Physics Study Material for exam preparation. Students can also go through AP Inter 2nd Year Physics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Physics Important Questions with Answers Chapter Wise 2022

Intermediate 2nd Year Physics Important Questions Chapter Wise 2022

TS AP Inter 2nd Year Physics Weightage Blue Print

We are providing well-organized Physics Important Questions Inter 2nd Year that will help students in their exam preparation. These Physics Inter 2nd Year Important Questions are designed by subject experts. So Students should utilise the Intermediate 2nd Year Physics Important Questions with Answers 2022 to pass their board exams with flying colours.

The questions given in the Inter Second Year Physics Important Questions are designed and laid out chronologically and as per the syllabus. Students can expect that these TS AP Intermediate Inter 2nd Year Physics Important Questions Chapter Wise Pdf 2022-2023 might be covered in the final exam paper.

AP Inter 2nd Year Physics Study Material Pdf | Intermediate 2nd Year Physics Textbook Solutions

September 3, 2022

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Physics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Physics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Physics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Physics Syllabus & AP Inter 2nd Year Physics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Physics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Physics Study Material Pdf Download | Sr Inter 2nd Year Physics Textbook Solutions

TS AP Inter 2nd Year Physics Weightage Blue Print 2022-2023

TS AP Inter 2nd Year Physics Weightage 2022-2023 | TS AP Inter 2nd Year Physics Blue Print 2022

TS AP Inter 2nd Year Physics Weightage Blue Print

Intermediate 2nd Year Physics Syllabus

TS AP Inter 2nd Year Physics Syllabus

Chapter 1 Waves

Introduction
Transverse and longitudinal waves
Displacement relation in a progressive wave
The speed of a traveling wave
The principle of superposition of waves
Reflection of waves
Beats
Doppler effect

Chapter 2 Ray Optics and Optical Instruments

Introduction
Reflection of Light by Spherical Mirrors
Refraction
Total Internal Reflection
Refraction at Spherical Surfaces and by Lenses
Refraction through a Prism
Dispersion by a Prism
Some Natural Phenomena due to Sunlight
Optical Instruments

Chapter 3 Wave Optics

Introduction
Huygens Principle
Refraction and reflection of plane waves using the Huygens Principle
Coherent and Incoherent Addition of Waves
Interference of Light Waves and Young’s Experiment
Diffraction
Polarisation

Chapter 4 Electric Charges and Fields

Introduction
Electric Charges
Conductors and Insulators
Charging by Induction
Basic Properties of Electric Charge
Coulomb’s Law
Forces between Multiple Charges
Electric Field
Electric Field Lines
Electric Flux
Electric Dipole
Dipole in a Uniform External Field
Continuous Charge Distribution
Gauss’s Law
Application of Gauss’s Law

Chapter 5 Electrostatic Potential and Capacitance

Introduction
Electrostatic Potential
Potential due to a Point Charge
Potential due to an Electric Dipole
Potential due to a System of Charges
Equipotential Surfaces
Potential Energy of a System of Charges
Potential Energy in an External Field
Electrostatics of Conductors
Dielectrics and Polarisation
Capacitors and Capacitance
The Parallel Plate Capacitor
Effect of Dielectric on Capacitance
Combination of Capacitors
Energy Stored in a Capacitor
Van de Graaff Generator

Chapter 6 Current Electricity

Introduction
Electric Current
Electric Currents in Conductors
Ohm’s law
The drift of Electrons and the Origin of Resistivity
Limitations of Ohm’s Law
The resistivity of various Materials
Temperature Dependence of Resistivity
Electrical Energy, Power
Combination of Resistors – Series and Parallel
Cells, emf, Internal Resistance
Cells in Series and in Parallel
Kirchhoff’s Laws
Wheatstone Bridge
Meter Bridge
Potentiometer

Chapter 7 Moving Charges and Magnetism

Introduction
Magnetic Force
Motion in a Magnetic Field
Motion in Combined Electric and Magnetic Fields
Magnetic Field due to a Current Element, Biot-SavartLaw
Magnetic Field on the Axis of a Circular Current Loop
Ampere’s Circuital Law
The Solenoid and the Toroid
The force between Two Parallel Currents, the Ampere
Torque on Current Loop, Magnetic Dipole
The Moving Coil Galvanometer

Chapter 8 Magnetism and Matter

Introduction
The Bar Magnet
Magnetism and Gauss’s Law
The Earth’s Magnetism
Magnetization and Magnetic Intensity
Magnetic Properties of Materials
Permanent Magnets and Electromagnets

Chapter 9 Electromagnetic Induction

Introduction
The Experiments of Faraday and Henry
Magnetic Flux
Faraday’s Law of Induction
Lenz’s Law and Conservation of Energy
Motional Electromotive Force
Energy Consideration: A Quantitative Study
Eddy Currents
Inductance
AC Generator

Chapter 10 Alternating Current

Introduction
AC Voltage Applied to a Resistor
Representation of AC Current and Voltage by Rotating Vectors – Phasors
AC Voltage Applied to an Inductor
AC Voltage Applied to a Capacitor
AC Voltage Applied to a Series LCR Circuit
Power in AC Circuit: The Power Factor
LC Oscillations
Transformers

Chapter 11 Electromagnetic Waves

Introduction
Displacement Current
Electromagnetic Waves
Electromagnetic Spectrum

Chapter 12 Dual Nature of Radiation and Matter

Introduction
Electron Emission Photoelectric Effect
Experimental Study of Photoelectric Effect
Photoelectric Effect and Wave Theory of Light
Einstein’s Photoelectric Equation: Energy Quantum of Radiation
Particle Nature of Light: The Photon
Wave Nature of Matter
Davisson and Germer Experiment

Chapter 13 Atoms

Introduction
Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom
Atomic Spectra
Bohr Model of the Hydrogen Atom
The Line Spectra of the Hydrogen Atom
DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation

Chapter 14 Nuclei

Introduction
Atomic Masses and Composition of Nucleus
Size of the Nucleus
Mass-Energy and Nuclear Binding Energy
Nuclear Force
Radioactivity
Nuclear Energy

Chapter 15 Semiconductor Electronics: Materials, Devices and Simple Circuits

Introduction
Classification of Metals, Conductors and Semiconductors
Intrinsic Semiconductor
Extrinsic Semiconductor
p-n Junction
Semiconductor diode
Application of Junction Diode as a Rectifier
Special Purpose p-n Junction Diodes
Junction Transistor
Digital Electronics and Logic Gates
Integrated Circuits

Chapter 16 Communication Systems

Introduction
Elements of a Communication System
Basic Terminology Used in Electronic Communication Systems
Bandwidth of Signals
The bandwidth of Transmission Medium
Propagation of Electromagnetic Waves
Modulation and its Necessity
Amplitude Modulation
Production of Amplitude Modulated Wave
Detection of Amplitude Modulated Wave

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Physics Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Physics Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

AP Inter 2nd Year Chemistry Notes

September 3, 2022

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Chemistry Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Chemistry Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Chemistry Study Material and AP Inter 2nd Year Chemistry Important Questions for exam preparation.

AP Intermediate 2nd Year Chemistry Notes

TS AP Inter 2nd Year Chemistry Weightage Blue Print

These TS AP Intermediate 2nd Year Chemistry Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Chemistry Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

AP Inter 2nd Year Chemistry Important Questions Chapter Wise Pdf 2022-2023 | Sr Inter Chemistry Important Questions

September 3, 2022

TS AP Intermediate 2nd Year Chemistry Important Questions Chapter Wise 2022: Here we have created a list of Telangana & Andhra Pradesh BIEAP TS AP Intermediate Sr Inter 2nd Year Chemistry Important Questions Chapter Wise with Answers 2022-2023 Pdf Download just for you. Those who are preparing for Inter exams should practice the Intermediate 2nd Year Chemistry Important Questions Pdf and do so will clear their doubts instantly. These Senior Inter Chemistry Important Questions with Answers enhance your conceptual knowledge and prepares you to solve different types of questions in the exam.

Students must practice these AP Inter 2nd Year Chemistry Important Questions Pdf to boost their preparation for the exam paper. These AP Intermediate 2nd Year Chemistry Important Questions with Answers are prepared as per the latest exam pattern. Each of these chapters contains a set of solved questions and additional questions for practice.

Students can also read AP Inter 2nd Year Chemistry Study Material for exam preparation. Students can also go through AP Inter 2nd Year Chemistry Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Chemistry Important Questions with Answers Chapter Wise 2022

Intermediate 2nd Year Chemistry Important Questions Chapter Wise 2022

TS AP Inter 2nd Year Chemistry Weightage Blue Print

We are providing well-organized Chemistry Important Questions Inter 2nd Year that will help students in their exam preparation. These Chemistry Inter 2nd Year Important Questions are designed by subject experts. So Students should utilise the Intermediate 2nd Year Chemistry Important Questions with Answers 2022 to pass their board exams with flying colours.

The questions given in the Inter Second Year Chemistry Important Questions are designed and laid out chronologically and as per the syllabus. Students can expect that these TS AP Intermediate Inter 2nd Year Chemistry Important Questions Chapter Wise Pdf 2022-2023 might be covered in the final exam paper.

AP Inter 2nd Year Chemistry Study Material Pdf | Intermediate 2nd Year Chemistry Textbook Solutions

September 3, 2022

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Chemistry Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Chemistry Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Chemistry Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Chemistry Syllabus & AP Inter 2nd Year Chemistry Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Chemistry Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Chemistry Study Material Pdf Download | Sr Inter 2nd Year Chemistry Textbook Solutions

Chapter 1 Solid State
Chapter 2 Solutions
Chapter 3 Electrochemistry and Chemical Kinetics
- Chapter 3(a) Electro Chemistry
- Chapter 3(b) Chemical Kinetics
Chapter 4 Surface Chemistry
Chapter 5 General Principles of Metallurgy
Chapter 6 P-Block Elements
Chapter 7 d and f Block Elements & Coordination Compounds
Chapter 8 Polymers
Chapter 9 Biomolecules
Chapter 10 Chemistry in Everyday Life
Chapter 11 Haloalkanes and Haloarenes
Chapter 12 Organic Compounds Containing C, H and O
- Chapter 12(a) Alcohols, Phenols, and Ethers
- Chapter 12(b) Aldehydes, Ketones, and Carboxylic Acids
Chapter 13 Organic Compounds Containing Nitrogen

TS AP Inter 2nd Year Chemistry Weightage Blue Print 2022-2023

TS AP Inter 2nd Year Chemistry Weightage 2022-2023 | TS AP Inter 2nd Year Chemistry Blue Print 2022

TS AP Inter 2nd Year Chemistry Weightage Blue Print

Intermediate 2nd Year Chemistry Syllabus

TS AP Inter 2nd Year Chemistry Syllabus

Chapter 1 Solid State

1.1 General characteristics of solid state
1.2 Amorphous and crystalline solids
1.3 Classification of crystalline solids based on different binding forces (molecular, ionic, metallic, and covalent solids)
1.4 Probing the structure of solids: X-ray crystallography
1.5 Crystal lattices and unit cells, Bravais lattices primitive and centered unit cells
1.6 Number of atoms in a unit cell (primitive, body-centered, and face-centered cubic unit cell)
1.7 Close packed structures: Close packing in one dimension, in two dimensions, and in three dimensions- tetrahedral and octahedral voids- formula of a compound and number of voids filled- locating tetrahedral and octahedral voids
1.8 Packing efficiency in simple cubic, bcc, and in hcp, ccp lattice.
1.9 Calculations involving unit cell dimensions- density of the unit cell.
1.10 Imperfections in solids-types of point defects-stoichiometric and non-stoichiometric defects
1.11 Electrical properties-conduction of electricity in metals, semiconductors, and insulators band theory of metals
1.12 Magnetic properties

Chapter 2 Solutions

2.1 Types of solutions
2.2 Expressing concentration of solutions-mass percentage, volume percentage, mass by volume percentage, parts per million, mole fraction, molarity, and molality
2.3 Solubility: Solubility of a solid in a liquid, solubility of a gas in a liquid, Henry’s law
2.4 Vapour pressure of liquid solutions: vapour pressure of liquid-liquid solutions. Raoult’s law is a special case of Henry’s law – vapour pressure of solutions of solids in liquids
2.5 Ideal and non-ideal solutions
2.6 Colligative properties and determination of molar mass-relative lowering of vapour pressure-elevation of boiling point-depression of freezing point-osmosis and osmotic pressure-reverse osmosis and water purification.
2.7 Abnormal molar masses-van’t Hoff factor

Chapter 3 Electrochemistry And Chemical Kinetics
Electrochemistry

3.1 Electrochemical cells
3.2 Galvanic cells measurement of electrode potentials
3.3 Nernst equation-equilibrium constant from Nernst equation-electrochemical cell and Gibbs energy of the cell reaction
3.4 Conductance of electrolytic solutions- measurement of the conductivity of ionic solutions- a variation of conductivity and molar conductivity with concentration-strong electrolytes and weak electrolytes-applications of Kohlrausch’s law
3.5 Electrolytic cells and electrolysis: Faraday’s laws of electrolysis-products of electrolysis
3.6 Batteries: primary batteries and secondary batteries
3.7 Fuel cells
3.8 Corrosion of metals-Hydrogen economy

Chemical Kinetics

3.9 Rate of a chemical reaction
3.10 Factors influencing the rate of a reaction: dependence of rate on concentration- rate expression and rate constant- order of a reaction, molecularity of a reaction
3.11 Integrated rate equations-zero order reactions-first order reactions-half life of a reaction
3.12 Pseudo first-order reaction
3.13 Temperature dependence of the rate of a reaction-effect of catalyst
3.14 Collision theory of chemical reaction rates

Chapter 4 Surface Chemistry

4.1 Adsorption and absorption: Distinction between adsorption and absorption-mechanism of adsorption-types of adsorption- characteristics of physisorption- characteristics of chemisorptions- adsorption isotherms- adsorption from solution phase- applications of adsorption
4.2 Catalysis: Catalysts, promoters, and poisons-auto catalysis- homogeneous and heterogeneous catalysis- adsorption theory of heterogeneous catalysis- important features of solid catalysts: (a) activity (b) selectivity- shape-selective catalysis by zeolites- enzyme catalysis- characteristics and mechanism- catalysts in industry
4.3 Colloids
4.4 Classification of colloids: Classification based on the physical state of the dispersed phase and dispersion medium- classification based on nature of the interaction between the dispersed phase and dispersion medium- classification based on the type of particles of the dispersed phase- multi molecular, macromolecular, and associated colloids-cleansing action of soaps-preparation of colloids-purification of colloidal solutions-properties of colloidal solutions: Tyndal effect, colour, Brownian movement-charge on colloidal particles, electrophoresis
4.5 Emulsions
4.6 Colloids Around us- application of colloids

Chapter 5 General Principles of Metallurgy

5.1 Occurance of metals
5.2 Concentration of ores- levigation, magnetic separation, froth floatation, leaching
5.3 Extraction of crude metal from concentrated ore-conversion to oxide, reduction of oxide to the metal
5.4 Thermodynamic principles of metallurgy-Ellingham diagram-limitations-applications-extraction of iron, copper, and zinc from their oxides
5.5 Electrochemical principles of metallurgy
5.6 Oxidation and reduction
5.7 Refining of crude metal distillation, liquation poling, electrolysis, zone refining, and vapour phase refining
5.8 Uses of aluminium, copper, zinc, and iron

Chapter 6 P-Block Elements
Group-15 Elements

6.1 Occurance- electronic configuration, atomic and ionic radii, ionization energy, electronegativity, physical and chemical properties
6.2 Dinitrogen- preparation, properties, and uses
6.3 Compounds of nitrogen-preparation and properties of ammonia
6.4 Oxides of nitrogen
6.5 Preparation and properties of nitric acid
6.6 Phosphorous-allotropic forms
6.7 Phosphine-preparation and properties
6.8 Phosphorous halides
6.9 Oxoacids of phosphorous

Group-16 Elements

6.10 Occurance- electronic configuration, atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, physical and chemical properties
6.11 Dioxygen-preparation, properties, and uses
6.12 Simple oxides
6.13 Ozone preparation, properties, structure and uses
6.14 Sulphur-allotropic forms
6.15 Sulphur dioxide preparation, properties and uses
6.16 Oxoacids of sulphur
6.17 Sulphuric acid-industrial process of manufacture, properties and uses

Group-17 Elements

6.18 Occurance, electronic configuration, atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, physical and chemical properties
6.19 Chlorine preparation, properties, and uses
6.20 Hydrogen chloride- preparation, properties and uses
6.21 Oxoacids of halogens
6.22 Interhalogen compounds

Group-18 Elements

6.23 Occurance, electronic configuration, ionisation enthalpy,atomic radii electron gain enthalpy, physical and chemical properties (a) Xenon-fluorine compounds – XeF₂, XeF₄ and XeF₆ – preparation,hydrolysis and formation of fluoro anions- structures of XeF₂, XeF₄ and XeF₆ (b) Xenon-oxygen compounds XeO3 and XeOF₄ – their formation and structures

Chapter 7 d and f Block Elements & Coordination Compounds
d And f Block Elements

7.1 Position in the periodic table
7.2 Electronic configuration of the d-block elements
7.3 General properties of the transition elements (d-block) -physical properties, variation in atomic and ionic sizes of transition series, ionization enthalpies, oxidation states, trends in the M²⁺/M and M³⁺/M²⁺ standard electrode potentials, trends in the stability of higher oxidation states, chemical reactivity and E^J values, magnetic properties, formation of colored ions, formation of complex compounds, catalytic properties, formation of interstitial compounds, alloy formation
7.4 Some important compounds of transition elements-oxides and oxoanions of metals-preparation and properties of potassium dichromate and potassium permanganate-structures of chromate, dichromate, manganate, and permanganate ions
7.5 Inner transition elements(f-block)-lanthanoids- electronic configuration-atomic and ionic sizes-oxidation states- general characteristics
7.6 Actinoids-electronic configuration atomic and ionic sizes, oxidation states, general characteristics, and comparison with lanthanoids
7.7 Some applications of d and f block elements

Coordination Compounds

7.8 Werner’s theory of coordination compounds
7.9 Definitions of some terms used in coordination compounds
7.10 Nomenclature of coordination compounds- IUPAC nomenclature
7.11 Isomerism in coordination compounds-(a)Stereo isomerism- Geometrical and optical isomerism (b)Structural isomerism- linkage, coordination, ionization, and solvate isomerism
7.12 Bonding in coordination compounds. (a)Valence bond theory – magnetic properties of coordination compounds-limitations of valence bond theory (b) Crystal field theory (i) Crystal field splitting in octahedral and tetrahedral coordination entities (ii) Colour in coordination compounds-limitations of crystal field theory
7.13 Bonding in metal carbonyls
7.14 Stability of coordination compounds
7.15 Importance and applications of coordination compounds

Chapter 8 Polymers

8.1 Classification of Polymers -Classification based on source, structure, mode of polymerization, molecular forces, and growth polymerization
8.2 Types of polymerization reactions- addition polymerization or chain growth polymerization- ionic polymerization, free radical mechanism-preparation of addition polymers- polythene, Teflon, and polyacrylonitrile-condensation polymerization or step growth polymerization-polyamides- preparation of Nylon 6,6 and nylon 6-poly esters- terylene- bakelite, melamine, formaldehyde polymer- copolymerization- Rubber- natural rubber-vulcanization of rubber-Synthetic rubbers- preparation of neoprene and buna-N
8.3 Molecular mass of polymers-number average and weight average molecular masses- polydispersity index(PDI)
8.4 Biodegradable polymers- PHBV, Nylon 2-nylon 6
8.5 Polymers of commercial importance- poly propene, polystyrene, polyvinyl chloride(PVC), urea-formaldehyde resin, glyptal, bakelite- their monomers, structures, and uses

Chapter 9 Biomolecules

9.1 Carbohydrates – Classification of carbohydrates- Monosaccharides: preparation of glucose from sucrose and starch- Properties and structure of glucose- D, L and (+), (-) configurations of glucose- Structure of fructose Disaccharides: Sucrose- preparation, structure-Invert sugar- Structures of maltose and lactose-Polysaccharides: Structures of starch cellulose and glycogen- Importance of carbohydrates
9.2 Aminoacids: Natural amino acids- classification of amino acids – structures and D and L forms-Zwitter ions Proteins: Structures, classification, fibrous and globular- primary, secondary, tertiary, and quaternary structures of proteins- Denaturation of proteins
9.3 Enzymes: Enzymes, mechanism of enzyme action
9.4 Vitamins: Explanation- names- classification of vitamins – sources of vitamins-deficiency diseases of different types of vitamins
9.5. Nucleic acids: chemical composition of nucleic acids, structures of nucleic acids, DNA fingerprinting biological functions of nucleic acids
9.6 Hormones: Definition, different types of hormones, their production, biological activity, diseases due to their abnormal activities.

Chapter 10 Chemistry in Everyday Life

10.1 Drugs and their classification: (a) Classification of drugs on the basis of pharmacological effect(b) Classification of drugs on the basis of drug action (c) Classification of drugs on the basis of chemical structure (d) Classification of drugs on the basis of molecular targets
10.2 Drug-Target interaction-Enzymes as drug targets (a) Catalytic action of enzymes (b) Drug-enzyme interaction Receptors as drug targets
10.3 Therapeutic action of different classes of drugs: antacids, antihistamines, neurologically active drugs: tranquilizers, analgesics- non-narcotic, narcotic analgesics, antimicrobials-antibiotics, antiseptics, and disinfectants- antifertility drugs
10.4 Chemicals in food- artificial sweetening agents, food preservatives, antioxidants in food
10.5 Cleansing agents-soaps and synthetic detergents

Chapter 11 Halo Alkanes and Halo Arenes

11.1 Classification and nomenclature
11.2 Nature of C-X bond
11.3 Methods of preparation: Alkyl halides and aryl halides- from alcohols, from hydrocarbons (a) by free radical halogenation (b) by electrophilic substitution (c) by replacement of diazonium group(Sand- Meyer reaction) (d) by the addition of hydrogen halides and halogens to alkenes-by halogen exchange(Finkelstein reaction)
11.4 Physical properties- melting and boiling points, density and solubility
11.5 Chemical reactions: Reactions of haloalkanes (i)Nucleophilic substitution reactions (a) SN2 mechanism (b) SN1 mechanism (c) stereochemical aspects of nucleophilic substitution reactions -optical activity (ii) Elimination reactions (iii) Reaction with metals-Reactions of haloarenes: (i) Nucleophilic substitution (ii)Electrophilic substitution and (iii) Reaction with metals
11.6 Polyhalogen compounds: Uses and environmental effects of dichloro methane, trichloromethane, triiodomethane, tetrachloro methane, freons, and DDT.

Chapter 12 Organic Compounds Containing C, H, and O (Alcohols, Phenols, Ethers, Aldehydes, Ketones, and Carboxylic Acids)
Alcohols, Phenols, and Ethers

12.1 Alcohols, phenols and ethers-classification
12.2 Nomenclature: (a) Alcohols, (b) phenols and (c) ethers
12.3 Structures of hydroxy and ether functional groups
12.4 Methods of preparation: Alcohols from alkenes and carbonyl compounds- Phenols from haloarenes, benzene sulphonic acid, diazonium salts, cumene
12.5 Physical properties of alcohols and phenols
12.6 Chemical reactions of alcohols and phenols (i) Reactions involving cleavage of O-H bond-Acidity of alcohols and phenols, esterification (ii) Reactions involving cleavage of C-O bond- reactions with HX, PX3, dehydration, and oxidation (iii) Reactions of phenols- electrophilic aromatic substitution, Kolbe’s reaction, Reimer – Tiemann reaction, reaction with zinc dust, oxidation
12.7 Commercially important alcohols (methanol, ethanol)
12.8 Ethers-Methods of preparation: By dehydration of alcohols, Williamson synthesis-Physical properties-Chemical reactions: Cleavage of C-O bond and electrophilic substitution of aromatic ethers.

Aldehydes and Ketones

12.9 Nomenclature and structure of carbonyl group 12.10Preparation of aldehydes and ketones-(1) by oxidation of alcohols (2) by dehydrogenation of alcohols (3) from hydrocarbons
12.10 Preparation of aldehydes (1) from acyl chlorides (2) from nitriles and esters (3) from hydrocarbons-Preparation of ketones (1) from acyl chlorides (2) from nitriles (3) from benzene or substituted benzenes
12.11 Physical properties of aldehydes and ketones
12.12 Chemical reactions of aldehydes and ketones-nucleophilic addition, reduction, oxidation, reactions due to Hydrogen, and other reactions (Cannizzaro reaction, electrophilic substitution reaction)
12.13 Uses of aldehydes and ketones

Carboxylic Acids

12.14 Nomenclature and structure of carboxyl group
12.15 Methods of preparation of carboxylic acids- (1)from primary alcohols and aldehydes (2) from alkylbenzenes(3)from nitriles and amides (4)from Grignard reagents (5) from acyl halides and anhydrides (6) from esters
12.16 Physical properties
12.17 Chemical reactions: (i) Reactions involving cleavage of O-H bond-acidity, reactions with metals and alkalies (ii) Reactions involving cleavage of C-OH bond-formation of anhydride, reactions with PCl₅, PCl₃, SOCl₂, esterification, and reaction with ammonia (iii) Reactions involving -COOH group-reduction, decarboxylation (iv) Substitution reactions in the hydrocarbon part – halogenation and ring substitution
12.18 Uses of carboxylic acids

Chapter 13 Organic Compounds Containing Nitrogen
I. Amines

13.1 Structure of amines
13.2 Classification
13.3 Nomenclature
13.4 Preparation of amines: reduction of nitro compounds, ammonolysis of alkyl halides, reduction of nitriles, reduction of amides, Gabriel phthalimide synthesis, and Hoffmann bromamide degradation reaction
13.5 Physical properties
13.6 Chemical reactions: the basic character of amines, alkylation, acylation, carbonyl amine reaction, reaction with nitrous acid, reaction with aryl sulphonyl chloride, the electrophilic substitution of aromatic amines-bromination, nitration and sulphonation

II. DIazonium Salts

13.7 Methods of preparation of diazonium salts (by diazotization)
13.8 Physical properties
13.9 Chemical reactions: Reactions involving

III. Cyanides and Isocyanides

13.10 Structure and nomenclature of cyanides and isocyanides Preparation, physical properties, and chemical reactions of cyanides and isocyanides

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Chemistry Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Chemistry Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Inter 2nd Year Maths 2B Differential Equations Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Differential Equations Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Differential Equations Important Questions

Question 1.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^{x – y} + x² e^-y [Mar. 06; May 05]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 1

Question 2.
x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y = 2x² sec²2x [May 11]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 2

Question 3.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y tan x = sin x. [T.S. Mar. 16]
Solution:
I.F. = \(e^{\int \tan x d x}\) = e^{log sec x} = sec x
y.sec x = \(\int\) sin x . sec x dx = \(\int\) tan x dx
= log sec x + c

Question 4.
cos x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = sec²x [Mar. 14]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + tan x . y = sec³x
I.F. = e^{\(\int\)tan x dx} = e^{log sec x} = sec x
y . sec x = \(\int\) sec⁴x dx = \(\int\) (1 + tan² x) sec² x
dx = tan x + \(\frac{\tan ^{3} x}{3}\) + c

Question 5.
(x + y + 1)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1.
Solution:
\(\frac{\mathrm{dx}}{\mathrm{dy}}\) = x + y + 1
\(\frac{\mathrm{dx}}{\mathrm{dy}}\) = x + y + 1
I.F. = e^{\(\int\) -dy} = e^-y
x . e^-y = \(\int\) e^-y (y + 1)dy
= – (y + 1) . e^-y + \(\int\) e^-y . dy
= – (y + 1) e^-y – e^-y
= – (y + 2) e^-y + c
x = – (y + 2) + c. e^-y

Question 6.
Find the order and degree of r
\(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) – 3 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) – e^x = 4. [Mar. 14]
Solution:
The equation is a polynomial in \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\).
The exponent of \(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) is 2.
Hence the degree is 2.
\(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) is the highest order derivative occuring in the equation.
Order of the equation is 3.

Question 7.
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} + x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 3
∴ The order of the equation is 2 and its degree is 1.

Question 8.
Find the order and degree of \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y [Mar. 16; May 11]
Solution:
Given equation is \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y
i.e., \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))³ = (6y)^{\(\frac{5}{6}\)}
Order = 2, degree = 1

Question 9.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x(2 \log x+1)}{\sin y+y \cos y}\) [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
\(\int\) sin y dy + \(\int\) y cos y dy = \(\int\) 2x log x dx + \(\int\) x dx
\(\int\) sin y dy + y sin y – \(\int\) sin y dy = x² log x – \(\int\) x² . \(\frac{1}{x}\) dx + \(\int\) x dx + c
y sin y = x² log x + c

Question 10.
(xy² + x) dx + (yx² + y) dy = 0. [A.P. Mar. 15, 07]
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x d x}{1+x^{2}}\) + \(\frac{y d x}{1+y^{2}}\) = 0
Integrating
\(\int \frac{x d x}{1+x^{2}}+\int \frac{y d y}{1+y^{2}}=0\)
\(\frac{1}{2}\)[(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is(1 + x²) (1 + y²) = k when k = c².

Question 11.
sin^-1 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = x + y [Mar. 07]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin (x + y)
x + y = t
1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) – 1 = sin t
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 1 + sin t
\(\frac{d t}{1+\sin t}\) = dx
Integrating both sides we get
\(\int \frac{d t}{1+\sin t}=\int d x\)
\(\int \frac{1-\sin t}{\cos ^{2} t} d t=x+c\)
\(\int\) sec² t dt = \(\int\) tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 12.
(x² – y²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy [May 11]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 4
= log y + c
\(\frac{-x^{2}}{2 y^{2}}\) = (log y + c)
-x² = 2y² (c + log y)
⇒ Solution is x² + 2y² (c + log y) = 0.

Question 13.
Solve : x dy = (y + x cos² \(\frac{y}{x}\)) dx.
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 5

Question 14.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0 [T.S. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 6
2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c

Question 15.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y tan x = cos³x [May 11]
Solution:
I.F. = e^{\(\int\) tan x dx} = e^{log sec x} = sec x
y . sec x = \(\int\) sec x. cos³ x dx
= \(\int\) cos²x dx
= \(\frac{1}{2}\) \(\int\) (1 + cos 2x) dx
= \(\frac{1}{2}\) (x + \(\frac{sin2x}{2}\)) + c
\(\frac{2 y}{\cos x}\) = x + sin x . cos x + c
Solution is 2y = x cos x + sin x . cos² x + c . cos x^

Question 16.
(1 + x²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = e^{tan-1 x} [May 07] [A.P. Mar. 16] [T.S. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 7

Question 17.
Solve (1 + y²)dx = (tan^-1y – x)dy. [A.P. Mar. 15]
Solution:
Given \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\tan ^{-1} y-x}{1+y^{2}}\)
Inter 2nd Year Maths 2B Differential Equations Important Questions 8

Question 18.
(x² – y²)dx – xy dy = 0 [May 06]
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy
Inter 2nd Year Maths 2B Differential Equations Important Questions 10
–\(\frac{1}{4}\) [log (x² – 2y²) – log x²] = log x + log c
–\(\frac{1}{4}\) log (x² – 2y²) + \(\frac{1}{4}\) . 2 log x = log x + log c
–\(\frac{1}{4}\) log (x² – 2y²) = \(\frac{1}{2}\) log x + log c
– log (x² – 2y²) = – 2 log x – 4 log c
log (x² – 2y²) = – 2 log x + log k where
k = \(\frac{1}{c^{4}}\) = log \(\frac{\mathrm{k}}{\mathrm{x}^{2}}\)
x² – 2y² = \(\frac{\mathrm{k}}{\mathrm{x}^{2}}\)
Solution is x² (x² – 2y²) = k

Question 19.
\(\frac{d y}{d x}=\frac{3 y-7 x+7}{3 x-7 y-3}\) [T.S. Mar. 16]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 11

= 3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10 ln (v + 1) – 4 ln (v – 1)
ln (v + 1)⁵ + ln (v – 1)² + ln x⁷ = ln c
(v + 1)⁵ . (v – 1)² . x⁷ = c
(\(\frac{y}{x}\) + 1)⁵ (\(\frac{y}{x}\) – 1)² . x⁷ = c
(y – x)² (y + x)⁵ = c
[y – (x – 1)]² (y + x – 1)⁵ = c
Solution is [y – x + 1]² (y + x – 1)⁵ = c.

Question 20.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) (x²y³ + xy) = 1 [Mar. 11]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy + x²y³
This is Bernoulli’s equation
x^-2 . \(\frac{\mathrm{dx}}{\mathrm{dy}}\) – \(\frac{1}{x}\) . y = y³
Inter 2nd Year Maths 2B Differential Equations Important Questions 13

Question 21.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [AP Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
= – 9(A cos 3x + B sin 3x)
= -9y .
is \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) + 9y = 0.
Alternate method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – 3A sin 3x + 3B sin cos 3x
Inter 2nd Year Maths 2B Differential Equations Important Questions 15
This is the required differential equation.

Question 22.
Solve (x² + y²) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 16
log cx(1 – v² = log 1
cx (1 – v²) = 1
cx (1 – \(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\)) = 1
c(x² – y²) = x is the required solution.

Question 23.
Give the solution of x sin² \(\frac{y}{x}\) dx = y dx – x dy which passes through the point (1, \(\frac{\pi}{4}\)). [Mar. 14]
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 17
The given curve passes through (1, \(\frac{\pi}{4}\))
cot \(\frac{\pi}{4}\) = log 1 + c
1 = 0 + c ⇒ c = 1
Solution is cot \(\frac{y}{x}\) = log x + 1

Question 24.
Find the order and degree of the differential equation \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = – p²y.
Solution:
The given equation is a polynomial equation in \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\)
Hence the degree is 1
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) is the highest order derivative occuring in the equation.
Its order is 2.

Question 25.
Find the order and degree of
(\(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\))² – 3 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))² – e^x = 4 [Mar. 14]
Solution:
The equation is a polynomial in \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\).
The exponent of \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\) is 2.
Hence the degree is 2.
\(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\) is the highest õrder derivative occuring in the equation.
Order of the equation is 3.

Question 26.
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} + x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} = -[x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y]
Cubing both sides
x^{\(\frac{3}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\)) = -[x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y]³
∴ The order of the equation is 2 and its degree is 1.

Question 27.
Find the order and degree of \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y [A.P. Mar. 16; May 11]
Solution:
Given equation is \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y
i.e., \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))³ = (6y)^{\(\frac{5}{6}\)}
Order = 2; degree = 1

Question 28.
Find the order of the differential equation corresponding to y = c(x – c)², where c is an arbitrary constant.
Solution:
The given differential equation is
y = c(x – c)²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2c(x – c)
∴ Order of the differential equation is 1.

Question 29.
Find the order of the differential equation corresponding to y = Ae^x + Be^3x + Ce^5x; (A, B, C being parameters) is a solution.
Solution:
Required differential equation is obtained by eliminating A, B, C from y,
\(\frac{\mathrm{dy}}{\mathrm{dx}}\), \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\), \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\)
Highest order deviation = \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\)
Order of the differential equation = 3.

Question 30.
Form the differential equation corresponding to y = cx – 2c², where c is a parameter.
Solution:
Given y = cx – 2c² ………………. (1)
Differentiating (1) w.r.to
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = c
Substituting in (1), required differential equation is
y = x . (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) – 2(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))²

Question 31.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [A.P. Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
= -9(A cos 3x + B sin 3x)
= -9y
is \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + 9y = 0

Alternate Method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x = 3B sin cos 3x
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
We get \(\left|\begin{array}{ccc}
y & -\cos 3 x & -\sin 3 x \\
\left(\frac{d y}{d x}\right) & 3 \sin 3 x & -3 \cos 3 x \\
\left(\frac{d^{2} y}{d x^{2}}\right) & 9 \cos 3 x & 9 \sin 3 x
\end{array}\right|\) = 0
y(27 sin² 3x + 27 cos² 3x) – (-9 sin 3x. cos 3x + 9 cos 3x. sin 3x) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + (3 cos² 3x + 3 sin² 3x) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = 0
= 27y + 3 .\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = 0 or \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + 9y = 0
This is the required differential equation.

Question 32.
Form the differential equation corresponding to the family of circles of radius r given by (x – a)² + (y – b)² = r², where a and b are parameters.
Solution:
We have (x – a)² + (y – b)² = r² ………………… (1)
Differentiating (1) w.r.to x
2(x – a) + 2(y – b) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 ……………….. (2)
Differentiating (2) w.r.to. x
1 + (y – b) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))² = 0 ……………… (3)
From (2) (x – a) = -(y – b) \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Substituting in (1), we get
Inter 2nd Year Maths 2B Differential Equations Important Questions 18
i.e., \(r^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3}\)
Which is the required differential equation.

Question 33.
Form the differential equation corresponding to the family of circles passing through the origin and having centres on Y-axis.
Solution:
The equation of the family of all circles passing through the origin and having centres on Y—axis is
x² + y² + 2hy = 0 …………………. (1)
Where h is a parameter
Differentiating (1) w.r. to x
2x + 2y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2h . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
or x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + h . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
-(x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = h. \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
h = \(\frac{-\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}\)
Substituting in (1)
We get x² + y² – 2y \(\frac{\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}\) = 0
x² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2xy – 2y² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
or (x² – y²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2xy = 0
This is the required differential equation.

Question 34.
Express the following differential equations in the form f(x) dx + g(y) dy = 0.
i) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{y}^{2}}{1+\mathrm{x}^{2}}\)
Solution:
⇒ \(\frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}\)
\(\frac{d x}{1+x^{2}}-\frac{d y}{1+y^{2}}\) = 0

ii) y – x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a (y² + \(\frac{\mathrm{dy}}{\mathrm{dx}}\))
Solution:
y – x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = ay² + a. \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
y – ay² = (x + a) . \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
\(\frac{d x}{x+a}=\frac{d y}{y-a y^{2}}\)

iii) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x-y + x² e^-y
Solution:
Multiplying in the e^y
e^y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x + x²
e^y . dy = (e^x + x²) dx
(e^x + x²) dx – e^y . dy = 0

iv) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + x² = x² e^3y
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x² . e^3y – x² = x²(e^3y – 1)
\(\frac{d y}{e^{3 y}-1}\) = x² dx ⇒ x² dx – \(\frac{d y}{e^{3 y}-1}\) = 0
or x²dx + \(\frac{1}{\left(1-e^{3 y}\right)}\) . dy = 0

Question 35.
Find the general solution of
x + y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0.
Solution:
Given equation is x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
x . dx + y . dy = 0
Integrating \(\frac{x^{2}}{2}\) + \(\frac{y^{2}}{2}\) = c
or x² + y² = 2 c = c’

Question 36.
Find the general solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x+y.
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x+y = e^x . e^y
\(\frac{d y}{e^{y}}\) = e^x dx
\(\int\) e^-y dy = \(\int\) e^x dx
e – e^-y = e^x
or e^x + e^-y = c is the required solution.

Question 37.
Solve y² – x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a(y + \(\frac{\mathrm{dy}}{\mathrm{dx}}\))
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 19

Question 38.
Solve \(\frac{d y}{d x}=\frac{y^{2}+2 y}{x-1}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 21

Question 39.
Solve \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\) [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
\(\int\) sin y dy + \(\int\) y cos y dy = \(\int\) 2x log x dx + \(\int\) x dx
\(\int\) sin y dy + y sin y – \(\int\) sin y dy = x² log x – \(\int\) x² . \(\frac{1}{x}\) dx + \(\int\) x dx + c
y sin y = x² log x + c

Question 40.
Find the equation of the curve whose slope, at any point, (x, y) is \(\frac{y}{x^{2}}\) and which satisfies the condition y = 1 when x = 3.
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 22

Question 41.
Solve y(1 + x) dx + x(1 + y) dy = 0
Solution:
The given equation can be written as
\(\frac{(1+x)}{x}\) dx + \(\frac{(1+y)}{y}\) . dy = 0
\(\int\) (1 + \(\frac{1}{x}\))dx + \(\int\) (1 + \(\frac{1}{y}\)) dy = 0
x + log x + y + log y = c
x + y + log (xy) = c is the required solution.

Question 42.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin (x + y) + cos (x + y)
Solution:
Put x + y = t
Inter 2nd Year Maths 2B Differential Equations Important Questions 23
x = log (1 + tan \(\frac{t}{2}\)) + c
But t = x + y
Solution is x = log (1 + tan \(\frac{x+y}{2}\)) + c

Question 43.
Solve (x – y)² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a²
Solution:
Put x – y = t
Inter 2nd Year Maths 2B Differential Equations Important Questions 24

Question 44.
Solve \(\sqrt{1+x^{2}} \sqrt{1+y^{2}}\) dx + xy dy = 0
Solution:
Given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 26

Question 45.
Solve \(\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}\)
Solution:
Put x – 2y = t
Inter 2nd Year Maths 2B Differential Equations Important Questions 29

Question 46.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\sqrt{y-x}\)
Solution:
Put y – x = t²
Inter 2nd Year Maths 2B Differential Equations Important Questions 30

Question 47.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 1 = e^x+y
Solution:
Put t = x + y
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^t
\(\int \frac{d t}{e^{t}}=\int d x\)
\(\int\) e^-t dt = \(\int\) dx
-e^-t = x + c
x + e^-t + c = 0
Solution is x + e^-(x+y) + c = 0

Question 48.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (3x + y + 4)²
Solution:
Put t = 3x + y + 4
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 3 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3 + t²
\(\frac{\mathrm{dt}}{\mathrm{t}^{3}+3}\) = dx
\(\int \frac{d t}{t^{2}+3}=\int d x\)
\(\frac{1}{\sqrt{3}}\) tan^-1 (\(\frac{t}{\sqrt{3}}\)) = x + c
Solution is \(\frac{1}{\sqrt{3}}\) tan^-1 (\(\frac{3 x+y+4}{\sqrt{3}}\)) = x + c

Question 49.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – x tan (y – x) = 1
Solution:
Put y – x = t
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 1 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = x tan t + 1 – 1 = x tan t
\(\frac{\mathrm{dt}}{\tan \mathrm{t}}\) = x dx
\(\int\) cot dt = \(\int\) x dx
log |sin | = \(\frac{x^{2}}{2}\) + c
Solution is log |sin (y – x)| = \(\frac{x^{2}}{2}\) + c

Question 50.
Show that f(x, y) = 1 + e^x/y is a homogeneous function of x and y.
Solution:
f(kx, xy) = 1 + e^kx/ky = 1 + e^x/y = f(x, y)
f(x, y) is a homogeneous function degree 0.

Question 51.
Show that f(x, y) = x\(\sqrt{x^{2}+y^{2}}\) – y² is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx\(\sqrt{k^{2} x^{2}+k^{2} y^{2}}\) – k²y².
= k² (x\(\sqrt{x^{2}+y^{2}}\) – y²) = k² f(x, y)
f(x, y) is a homogeneous function of degree 2.

Question 52.
Show that f(x, y) = x – y log y + y log x is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx – ky. log ky + ky log (kx)
= k(x – y log (ky) + y log kx)
= k(x – y log k – y log y + y log k + y log x)
= k(x – y log y + y log x)
= k. f(x, y)
f(x, y) is a homogeneous function of degree 1.

Question 53.
Express (1 + e^x/y)dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0 in the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) = F (\(\frac{x}{y}\))
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 31

Question 54.
Express (x\(\sqrt{x^{2}+y^{2}}\) – y²) dx + xy dy = 0 in the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = F (\(\frac{x}{y}\))
Solution:
Given equation is
Inter 2nd Year Maths 2B Differential Equations Important Questions 32

Question 55.
Express \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y}{x+y e^{\frac{-2 x}{y}}}\) in the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) = F (\(\frac{x}{y}\))
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 33

Question 56.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y^{2}-2 x y}{x^{2}-x y}\)
Solution:
The given equation is a homogeneous equation.
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Important Questions 34
\(\log v \sqrt{2 v-3}=-3 \log \frac{x}{c}=\log \frac{c^{3}}{x^{3}}\)

Question 57.
Solve (x² + y²) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^{2}+y^{2}}{2 x y}\)
This is a homogeneous function
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Important Questions 36
-log (1 – v²) = log x + log c
= log cx
log cx + log (1 – v²) = 0
log cx(1 – v²) = log 1
cx (1 – v²) = 1
cx (1 – \(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\)) = 1
c(x² – y²) = x is the required solution.

Question 58.
Solve xy² dy – (x³ + y³) dx = 0.
Solution:
Given equation is xy² dy = (x³ + y²) dx
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^{3}+y^{3}}{x y^{2}}\)
This is a homogeneous equation.
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Important Questions 37

Question 59.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^{2}+y^{2}}{2 x^{2}}\)
Solution:
This is a homogeneous equation.
Inter 2nd Year Maths 2B Differential Equations Important Questions 38

Question 60.
Solve x sec (\(\frac{y}{x}\)) . (y dx + x dy) = y cosec (\(\frac{y}{x}\)) . (x dy – y dx)
Solution:
This given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 39

log(\(\frac{\sin v}{v}\)) = log cx²
\(\frac{\sin v}{v}\) = cx²
\(\frac{x}{y}\) sin (\(\frac{y}{x}\)) = cx²
Solution is sin(\(\frac{y}{x}\)) = cxy.

Question 61.
Give the solution of x sin² \(\frac{y}{x}\) dx = y dx – x dy which passes through the point (1, \(\frac{\pi}{4}\)). [Mar. 14]
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 41
The given curve passes through (1, \(\frac{\pi}{4}\))
cot \(\frac{\pi}{4}\) = log 1 + c
1 = 0 + c ⇒ c = 1
Solution is cot \(\frac{y}{x}\) = log x + 1

Question 62.
Solve (x³ – 3xy²) dx + (3x²y – y³) dy = 0
Solution:
(x³ – 3xy²) dx = -(3x²y – y³) dy
Inter 2nd Year Maths 2B Differential Equations Important Questions 42

Question 63.
Transform the following two differential equations Into linear form.
x log x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 2 log x
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + \(\frac{1}{x \log x}\) . y = \(\frac{2}{x}\)
This is of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + Py = Q

Question 64.
(x + 2y³) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x+2 y^{3}}{y}\) = \(\frac{x}{y}\) + 2y²
\(\frac{\mathrm{dx}}{\mathrm{dy}}\) – \(\frac{1}{y}\) . x = 2y²
This is of the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) + Px = Q.

Question 65.
(cos x) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = tan x
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y(tan x) = (tan x) (sec x)
P = tan x ⇒ \(\int\) P dx = \(\int\) tan x dx = log sec x
I.F. = e^{\(\int\) log sec x} = sec x

Question 66.
Solve (2x – 10y³) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = o
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 44

Question 67.
Solve (1 + x²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2xy – 4x² = 0
Solution:
Given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 45

Question 68.
Solve \(\frac{1}{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y . e^x = e^{(1 – x)ex}
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + (x.e^x) . y = x . e^{(1 – x)ex}
I.F. = e^{\(\int\).ex dx} = e^{(x – 1) ex}
y . e^{(x – 1) ex} = \(\int\) x dx
= \(\frac{\mathrm{x}^{2}}{2}\) + c
2y. e^{(x – 1) ex} = x² + 2c is the required solution.

Question 69.
Solve sin² x. \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = cot x
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y cosec² x = cot x . cosec² x
I.F. = e^{\(\int\) cosec2 x dx} = e^{-cot x}
y . e^{-cot x} = \(\int\) e^{-cot x} . cosec² x. cot x dx ……………… (1)
Consider \(\int\) e^{-cot x} . cosec² x . cot x dx
Put -cot x = t ⇒ cosec² x dx = dt
(1) becomes y . e^t = \(\int\) -t. e^t dt
= -(t – 1) e^t + c
y . e^{-cot x} = -(-cot x – 1) e^{-cot x} + 1
= (cot x + 1) e^{-cot x} + c is the required solution.

Question 70.
Find the solution of the equation
x(x – 2) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2(x – 1)y = x³(x – 2)
which satisfies the condition that y = 9 when x = 3.
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Important Questions 46

Question 71.
Solve (1 + y²)dx = (tan^-1y – x)dy. [A.P. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Differential Equations Important Questions 48
is the solution.

Inter 2nd Year Maths 2B Definite Integrals Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Definite Integrals Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Definite Integrals Important Questions

Question 1.
\(\int_{2}^{3} \frac{2 x}{1+x^{2}} d x\) [T.S. Mar. 16; May 06]
Solution:
I = \(\left[\ln \left|1+x^{2}\right|\right]_{2}^{3}\)
= ln 10 – ln 5
= ln (10/5)
= ln 2

Question 2.
\( \int_{0}^{\pi} \sqrt{2} \cdot \sqrt{2} \sqrt{\cos ^{2} \frac{\theta}{2} d \theta}\) [A.P. Mar. 16; Mar. 05]
Solution:
I = \( \int_{0}^{\pi} \sqrt{2} \cdot \sqrt{2} \sqrt{\cos ^{2} \frac{\theta}{2} d \theta}\)
= \(\int_{0}^{\pi} 2 \cdot \cos \theta / 2 d \theta\)
= \([4 \sin \theta / 2]_{0}^{\pi}\)
= 4 (sin \(\frac{\pi}{2}\) – sin 0)
= 4

Question 3.
\(\int_{0}^{2}|1-x| d x\) [A.P. Mar. 15; May 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 1

Question 4.
I = \(\int_{1}^{5} \frac{d x}{\sqrt{2 x-1}}\) [T.S. Mar. 15]
Solution:
Let 2x – 1 = t²
2 dx = 2t dt
dx = t dt
UL : t = 3
LL : t = 1
I = \(\int_{1}^{3} \frac{t d t}{t}\)
= \(\int_{1}^{3} d t\)
= \([\mathrm{t}]_{1}^{3}\) = 3 – 1
= 2

Question 5.
I = \(\int_{0}^{1} \frac{x^{2}}{x^{2}+1} d x\) [Mar. 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 2

Question 6.
\(\int_{0}^{2 \pi}\) sin²x cos⁴; x dx [T.S. Mar. 15; Mar 14]
Solution:
sin²x cos⁴x is even function.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 3

Question 7.
Evaluate \(\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x\) [T.S. Mar. 16]
Solution:
Put x = a sin θ ⇒ dx = a cos θ . dθ
θ = 0 ⇒ x = 0, x = a ⇒ θ = \(\frac{\pi}{2}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 4

Question 8.
Find \(\int_{-\pi / 2}^{\pi / 2}\) sin² x cos⁴ x dx [A.P. Mar. 16]
Solution:
f(x) is even
∴ \(\int_{-\pi / 2}^{\pi / 2}\) sin² x cos⁴ x dx = 2 \(\int_{0}^{\pi / 2} \sin ^{2} x \cdot \cos ^{4} x d x\)
= 2 . \(\frac{3}{6}\) . \(\frac{1}{4}\) . \(\frac{1}{4}\) . \(\frac{\pi}{4}\) = \(\frac{\pi}{16}\)

Question 9.
\(\int_{0}^{\pi / 2} \cdot \frac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x\) [Mar. 14, 08]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 5

Question 10.
I = \(\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\) [Mar. 08]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 6
= –\(\frac{1}{40} \ln \left[\frac{1 / 4}{9 / 4}\right]\) = \(\frac{1}{40}\) . 2ln . 3 = \(\frac{1}{20}\) ln 3

Question 11.
y = x³ + 3, y = 0, x = -1, x = 2 [Mar. 05]
Solution:
Required area PABQ
Inter 2nd Year Maths 2B Definite Integrals Important Questions 7

Question 12.
x = 4 – y², x = 0. [Mar. 11]
Solution:
The given parabola x = 4 – y² meets, the x – axis at A (4, 0) and Y – axis at P(0, 2) and Q(6, -2).
The parabola is symmetrical about X – axis
Inter 2nd Year Maths 2B Definite Integrals Important Questions 8
Required area = 2 Area of OAP

Question 13.
Evaluate the following definite integrals.
(i) \(\int_{0}^{\pi / 2} \sin ^{4} x \cos ^{5} x d x\)
(ii) \(\int_{0}^{\pi / 2} \sin ^{5} x \cos ^{4} x d x\)
(iii) \(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{4} x d x\)
Solution:
(i) \(\int_{0}^{\pi / 2} \sin ^{4} x \cos ^{5} x d x\)
= \(\frac{4}{9}\) . \(\frac{2}{7}\) . \(\frac{1}{5}\) = \(\frac{8}{315}\)

(ii) \(\int_{0}^{\pi / 2} \sin ^{5} x \cos ^{4} x d x\)
= \(\frac{3}{9}\) . \(\frac{1}{7}\) . \(\frac{4}{5}\) . \(\frac{2}{3}\) = \(\frac{8}{315}\)

(iii) \(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{4} x d x\)
= \(\frac{3}{10}\) . \(\frac{1}{8}\) . \(\frac{5}{6}\) . \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\)
= \(\frac{3}{512}\) π

Question 14.
\(\int_{0}^{\pi / 2} \frac{d x}{4+5 \cos x}\) [A.P. Mar. 16, 15]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 10
= \(\frac{1}{3}\left[\ln \frac{4}{2}\right]=\frac{1}{3} \ln 2\)

Question 15.
\(\int_{0}^{\pi} \frac{x}{1+\sin x} d x\) [May 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 11

Question 16.
\(\int_{0}^{\pi} \frac{x \sin ^{3} x}{1+\cos ^{2} x} d x\) [T.S. Mar. 15; Mar. 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 13

Question 17.
\(\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x\) [Mar. 07, 05]
Solution:
Put x = tan θ
dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = 1 ⇒ θ = \(\frac{\pi}{4}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 15

Question 18.
\(\int_{0}^{\pi / 4} \log (1+\tan x) d x\) [A.P. Mar. 16]
Solution:
I = \(\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 17

Question 19.
y = 4x – x², y = 5 – 2x. [T.S. Mar. 16]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 18
y = 4x – x² …………….. (i)
y = 5 – 2x ……………….. (ii)
y = -([x – 2]²) + 4
y – 4 = (x – 2)²
Solving equations (i) and (ii) we get
4x – x² = 5 – 2x
x² – 6x + 5 = 0
(x – 5) (x – 1) = 0
x = 1, 5
Required area = \(=\int_{1}^{5}\left(4 x-x^{2}-5+2 x\right) d x\)
= \(\int_{-1}^{5}\left(6 x-x^{2}-5\right) d x\)
= \(\left(3 x^{2}-\frac{x^{3}}{3}-5 x\right)_{1}^{5}\)
= (75 – \(\frac{125}{3}\) – 25) – (3 – \(\frac{1}{3}\) – 5)
= 50 – \(\frac{125}{3}\) + 2 + \(\frac{1}{3}\)
= \(\frac{150-125+6+1}{3}\) = \(\frac{32}{3}\) sq. units.

Question 20.
y² = 4x, y² = 4(4 – x) [May 11]
Solution:
Equations of the curves are y² = 4x ………………… (1)
y² = 4(4 – x) …………………. (2)
Eliminating y, we get
4x = 4 (4 – x)
2x = 4 ⇒ x = 2
Substituting in equation (1), y² = 8
Inter 2nd Year Maths 2B Definite Integrals Important Questions 19
= 2[\(\frac{4}{3}\)(2\(\sqrt{2}\)) – \(\frac{4}{3}\)(-2\(\sqrt{2}\))]
= 2(\(\frac{8 \sqrt{2}}{3}\) + \(\frac{8 \sqrt{2}}{3}\))
= 2(\(\frac{16 \sqrt{2}}{3}\)) = \(\frac{32 \sqrt{2}}{3}\) sq. units

Question 21.
Show that the area of the region bounded by \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (ellipse) is it ab. also deduce the area of the circle x² + y² = a². [Mar. 14, May 05]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 20
The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4 . \(\frac{\pi}{4}\) ab

(from Prob. 8 in ex 10(a))
= πab
Substituting b = a, we get the circle
x² + y² = a²
Area of the circle = πa(a) = πa² sq. units.

Question 22.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) [Mar. 14]
Solution:
Let A = \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
Put x = \(\frac{\pi}{2}\) – t, dx = – dt
Inter 2nd Year Maths 2B Definite Integrals Important Questions 22
= \(\int_{\pi / 6}^{\pi / 3} d x=(x)_{\pi / 6}^{\pi / 3}\)
= \(\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)
A = \(\frac{\pi}{12}\)

Question 23.
Find \(\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\sin x} d x\) [T.S. Mar. 16] [A.P. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 23

= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 2 . tan \(\frac{\pi}{2}\) – 2 . 0
= 2 – 1 = 2
2A = \(\pi(x)_{0}^{\pi}\) – 2π = π(π) – 2 = π² – 2π
A = \(\frac{\pi^{2}}{2}\) – π

Question 24.
Find \(\int^{\pi} x \sin ^{7} x \cos ^{6} x d x .\) [May 05] [T.S. Mar. 19]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 25
A = \(\pi \int_{0}^{\pi / 2} \sin ^{7} x \cos ^{6} x d x\)
= π . \(\frac{6}{17}\) . \(\frac{3}{11}\) . \(\frac{1}{9}\) . \(\frac{6}{7}\) . \(\frac{4}{5}\) . \(\frac{2}{3}\)
= π \(\frac{16}{3003}\)

Question 25.
Evaluate \(\int_{1}^{2} x^{5} d x\) dx
Solution:
\(\int_{1}^{2} x^{5} \cdot d x=\left[\frac{x^{6}}{6}\right]_{1}^{2}\)
= \(\frac{2^{6}}{6}-\frac{1}{6}=\frac{63}{6}=\frac{21}{2}\)

Question 26.
Evaluate \(\int_{0}^{\pi} \sin x d x\)
Solution:
\(\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}\)
= – cos π – (- cos 0)
= + 1 + 1= 2

Question 27.
Evaluate \(\int_{0}^{a} \frac{d x}{x^{2}+a^{2}}\)
Solution:
\(\int_{0}^{a} \frac{d x}{x^{2}+a^{2}}=\left[\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}\)
= \(\frac{1}{a}\) [tan^-1 (1) – tan^-1 (0)]
= \(\frac{1}{a}\) (\(\frac{\pi}{4}\) – 0) = \(\frac{\pi}{4 a}\)

Question 28.
Evaluate \(\int_{1}^{4} x \sqrt{x^{2}-1} d x\)
Solution:
g(x) = x² – 1
f(t) = \(\sqrt{t}\)
g'(x) = 2x
Inter 2nd Year Maths 2B Definite Integrals Important Questions 26

Question 29.
Evaluate \(\int_{0}^{2} \sqrt{4-x^{2}} d x\)
Solution:
Let g(θ) = 2 sin θ ⇒ g'(θ) = 2 cos θ
f(x) = \(\sqrt{4-x^{2}}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 27

Question 30.
Evaluate \(\int_{0}^{16} \frac{x^{1 / 4}}{1+x^{1 / 2}} d x\)
Solution:
Put t⁴ = x ⇒ dx = 4t³ . dt
Inter 2nd Year Maths 2B Definite Integrals Important Questions 28

Question 31.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin |x| d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 29

Question 32.
Show that \(\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x\)
Solution:
f(x) = sinⁿx.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 31

Question 33.
Evaluate \(\int_{0}^{\pi / 2} \frac{\cos ^{5 / 2} x}{\sin ^{5 / 2} x+\cos ^{5 / 2} x} d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 32
= \(\int_{0}^{\pi / 2} d x=(x)_{0}^{\pi / 2}=\frac{\pi}{2}\)
A = \(\int_{0}^{\pi / 2} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x} d x=\frac{\pi}{4}\)

Question 34.
Show that \(\int_{0}^{\pi / 2} \frac{x}{\sin x+\cos x} d x=\frac{\pi}{2 \sqrt{2}} \log (\sqrt{2}+1)\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 33

= \(\frac{\pi}{4 \sqrt{2}}\) log (\(\sqrt{2}\) + 1)²
= \(\frac{\pi}{4 \sqrt{2}}\) 2 log (\(\sqrt{2}\) + 1)
= \(\frac{\pi}{4 \sqrt{2}}\) log (\(\sqrt{2}\) + 1)

Question 35.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) [Mar 14]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 35

Question 36.
Find \(\int_{-a}^{a}\left(x^{2}+\sqrt{a^{2}-x^{2}}\right) d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 37
= 2(\(\frac{a^{3}}{3}\) – 0) + 2(0 + \(\frac{a^{2}}{3}\) sin^-1 (1) – 0 – 0)
= \(\frac{2a^{3}}{3}\) + a² . \(\frac{\pi}{2}\)

Question 37.
Find \(\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\sin x} d x\) [T.S. Mar. 16] [A.P. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 38

= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 2 . tan \(\frac{\pi}{2}\) – 2 . 0
= 2 – 1 = 2
2A = \(\pi(x)_{0}^{\pi}\) – 2π = π(π) – 2 = π² – 2π
A = \(\frac{\pi^{2}}{2}\) – π

Question 38.
Evaluate \(\int_{0}^{\pi / 2} x \sin x d x\)
Solution:
\(\int_{0}^{\pi / 2} x \cdot \sin x d x=(-x \cdot \cos x)_{0}^{\pi / 2}+\int_{0}^{\pi / 2} \cos x d x\)
= (0 – 0) + \((\sin x)_{0}^{\pi / 2}\)
= sin \(\frac{\pi}{2}\) – sin 0 = 1 – 0 = 1

Question 39.
Evaluate \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{n}\left[\frac{n-i}{n+i}\right]\) by using the method of finding definite integral as the limit of a sum.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 40

Question 40.
Evaluate \(\lim _{n \rightarrow \infty} \frac{2^{k}+4^{k}+6^{k}+\ldots .+(2 n)^{k}}{n^{k+1}}\) by using the method of finding definite integral as the limit of a sum.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 41

Question 41.
Evaluate \(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}}\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 42

Question 42.
Let f: R → R be a continuous periodic function and T be the period of it. Then prove that for any positive integer n,
\(\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x\) ………………. (1)
Solution:
Let k be an integer arid define
g : [kT, (k + 1)T] → [0, T] as g(t) = t – kT.
Then g'(t) = 1 for all t ∈ [kT, (k + 1)T].
Hence by \(\int_{g(c)}^{g(d)} f(t) d t=\int_{c}^{d} f(g(x)) g^{\prime}(x) d x,(f \circ g)\)
g’ is integrable on [kT, (k + 1 )T] and
\(\int_{k T}^{(k+1) T} f(g(t)) g^{\prime}(t) d t=\int_{0}^{T} f(x) d x\) ………… (2)
We have f(g(t)) g'(t) = f(t – kT), 1 = f(t),
since f is periodic with T as the period.
Hence \(\int_{k T}^{(k+1) T} f(g(t)) g^{\prime}(t) d t=\int_{k T}^{(k+1) T} f(t) d t\) ………….. (3)
Thus from (2) and (3),
\(\int_{k T}^{(k+1) T} f(t) d t=\int_{0}^{T} f(t) d t\) ………………. (4)
Let us now prove eq. (1) by using the principle of mathematical induction.
For n = 1, clearly (1) is true.
Assume (1) is true for a positive integer m.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 43
Hence eq. (1) is true for n = m + 1
Thus, eq. (1) ¡s true for any positive integer n, by the principle of mathematical induction.

Question 43.
Find
(i) \(\int_{0}^{\pi / 2} \sin ^{4} x d x\)
(ii) \(\int_{0}^{\pi / 2} \sin ^{7} x d x\)
(iii) \(\int_{0}^{\pi / 2} \cos ^{8} x d x\)
Solution:
(i) \(\int_{0}^{\pi / 2} \sin ^{4} x d x\)
= \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\) = \(\frac{3\pi}{16}\)

(ii) \(\int_{0}^{\pi / 2} \sin ^{7} x d x\)
= \(\frac{6}{7}\) . \(\frac{4}{5}\) . \(\frac{2}{3}\) = \(\frac{16}{35}\)

(iii) \(\int_{0}^{\pi / 2} \cos ^{8} x d x\)
= \(\frac{7}{8}\) . \(\frac{5}{6}\) . \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\) = \(\frac{35\pi}{256}\)

Question 44.
Evaluate \(\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x\) [T.S. Mar. 16]
Solution:
Put x = a sin θ ⇒ dx = a cos θ . dθ
θ = 0 ⇒ x = 0, x = a ⇒ θ = \(\frac{\pi}{2}\)
\(\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x=\int_{0}^{\pi / 2}(a \cos \theta)(a \cos \theta) d \theta\)
= a² \(\int_{0}^{\pi / 2}\) cos² θ dθ
= a² . \(\frac{1}{2}\) . \(\frac{\pi}{2}\) = \(\frac{\pi a^{2}}{4}\)

Question 45.
Evaluate the following definite integrals.
(i) \(\int_{0}^{\pi / 2}\) sin⁴ x . cos⁵ x dx
(ii) \(\int_{0}^{\pi / 2}\) sin⁵ x . cos⁴ x dx
(iii) \(\int_{0}^{\pi / 2}\) sin⁶ x . cos⁴ x dx
Solution:
(i) \(\int_{0}^{\pi / 2}\) sin⁴ x . cos⁵ x dx
= \(\frac{4}{9}\) . \(\frac{2}{7}\) . \(\frac{1}{5}\) = \(\frac{8}{315}\)

(ii) \(\int_{0}^{\pi / 2}\) sin⁵ x . cos⁴ x dx
= \(\frac{3}{9}\) . \(\frac{1}{7}\) . \(\frac{4}{5}\) . \(\frac{2}{3}\) = \(\frac{8}{315}\)

(iii) \(\int_{0}^{\pi / 2}\) sin⁶ x . cos⁴ x dx
= \(\frac{3}{10}\) . \(\frac{1}{8}\) . \(\frac{5}{6}\) . \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\)
= \(\frac{3}{512}\) π

Question 46.
Find \(\int_{0}^{2 \pi}\) sin⁴ x . cos⁶ x dx [T.S. Mar. 19]
Solution:
f(x) = sin⁴ x . cos⁶ x dx
f(2π – x) = f(π – x) = f(x)
\(\int_{0}^{2 \pi}\) sin⁴ x . cos⁶ x dx = 2 \(\int_{0}^{2 \pi}\) sin⁴ x . cos⁶ x dx
= 4 \(\int_{0}^{2 \pi}\) sin⁴ x . cos⁶ x dx
= 4 . \(\frac{5}{10}\) . \(\frac{3}{8}\) . \(\frac{1}{6}\) . \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\)
= \(\frac{3}{128}\) π

Question 47.
Find \(\int_{-\pi / 2}^{\pi / 2}\) sin² x . cos⁴ x dx [A.P. Mar. 16, 19]
Solution:
f(x) is even
∴ \(\int_{-\pi / 2}^{\pi / 2}\) sin² x . cos⁴ x dx = 2 \(\int_{-\pi / 2}^{\pi / 2}\) sin² x . cos⁴ x dx
= 2 . \(\frac{3}{6}\) . \(\frac{1}{4}\) . \(\frac{1}{4}\) . \(\frac{\pi}{4}\)
= \(\frac{\pi}{16}\)

Question 48.
Find \(\int_{0}^{\pi}\) x sin⁷ x . cos⁶ x dx [May 05]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 44
\(\int_{0}^{\pi}\) x sin⁷ x . cos⁶ x dx = 2 \(\int_{0}^{\pi}\) sin⁷ x . cos⁶ x dx
A = π \(\int_{0}^{\pi}\) x sin⁷ x . cos⁶ x dx
= π . \(\frac{6}{17}\) . \(\frac{3}{11}\) . \(\frac{1}{9}\) . \(\frac{6}{7}\) . \(\frac{4}{5}\) . \(\frac{2}{3}\)
= π . \(\frac{16}{3003}\)

Question 49.
Find \(\int_{-a}^{a}\) a² (a² – x²)^3/2 dx
Solution:
f(x) = x² (a² – x²)
f(x) is even
Inter 2nd Year Maths 2B Definite Integrals Important Questions 45

Question 50.
Find \(\int_{0}^{1} x^{3 / 2} \sqrt{1-x} d x\)
Solution:
Put x = sin² θ
dx = 2 sin θ . cos θ . dθ
x = 0 ⇒ θ = 0, x = 1 ⇒ θ = \(\frac{\pi}{2}\)
\(\int_{0}^{1} x^{3 / 2} \sqrt{1-x} d x\)
= \(\int_{0}^{\pi / 2} \sin ^{3} \theta \cdot \cos \theta .2 \sin \theta \cos \theta d \theta\)
= \(2 \int_{0}^{\pi / 2} \sin ^{4} \theta \cdot \cos ^{2} \theta d \theta\)
= 2 . \(\frac{3}{6}\) . \(\frac{1}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\) = \(\frac{\pi}{16}\)

Question 51.
Find the area under the curve f(x) = sin x in [0, 2π].
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 46
f(x) = sin x,
We know that in [0, π], sin x ≥ 0 and [π, 2π], sin x ≤ 0
Required area = \(\int_{1}^{\pi}\) sinx dx + \(\int_{\pi}^{2 \pi}\) (-sinx) dx
= \((-\cos x)_{0}^{\pi}[\cos x]_{\pi}^{2 \pi}\)
= – cos π + cos 0 + cos 2π – cos π
= -(-1) + 1 + 1-(-1) = 1 + 1 + 1 + 1
= 4.

Question 52.
Find the area under the curve f(x) = cos x in [0, 2π].
Solution:
We know that cos x ≥ 0 in (0, \(\frac{\pi}{2}\)) ∪ (\(\frac{3\pi}{2}\), π) and ≤ 0 in \(\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 47
= sin \(\frac{\pi}{2}\) – sin 0 – sin \(\frac{3\pi}{2}\) + sin \(\frac{\pi}{2}\) + sin 2π – sin \(\frac{3\pi}{2}\)
= 1 – 0 – (-1) + 1 + 0 – (-1)
= 1 + 1 + 1 + 1 = 4.

Question 53.
Find the area bounded by the parabola y = x², the X-axis and the lines x = -1, x = 2.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 48

Question 54.
Find the area cut off between the line y = 0 and the parabola y = x² – 4x + 3.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 49
Equation of the parabola is
y = x² – 4x + 3
Equation of the line is y = 0
x² – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1, 3
The curve takes negative values for the values of x between 1 and 3.
Required area = \(\int_{1}^{3}\) -(x² – 4x + 3)dx
= \(\int_{1}^{3}\) (-x² + 4x – 3) dx
= \(\left(-\frac{x^{3}}{3}+2 x^{2}-3 x\right)_{1}^{3}\)
= (-9 + 18 – 9) – (-\(\frac{1}{3}\) + 2 – 3)
= \(\frac{1}{3}\) – 2 + 3 = \(\frac{4}{3}\)

Question 55.
Find the area bounded by y = sin x and y = cos x between any two consecutive points of intersection.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 50
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = 4\(\frac{1}{\sqrt{2}}\) = 2\(\sqrt{2}\)

Question 56.
Find the area of one of the curvilinear triangles bounded by y = sin x, y = cos x and X – axis.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 51
ln (0, \(\frac{\pi}{4}\)) cos x ≥ sin x and (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\)), cos x ≤ sin x.
Required area = \(\int_{0}^{\pi / 4}\) sin x dx + \(\int_{\pi / 4}^{\pi / 2}\) cos x dx
= \((-\cos x)_{0}^{\pi / 4}+(\sin x)_{\pi / 4}^{\pi / 2}\)
= – cos \(\frac{\pi}{4}\) + cos 0 + sin \(\frac{\pi}{2}\) – sin \(\frac{\pi}{4}\)
= – \(\frac{1}{\sqrt{2}}\) + 1 + 1 – \(\frac{1}{\sqrt{2}}\)
= 2(1 – \(\frac{1}{\sqrt{2}}\)) = 2 – \(\sqrt{2}\)

Question 57.
Find the area of the right angled triangle with base b and altitude h, using the fundamental theorem of integral calculus.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 52
OAB is a right angled triangle and ∠B = 90° take ‘O’ as the origin and OB as positive X-axis
If OB = band AB = h, then Co-ordinates of A are (b, h)
Equation of OA is y = \(\frac{h}{b}\) x
Area of the triangle OAB = \(\int_{0}^{b} \frac{h}{b} x d x\)
= \(\frac{h}{b}\left(\frac{x^{2}}{2}\right)_{0}^{b}=\frac{h}{b} \cdot \frac{b^{2}}{2}=\frac{1}{2} b h .\)

Question 58.
Find the area bounded between the curves y² – 1 = 2x and x = 0.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 53
= \(\left(-\frac{y^{3}}{3}+y\right)_{0}^{1}\) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 59.
Find the area enclosed by the curves y = 3x and y = 6x – x².
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 54
y = 6x – x²
The straight line y = 3x meets the parabola
y = 6x – x²
3x = 6x = x²
x² – 3x = 0
x(x – 3) = 0
x = 0 or 3
Required area = \(\int_{0}^{3}\left(6 x-x^{2}-3 x\right) d x\)
= \(\int_{0}^{3}\left(3 x-x^{2}\right) d x=\left(\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{3}\)
= \(\frac{27}{2}\) – \(\frac{27}{3}\) = \(\frac{27}{6}\) = \(\frac{9}{2}\)

Question 60.
Find the area enclosed between y = x² – 5x and y = 4 – 2x.
Solution:
Equations of the curves are
y = x² – 5x ………. (1)
y = 4 – 2x …………….. (2)
x² – 5x = 4 – 2x
x² – 3x – 4 = 0
(x + 1)(x – 4) = 0
x = -1, 4
Inter 2nd Year Maths 2B Definite Integrals Important Questions 55
= 44 – \(\frac{64}{3}\) – \(\frac{3}{2}\) – \(\frac{1}{3}\)
= \(\frac{264-128-9-2}{6}\) = \(\frac{125}{6}\)

Question 61.
Find the area bounded between the curves y = x², y = \(\sqrt{x}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 56
= \(\left(\frac{2}{3} x \sqrt{x}-\frac{x^{3}}{3}\right)_{0}^{1}\)
= \(\frac{2}{3}\) – \(\frac{1}{3}\) = \(\frac{1}{3}\)

Question 62.
Find the area bounded between the curves y² = 4ax, x² = 4by (a > 0, b > 0).
Solution:
Equations of the given curves are
y² = 4ax …………………… (1)
x² = 4by ……………………. (2)
From equation (2), y = \(\frac{x^{2}}{4 b}\)
Substituting in (1) \(\left(\frac{x^{2}}{4 b}\right)^{2}\) = 4ax
x⁴ = (16 b²) |4ax|
Inter 2nd Year Maths 2B Definite Integrals Important Questions 57

Inter 2nd Year Maths 2B Integration Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Integration Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Integration Important Questions

Question 1.
\(\int\left(\frac{1}{1-x^{2}}+\frac{2}{1+x^{2}}\right)\) [May 11]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 1

Question 2.
\(\int\) sec²x cosec²x dx on I ⊂ R \ ({nπ : n ∈ Z} ∪ {(2n + 1) \(\frac{\pi}{2}\) : n ∈ Z}) [T.S. Mar. 16; Mar, May 07]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 2

Question 3.
\(\int \frac{1+\cos ^{2} x}{1-\cos 2 x}\) dx on I ⊂ R \ {nπ : n ∈ Z} [Mar. 13]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 3

Question 4.
\(\int \sqrt{1-\cos 2 x}\) dx on I ⊂ [2nπ, (n + 1) π], n ∈ Z [May 06]
Solution:
\(\int \sqrt{1-\cos 2 x}\) dx = \(\int \sqrt{2}\) sin x dx
= –\(\sqrt{2}\) cos x + C

Question 5.
\(\int \frac{1}{\cosh x+\sinh x}\) dx on R. [A.P. Mar. 16]
Solution:
\(\int \frac{1}{\cosh x+\sinh x}\) dx
= \(\int \frac{\cosh x-\sinh x}{\cosh ^{2} x-\sinh ^{2} x}\) dx
= \(\int\) (cosh x – sinh x) dx
= sinh x – cosh x + C

Question 6.
\(\int \frac{1}{1+\cos x}\) dx on I ⊂ R \ {(2n + 1)π : n ∈ Z} [T.S. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 4

Question 7.
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx, x ∈ R. [A.P. Mar. 15]
Solution:
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx
t = tan^-1 x ⇒ dt = \(\frac{d x}{1+x^{2}}\)
\(\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}}\)dx = \(\int\) sin t dt
= – cos t + t
= -cos (tan^-1 x) + C

Question 8.
\(\int \frac{\log (1+x)}{1+x}\)dx on (-1, ∞) [T.S. Mar. 15]
Solution:
\(\int \frac{\log (1+x)}{1+x}\)dx
t = 1 + x ⇒ dt = dx
Inter 2nd Year Maths 2B Integration Important Questions 5

Question 9.
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\) dx on I = (-1, 1). [May 05]
Solution:
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\)
t = x³ ⇒ dt = mx² dx
\(\int \frac{x^{2}}{\sqrt{1-x^{6}}}\) = \(\frac{1}{3} \int \frac{d \mathrm{t}}{\sqrt{1-\mathrm{t}^{2}}}\)
= \(\frac{1}{3}\) sin^-1 t + C
= \(\frac{1}{3}\) sin^-1 (x³) + C

Question 10.
\(\int \frac{x^{8}}{1+x^{18}}\) dx on R. [A.P. Mar. 16]
Solution:
t = x⁹ ⇒ dt = 9x⁸ dx
\(\int \frac{x^{8} d x}{1+x^{18}}=\int \frac{x^{8}}{1+\left(x^{9}\right)^{2}} d x\)
= \(\frac{1}{9} \int \frac{d t}{1+t^{2}}\) = = \(\frac{1}{9}\) tan^-1 t + C
= \(\frac{1}{9}\) tan^-1 (x⁹) + C

Question 11.
\(\int \frac{1}{x \log x[\log (\log x)]}\) dx on (1, ∞) [Mar. 11]
Solution:
t = log (log x)
dt = \(\frac{1}{\log x} \cdot \frac{1}{x}\) dx
\(\int \frac{1}{x \log x[\log (\log x)]}\) dx = \(\int \frac{d t}{t}\)
= log |t| + C
= log |log(log x)| + C

Question 12.
\(\int \frac{1}{(x+3) \sqrt{x+2}}\)dx on I ⊂ (-2, ∞) [Mar. 14]
Solution:
x + 2 = t²
dx = 2t dt
\(\int \frac{d x}{(x+3) \sqrt{x+2}}=\int \frac{2 t d t}{t\left(t^{2}+1\right)}\)
= \(2 \int \frac{d t}{t^{2}+1}\)
= 2 tan^-1 (t) + C
= 2 tan^-1 (\(\sqrt{x+2}\)) + C

Question 13.
\(\int \frac{\cot (\log x)}{x}\) dx, x ∈ I ⊂ (0, ∞) \ {e^nπ : n ∈ Z). [Mar. 05]
Solution:
t = log x ⇒ dt = \(\frac{\mathrm{dx}}{\mathrm{x}}\) \(\int\)
\(\int \frac{\cot (\log x)}{x}\) dx = \(\int\) cot t dt = log (sin t) + C
= log |sin (log x)| + C

Question 14.
\(\int\) (tan x + log sec x)e^x dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π) n ∈ Z. [May 07, Mar. 08]
Solution:
t = log |sec x| ⇒ dt = \(\frac{1}{\sec x}\) . sec x . tan x dx
= tan x dx
\(\int\) (tan x + log sec x)e^x dx = e^x . log|sec x| + C

Question 15.
\(\int \sqrt{x}\) log x dx on (0, ∞) [T.S. Mar. 16]
Solution:
\(\int \sqrt{x}\) log x dx
= (log x) \(\frac{2}{3}\) x^3/2 – \(\frac{2}{3}\) \(\int\) x^3/2 \(\frac{1}{x}\) dx
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{2}{3}\) \(\int\) x^1/2 dx
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{2}{3}\) \(\frac{x^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{3}\) x^3/2 (log x) – \(\frac{4}{9}\) x^3/2 + C

Question 16.
\(\int\) e^x (tan x + sec² x)dx on I ⊂ R \ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z} [Mar 06]
Solution:
f(x) = tanx = f'(x) ⇒ sec² x dx
I = \(\int\) e^x [f(x) + f'(x)] dx = e^x. f(x) + C
= e^x . tan x + C

Question 17.
\(\int e^{x}\left(\frac{1+x \log x}{x}\right)\) dx on (0, ∞) [A.P. Mar. 15, Mar. 13]
Solution:
\(\int e^{x}\left(\frac{1+x \log x}{x}\right)\) dx = \(\int\) e^x (log x + \(\frac{1}{x}\))dx
= e^x . log x + C

Question 18.
\(\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5}\)dx [Mar. 07]
Solution:
t = sin x ⇒ dt = cos x dx
I = \(\int \frac{d t}{t^{2}+4 t+5}=\int \frac{d t}{(t+2)^{2}+1}\)
= tan^-1(t + 2) + C
= tan^-1(sin x + 2) + C

Question 19.
\(\int \frac{d x}{(x+1)(x+2)}\) [Mar. 14, May 11]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 6

Question 20.
\(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx on I ⊂ R \ {x ∈ R : cos (xe^x) = 0} [T.S. Mar. 17]
Solution:
t = x . e^x
dt = (x . e^x + e^x) dx = e^x (1 + x) dx
\(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx = \(\int \frac{d t}{\cos ^{2} t}=\int \sec ^{2} t d t\)
= tan t + C
= tan (x . e^x) + C

Question 21.
\(\int\) x tan^-1 x dx, x∈ R [Mar. 05]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 7

Question 22.
\(\int \sqrt{1+3 x-x^{2}} d x\) [May 11]
Solution:
\(\int \sqrt{1+3 x-x^{2}} d x=\int \sqrt{1-\left(x^{2}-3 x\right)} d x\)
= \(\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x\)
Inter 2nd Year Maths 2B Integration Important Questions 8

Question 23.
\(\int \frac{9 \cos x-\sin x}{4 \sin x+5 \cos x} d x\) [T.S. Mar. 17; Mar. 08]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 9

Question 24.
\(\int \frac{d x}{5+4 \cos 2 x}\) [Mar. 11]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 10

Question 25.
Obtain reduction formula for I_n = \(\int\) cotⁿ x dx, n being a positive integer. n ≥ 2 and deduce the value of \(\int\) cot⁴ x dx. [T.S. Mar. 19] [A.P. Mar. 16; May 11]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 11

Question 26.
Obtain reduction formula for I_n = \(\int\) cosecⁿ x dx, n being a positive integer. n ≥ 2 and deduce the value of \(\int\) cosec⁵ x dx. [T.S. Mar. 16]
Solution:
I_n = \(\int\) cosecⁿ x dx = \(\int\) cosec^{n – 2}x . cosec² x dx
= cosec^{n – 2}x (-cot x) + \(\int\) cot x . (n – 2) cosec^{n – 3} . (cot x) dx
= – cosec^{n – 2}x . cot x + (n – 2) \(\int\) cosec^{n – 2} x . (cosec² x -1) dx
= – cosec^{n – 2}x . cot x + (n – 2) I_{n – 2} – (n – 2)I_n
I_n (1 + n – 2) = – cosec^{n – 2} x . cot x + (n – 2) I_{n – 2}
Inter 2nd Year Maths 2B Integration Important Questions 12

Question 27.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\) dx. [T.S. Mar. 15]
Solution:
We write
2x + 5 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 2x + 10) + B
= A(2x – 2) + B
On comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
Thus 2x + 5 = (2x – 2) + 7
Hence \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\) dx
= \(\int \frac{2 x-2}{\sqrt{x^{2}-2 x+10}} d x+7 \int \frac{d x}{\sqrt{x^{2}-2 x+10}}+C\)
= \(2 \sqrt{x^{2}-2 x+10}+7 \int \frac{d x}{\sqrt{(x-1)^{2}+3^{2}}}\) + C
= \(2 \sqrt{x^{2}-2 x+10}+7 \sinh ^{-1}\left(\frac{x-1}{3}\right)\) + C

Question 28.
Evaluate \(\int\) sin⁴ x dx. [Mar. 14]
Solution:
I_n = sinⁿ x dx = \(-\frac{\sin ^{n-1} x \cdot \cos x}{n}+\frac{n-1}{n} \cdot I_{n-2}\)
Inter 2nd Year Maths 2B Integration Important Questions 13

Question 29.
\(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\) [A.P. Mar. 15]
Solution:
Let 2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
Equating the coefficient of sin x and cos x we get
4A + 5B = 2
5A – 4B = 3
Inter 2nd Year Maths 2B Integration Important Questions 14

Question 30
\(\int \frac{1}{1+\sin x+\cos x}\) dx [T.S. Mar. 15]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 15

Question 31.
\(\int\) (6x + 5) \(\sqrt{6-2 x^{2}+x}\) dx [May 06]
Solution:
Let 6x + 5 = A(1 – 4x) + B
Equating the co-efficients of x
6 = -4 A ⇒ A = \(\frac{-3}{2}\)
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
Inter 2nd Year Maths 2B Integration Important Questions 16

Question 32.
\(\int \frac{d x}{4+5 \sin x}\) [Mar. 05]
Solution:
t = tan \(\frac{x}{2}\) ⇒ dt = sec² \(\frac{x}{2}\) . \(\frac{1}{2}\) dx
Inter 2nd Year Maths 2B Integration Important Questions 18

Question 33.
\(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^{2}}}\) [May 05]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 20

Question 34.
\(\int \frac{d x}{4 \cos x+3 \sin x}\) [Mar. 06]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 22

Question 35.
\(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\) [Mar. 14, 11] [A.P. & T.S. Mar. 16]
Solution:
Let 2 sin x + 3 cos x + 4
= A(3 sin x+4 cos x + 5) + 3(3 cos x – 4sin x) + C
Equating the co-efficients of
sin x. we get 3A – 4B = 2
cos x, we get 4A + 3B = 3
Inter 2nd Year Maths 2B Integration Important Questions 24

Substituting in (1)
I = \(\frac{18}{25}\) . x + \(\frac{1}{25}\) log |3 sin x + 4 cos x + 5| – \(\frac{4}{5\left(3+\tan \frac{x}{2}\right)}\) + C

Question 36.
\(\int \frac{x+3}{(x-1)\left(x^{2}+1\right)} d x\) dx [May 07]
Solution:
Let \(\frac{x+3}{(x-1)\left(x^{2}+1\right)}\) = \(\frac{A}{x-1}\) + \(\frac{B x+C}{x^{2}+1}\)
⇒ (x + 3) = A(x² + 1) + (Bx + C)(x – 1) ………………… (1)
Put x = 1 in (1)
Then 4 = A(1 + 1) + 0 ⇒ A = 2
Put x = 0 in (1)
3 = A(1) + C(-1)
⇒ A – C = 3 ⇒ C = A – 3 = 2 – 3 = -1
Equating coefficient of x² in (1)
0 = A + B
⇒ B = -A = -2
Inter 2nd Year Maths 2B Integration Important Questions 26

Question 37.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 27

Question 38.
Find \(\int\) 2x⁷ dx on R.
Solution:
\(\int\) 2x⁷ dx = 2 \(\int\) x⁷ dx
= 2 . \(\frac{x^{8}}{8}\) + C
= \(\frac{x^{8}}{4}\) + C

Question 39.
Evaluate \(\int\) cot²x dx on I ⊂ R \ {nπ : n ∈ Z}.
Solution:
\(\int\) cot²x dx = \(\int\) (cosec²x – 1) dx
= \(\int\) cosec² x dx – \(\int\) dx
= -cot x – x + C

Question 40.
Evaluate \(\int\left(\frac{x^{6}-1}{1+x^{2}}\right)\) dx for x ∈ R.
Solution:
\(\int\left(\frac{x^{6}-1}{1+x^{2}}\right)\) dx = \(\int\)[(x⁴ – x² + 1) – \(\frac{2}{1+x^{2}}\)] dx
= \(\int\) x⁴ dx – \(\int\) x² dx + \(\int\) dx – 2 \(\int \frac{d x}{1+x^{2}}\)
= \(\frac{x^{5}}{5}\) – \(\frac{x^{3}}{3}\) + x – 2 tan^-1 x + C.

Question 41.
Find \(\int\) (1 – x) (4 – 3x) (3 + 2x) dx, x ∈ R.
Solution:
(1 – x) (4 – 3x) (3 + 2x) = 6x³ – 5x² – 13x + 12
\(\int\)(1 – x) (4 – 3x) (3 + 2x) dx
= \(\int\) (6x³ – 5x² – 13x + 12) dx
= 6\(\int\) x³dx – 5 \(\int\) x² dx – 13 \(\int\) x dx + 12 \(\int\) dx
= \(\frac{6 x^{4}}{4}\) – 5 \(\frac{x^{3}}{3}\) – \(\frac{13 x^{2}}{2}\) + 12x + C
= \(\frac{3}{2}\)x⁴ – \(\frac{5}{3}\)x³ – \(\frac{13}{2}\)x² + 12x + C.

Question 42.
Evaluate \(\int\left(x+\frac{1}{x}\right)^{3}\) dx, x > 0.
Solution:
(x + \(\frac{1}{x}\))³ = x³ + 3x + \(\frac{3}{x}\) + \(\frac{1}{x^{3}}\)
\(\int\) (x + + \(\frac{1}{x}\))³ dx = \(\int\) (x³ + 3x + \(\frac{3}{x}\) + \(\frac{1}{x^{3}}\)) dx
= \(\int\) x³ dx + 3 \(\int\) x dx + 3 \(\int\) \(\frac{\mathrm{dx}}{\mathrm{x}}\) + \(\int\) \(\frac{d x}{x^{3}}\)
= \(\frac{x^{4}}{4}\) + \(\frac{3 x^{2}}{2}\) + 3 log x – \(\frac{1}{2 x^{2}}\) + C

Question 43.
Find \(\int \sqrt{1+\sin 2 x}\) dx on R.
Solution:
1 + sin 2x = sin² x + cos² x + 2 sin x . cos x
= (sin x + cos x)²
\(\sqrt{1+\sin 2 x}\) = sin x + cos x
If 2nπ – \(\frac{\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{3\pi}{4}\)
= -(sin x + cos x). otherwise
If 2nπ – \(\frac{\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{3\pi}{4}\), then
\(\int \sqrt{1+\sin 2 x}\) dx = \(\int\) (sin x + cos x)dx
= -cos x + sin x + C
If 2nπ + \(\frac{3\pi}{4}\) ≤ x ≤ 2nπ + \(\frac{7\pi}{4}\)
\(\int \sqrt{1+\sin 2 x}\)
= \(\int\) -(sin x + cos x)dx
= –\(\int\) sin x dx – \(\int\) cos x dx
= cos x – sin x + c

Question 44.
Evaluate \(\int \frac{2 x^{3}-3 x+5}{2 x^{2}}\) dx for x > 0 and verify the result by differentiation.
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 29
and it is the given expression and hence the result is correct.

Question 45.
Evaluate \(\int \frac{x^{5}}{1+x^{12}}\) dx on R.
Solution:
We define f : R → R by f(t) = \(\frac{1}{1+t^{2}}\)
g : R → R by g(x) = x⁶
Then g'(x) = 6x⁵
Define F : R → R by F(t) = tan^-1 t
F is the primitive of f
\(\int \frac{x^{5}}{1+x^{12}}\) dx = \(\frac{1}{6}\) \(\int\) f(g(x)) g'(x) dx
= \(\frac{1}{6}\) (F(t) + C)_t=g(x)
= \(\frac{1}{6}\) [tan^-1 t + C]_t=x⁶
= \(\frac{1}{6}\) tan^-1 x⁶ + C

Question 46.
Evaluate \(\int\) cos³ x sin x dx on R.
Solution:
We define : f = R → R by f(x) = cosx
∴ f'(x) = – sin x
\(\int\) cos³ x sin x dx = \(\int\) (f(x))³ [-f'(x)] dx
= \(\frac{-[f(x)]^{4}}{4}\) + C
= \(\frac{-\cos ^{4} x}{4}\) + C

Question 47.
Find \(\int\) (1 – \(\frac{1}{x^{2}}\)) e^{(x + \(\frac{1}{x}\))} dx on I where I = (0, ∞)
Solution:
Let J = I = (0, ∞)
Define f : I → R by f(t) = e^t and g : J → R by g(x) = x + \(\frac{1}{x}\)
Then g(J) ⊂ I, g'(x) = 1 – \(\frac{1}{x^{2}}\)
\(\int\) (1 – \(\frac{1}{x^{2}}\)) e^{(x + \(\frac{1}{x}\))} dx = \(\int\) f (g(x)) g'(x) dx
= \(\int\) [f(t)dt]_{t = g(x)}
= [\(\int\)e^t dt]_{t = g(x)}
= [e^t + c]_{t=x+\(\frac{1}{x}\)}
= e^{(x+\(\frac{1}{x}\))} + C

Question 48.
Evaluate \(\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^{2}}}\) dx on I = (0, 1).
Solution:
We define f : I → R by f(x) = sin^-1x
f'(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
\(\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^{2}}}\) dx = \(\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x\)
= 2\(\sqrt{f(x)}\) + C
= 2 \(\sqrt{\sin ^{-1} x}\) + C

Question 49.
Evaluate \(\int \frac{\sin ^{4} x}{\cos ^{6} x}\) dx, x ∈ I ⊂ R \ {\(\frac{(2 n+1) \pi}{2}\) : n ∈ z}
Solution:
\(\int \frac{\sin ^{4} \dot{x}}{\cos ^{6} x}\) dx = \(\int\) tan⁴ x . sec² x dx
We define f : I → R by f(x)
f(x) = tan x, then f'(x) = sec²x
\(\int \frac{\sin ^{4} \dot{x}}{\cos ^{6} x}\) dx = \(\int\) [f(x)]⁴ . f'(x) dx = \(\frac{[f(x)]^{5}}{5}\) + C.
= \(\frac{1}{5}\) tan⁵x + C.

Question 50.
Evaluate \(\int\) sin² x dx on R.
Solution:
\(\int\) sin²x dx = \(\int \frac{(1-\cos 2 x)}{2}\) dx
= \(\frac{1}{2}\) \(\int\) dx – \(\frac{1}{2}\) \(\int\) cos 2x dx
= \(\frac{1}{2}\) x – \(\frac{1}{4}\) sin 2x + C.
(since \(\int\) cos 2x dx = \(\frac{1}{2}\) sin 2x + C)

Question 51.
Evaluate \(\int \frac{1}{a \sin x+b \cos x}\) dx where a, b ∈ R and a² + b² ≠ 0 on R.
Solution:
We can find real numbers r and θ such that
a = r cos θ, b = r sin θ
Then r = \(\sqrt{a^{2}+b^{2}}\), cos θ = \(\frac{a}{r}\) and sin θ = \(\frac{b}{r}\)
a sin x + b cos x = r . cos θ sin x + r sin θ cos x
= r[cos θ sin x + sin θ cos x]
= r sin (x + θ)
\(
= [latex]\frac{1}{r}\) (cosec (x + θ) dx
= \(\frac{1}{r}\) log |tan \(\frac{1}{2}\)(x + θ)| + C
= \(\frac{1}{\sqrt{a^{2}+b^{2}}} \log \left|\tan \frac{1}{2}(\tilde{x}+\theta)\right|+c\)
For all x ∈ I where I is an interval disjoint with {nπ – θ : n ∈ z}.

Question 52.
Find \(\int \frac{x^{2}}{\sqrt{x+5}}\) dx on (-5, ∞)
Solution:
Put t = x + 5 so that t > 0 on (-5, ∞)
dx = dt and x = t – 5
Inter 2nd Year Maths 2B Integration Important Questions 30

Question 53.
Find \(\int \frac{x}{\sqrt{1-x}}\) dx, x ∈ I = (0, 1)
Solution:
We define f : I → R by f(x) = \(\frac{x}{\sqrt{1-x}}\)
Let J = (0, \(\frac{\pi}{2}\))
Define Φ : J → I by Φ(θ) = sin² θ
Then Φ is a bijective mapping from J to I Further Φ and Φ^-1 are differentiable on their respective domains.
put x = Φ(θ) = sin²θ
dx = 2 sin θ . cos θ dθ
Inter 2nd Year Maths 2B Integration Important Questions 31

Question 54.
Evaluate \(\int \frac{d x}{(x+5) \sqrt{x+4}}\) on (-4, ∞)
Solution:
Let I = (-4, ∞)
Define f on I as f(x) = \(\frac{d x}{(x+5) \sqrt{x+4}}\)
Let J = (0, ∞)
We define Φ : J → I by Φ(t) = t² – 4
Then Φ is differentiable and is a bijection
Put x = Φ(t) = t² – 4
then t = \(\sqrt{x+4}\) ⇒ dx = 2t dt
Thus \(\int \frac{d x}{(x+5) \sqrt{x+4}}\) = \(\int \frac{2 t d t}{\left(t^{2}+1\right) t}\)
= \(\int \frac{2 d t}{t^{2}+1}\)
= 2 tan^-1 t + C
= 2 tan^-1 (\(\sqrt{x+4}\)) + C

Question 55.
Evaluate \(\int \frac{d x}{\sqrt{4-9 x^{2}}}\) on I = (-\(\frac{2}{3}\), \(\frac{2}{3}\))
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 32

Question 56.
Evaluate \(\int \frac{1}{a^{2}-x^{2}}\) dx for x ∈ I = (-a, a)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 33

Question 57.
Evaluate \(\int \frac{1}{1+4 x^{2}}\) dx on R.
Solution:
\(\int \frac{1}{1+4 x^{2}}\) dx = \(\int \frac{d x}{4\left(\left(\frac{1}{2}\right)^{2}+x^{2}\right)}\)
= \(\frac{1}{4}\) (2 tan^-1 2x) + C
= \(\frac{1}{2}\) tan^-1 2x + C

Question 58.
Find \(\int \frac{1}{\sqrt{4-x^{2}}}\) dx on (-2, 2).
Solution:
\(\int \frac{1}{\sqrt{4-x^{2}}}\) dx = \(\int \frac{1}{\sqrt{2^{2}-x^{2}}}\) dx = sin^-1(\(\frac{x}{2}\) dx) + C

Question 59.
Evaluate \(\int \sqrt{4 x^{2}+9}\) dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 34

Question 60.
Evaluate \(\int \sqrt{9 x^{2}-25} d x\) on [\(\frac{5}{3}\), ∞)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 35

Question 61.
Evaluate \(\int \sqrt{16-25 x^{2}}\) dx on (\(\frac{-4}{5}\), \(\frac{4}{5}\))
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 36

Question 62.
Evaluate \(\int\) x sin^-1x dx on (-1, 1).
Solution:
Let u(x) = sin^-1 x and v(x) = \(\frac{x^{2}}{2}\) so that
v'(x) = x
∴ u(x) v'(x) = x sin^-1x
Even though the domain of u is (-1, 1) the function u ¡s differentiable only on (-1, 1).
From the same formula, we have
Inter 2nd Year Maths 2B Integration Important Questions 37

Question 63.
Evaluate \(\int\) x² cos x dx.
Solution:
Let us take u(x) = x², v(x) = sin x
so that v'(x) = cos x
u(x) v'(x) = x² cosx
By using the formula for integration by parts, we have
\(\int\) x² cos x dx = x² sin x – \(\int\) sin x (x²)’ dx
= x² sin x – 2 \(\int\) x sin x dx + C.
Again, by applying the formula for integration by parts to
\(\int\) x sin x dx, we get
\(\int\) x. sin x dx = -x cos x – \(\int\) (-cos x) dx
= -x cos x + sin x + C₂
\(\int\) x² cos x dx = x² sin x – 2(sin x – x cos x) + C
= x² sin x – 2 sin x + 2x cosx + C
= (x² – 2) sin x + 2x cos x + C
In evaluating certain integrals by using the formula for integration by parts,, twice or more than twice, we come across the given integral with change of sign. This enables us to evaluate the given integral.

Question 64.
Evaluate \(\int\) e^x sin x dx on R.
Solution:
Let A = e^x sin x dx on R
A = \(\int\) e^x . sin x dx = \(\int\) e^x (-cos x)’ dx.
= e^x (-cos x) – \(\int\) (-cos x) (e^x)’ dx
= – e^x cos x + \(\int\) e^x cos x dx + C₁ ………(1)
\(\int\) e^x cos x dx = e^x. sin x – \(\int\) e^x . sin x dx
= e^x sinx – A …………….. (2)
From (1) and (2)
A = – e^x cos x + e^x sin x – A + C₁
2A = e^x (sin x – cos x) + C₁
A = \(\frac{1}{2}\) e^x (sin x – cos x) + C where
C = \(\frac{C_{1}}{2}\)
i.e., \(\int\) e^x sin x dx = \(\frac{1}{2}\) e^x (sin x – cos x) + C.

Question 65.
Find \(\int\) e^ax cos (bx + c) dx on R, where a. b, c are real numbers and b ≠ 0.
Solution:
Let A = \(\int\) e^ax cos (bx + c)dx
Then from the formula for integration by parts
A = e^ax [latex]\frac{\sin (b x+c)}{b}[/latex] – \(\int\) a e^ax [latex]\frac{\sin (b x+c)}{b}[/latex] dx
= \(\frac{1}{b}\) e^ax sin(bx + c) – \(\frac{a}{b}\) \(\int\) e^ax . sin(bx + c) dx
Inter 2nd Year Maths 2B Integration Important Questions 39
By taking c = 0, we get
\(\int\) e^ax . cos bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) [a cos bx + b sin bx] + K

Question 66.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) dx, on (-1, 1)
Solution:
Put x = cos θ, θ ∈ (0, π) dx = -sin θ . dθ
Inter 2nd Year Maths 2B Integration Important Questions 40

Question 67.
Evaluate \(\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\) on I ⊂ R \ {2nπ : n ∈ z}.
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 41
= -e^x . cot \(\frac{x}{2}\) + C

Question 68.
Evaluate \(\int \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) d x\) on I ⊂ R \ {-1, 1}
Solution:
Let x = tan θ ⇒ dx = sec² θ dθ
\(\frac{2 x}{1-x^{2}}=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\) = tan 2θ
tan^-1 \(\left(\frac{2 x}{1-x^{2}}\right)\) = tan^-1 (tan 2θ ) = 2θ + nπ
Where n = 0 if |x| < 1 = -1 if x > 1
= 1 if x < -1
We have dθ = \(\frac{1}{1+x^{2}}\) dx and
1 + x² = 1 + tan² θ = sec²θ
∴ \(\int \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) d x\)
= \(\int\left(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)\left(1+x^{2}\right) \frac{1}{1+x^{2}} d x\)
= \(\int\) (2θ + nπ) \(\int\) sec²θ dθ
= 2 \(\int\) θ sec² θ dθ + nπ ) \(\int\) sec² θ dθ + c
= 2 (θ tan θ – \(\int\) tan θ dθ) nπ tan θ + c
= 2 (θ tan θ + log |cos θ| + nπ tan θ + c
= (2θ + nπ) tan θ + 2 log cos θ + c
= (2θ + nπ) tan θ + log cos² θ + c
= (2θ + nπ) tan θ + log sec² θ + c
= x tan^-1 \(\left(\frac{2 x}{1-x^{2}}\right)\) – log (1 + x²) + c

Question 69.
Find \(\int x^{2} \cdot \frac{\exp \left(m \sin ^{-1} x\right)}{\sqrt{1-x^{2}}}\)dx on (-1, 1) where m is a real number. (Here for y ∈ R, exp. {y} stands for e^y).
Solution:
Let t = sin^-1x, then
Inter 2nd Year Maths 2B Integration Important Questions 42

Question 70.
Evaluate \(\int \frac{d x}{4 x^{2}-4 x-7}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 45

Question 71.
Find \(\int \frac{d x}{5-2 x^{2}+4 x}\)
Solution:
5 – 2x² + 4x = -2 (x² – 2x – \(\frac{5}{2}\))
= -2 ((x – 1)² – \(\frac{5}{2}\) – 1)
= -2 ((x – 1)² – \(\left(\sqrt{\frac{7}{2}}\right)^{2}\))
\(\int \frac{d x}{5-2 x^{2}+4 x}\)
= \(-\frac{1}{2} \int \frac{1}{\left((x-1)^{2}-\sqrt{\frac{7}{2}}\right)^{2}} d x+C\)
Inter 2nd Year Maths 2B Integration Important Questions 46

Question 72.
Evaluate \(\int \frac{d x}{x^{2}+x+1}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 47

Question 73.
Evaluate \(\int \frac{d x}{\sqrt{x^{2}+2 x+10}}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 48

Question 74.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^{2}}}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 49

Question 75.
Evaluate \(\int \sqrt{3+8 x-3 x^{2}} d x\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 50

Question 76.
Evaluate \(\int \frac{x+1}{x^{2}+3 x+12}\) dx.
Solution:
We write x + 1 = A(2x + 3) + B
Equating the co-efficients of x; we get 1 = 2A.
A = \(\frac{1}{2}\)
Equating the constants 3A + B = 1
B = 1 – 3A = 1 – \(\frac{3}{2}\) = –\(\frac{1}{2}\)
x + 1 = \(\frac{1}{2}\) (2x + 3) – \(\frac{1}{2}\)
Inter 2nd Year Maths 2B Integration Important Questions 53

Question 77.
Evaluate \(\int(3 x-2) \sqrt{2 x^{2}-x+1} d x\)
Solution:
Let (3x – 2) = A(4x – 1) + B
Equating the co-efficients of x, we get 3 = 4A
A = \(\frac{3}{4}\)
Equating the constants -2 = -A + B
B = -2 + A = -2 + \(\frac{3}{4}\)
= \(\frac{-5}{4}\)
Inter 2nd Year Maths 2B Integration Important Questions 54

Question 78.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\)dx. [T.S. Mar. 15]
Solution:
We write
2x + 5 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 2x + 10) + B
= A (2x – 2) + B
On comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
Thus 2x + 5 = (2x – 2) + 7
Hence \(\int \frac{2 x+5}{\sqrt{x^{2}-2 x+10}}\)dx
= \(\int \frac{2 x-2}{\sqrt{x^{2}-2 x+10}} d x+7 \int \frac{d x}{\sqrt{x^{2}-2 x+10}}+C\)
Inter 2nd Year Maths 2B Integration Important Questions 56

Question 79.
Evaluate \(\int \frac{d x}{(x+5) \sqrt{x+4}}\)
Solution:
Put t = \(\sqrt{x+4}\)
dt = \(\frac{1}{2 \sqrt{x+4}}\) dx
We have t² = x + 4
x + 5 = t² + 1
\(\int \frac{d x}{(x+5)(\sqrt{x+4})}=\int \frac{2}{t^{2}+1} d t\)
= 2 tan^-1 t + C
= 2 tan^-1 (\(\sqrt{x+4}\)) + C.

Question 80.
Evaluate \(\int \frac{d x}{5+4 \cos x}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 57

Question 81.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:
t = tan \(\frac{x}{2}\) ⇒ dx = \(\frac{2 d t}{1+t^{2}}\)
Inter 2nd Year Maths 2B Integration Important Questions 58

Question 82.
Find \(\int \frac{d x}{d+e \tan x}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 60

Question 83.
Evaluate \(\int \frac{\sin x}{d \cos x+e \sin x} d x\) and \(\int \frac{\cos x}{d \cos x+e \sin x} d x\).
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 62

Question 84.
Evaluate \(\int \frac{\cos x+3 \sin x+7}{\cos x+\sin x+1} d x\)
Solution:
Let cos x + 3 sin x + 7 = A(cos x + sin x + 1)’ + B(cos x + sin x + 1) + C
Comparing the coefficients
A + B = 1, A – B = 3, B + C = 7
A = -1, B = 2, C = 5
Inter 2nd Year Maths 2B Integration Important Questions 64

Question 85.
Find \(\int \frac{x^{3}-2 x+3}{x^{2}+x-2}\) dx.
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 66

Question 86.
Find \(\int \frac{d x}{x^{2}-81}\)
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 68

Question 87.
Find \(\int \frac{2 x^{2}-5 x+1}{x^{2}\left(x^{2}-1\right)}\) dx.
Solution:
Let \(\frac{2 x^{2}-5 x+1}{x^{2}\left(x^{2}-1\right)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{x+1}\)
2x² – 5x + 1 = Ax(x² – 1) + B(x² – 1) + Cx² (x + 1) + Dx² (x – 1)
x = 1 ⇒ 2 – 5 + 1 = C (1 + 1) ⇒ 2C = -2
⇒ C = -1
x = -1 ⇒ 2 + 5 + 1 = D (-1 -1)
⇒ 8 = -2B ⇒ D = -4
x = 0 ⇒ 1 = B(-1) ⇒ B = -1
Equating the coefficients of x³
0 = A + C + D ⇒ A = -C – D = 1 + 4 = 5
Inter 2nd Year Maths 2B Integration Important Questions 69

Question 88.
Find \(\int \frac{3 x-5}{x\left(x^{2}+2 x+4\right)}\) dx.
Solution:
\(\frac{3 x-5}{x\left(x^{2}+2 x+4\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+2 x+4}\)
3x – 5 = A(x² + 2x + 4) + (Bx + C) . x
x = 0 ⇒ -5 = 4 A ⇒ A = –\(\frac{5}{4}\)
Equating the coefficients of x²
A + B = 0 ⇒ B = -A = \(\frac{5}{4}\)
Equating the coefficient of x
3 = 2 A + C
C = 3 – 2 A = 3 + 2 . \(\frac{5}{4}\) = \(\frac{11}{2}\)
Inter 2nd Year Maths 2B Integration Important Questions 70

Question 89.
Find \(\int \frac{2 x+1}{x\left(x^{2}+4\right)^{2}} d x\)
Solution:
Let \(\frac{2 x+1}{x\left(x^{2}+4\right)^{2}}=\frac{A}{x}+\frac{B x+C}{x^{2}+4}+\frac{D x+E}{\left(x^{2}+4\right)^{2}}\)
2x + 1 = A (x² + 4)² + (Bx + C) + x (x² + 4) + (Dx + E)x
Equating the coefficients of like power of x, we obtain
A + B = 0, C = 0, 8A + 4B + D = 0,
4C + E = 2, A = \(\frac{1}{16}\)
Solving these equation, we obtain
Inter 2nd Year Maths 2B Integration Important Questions 72

Question 90.
Evaluate \(\int\) x³ . e^5x dx.
Solution:
We take a = 5, use the reduction formula
Inter 2nd Year Maths 2B Integration Important Questions 74

Question 91.
Evaluate \(\int\) sin⁴ x dx. [Mar. 14]
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 76

Question 92.
Evaluate \(\int\) tan⁶ x dx.
Solution:
Inter 2nd Year Maths 2B Integration Important Questions 78

Question 93.
\(\int\) sec⁵ x dx.
Solution:
Reduction formula is
Inter 2nd Year Maths 2B Integration Important Questions 79

Inter 2nd Year Maths 2B Hyperbola Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Hyperbola Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Hyperbola Important Questions

Question 1.
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8. [T.S. Mar. 16; May 11]
Solution:
Foci are S(±5,0) ∴ ae = 5
Length of transverse axis = 2a = 8
a = 4
e = \(\frac{5}{4}\)
b² = a²(e² – 1) = 16(\(\frac{25}{16}\) – 1) = 9
Equation of the hyperbola is \(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1
9x² – 16y² = 144.

Question 2.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate-hyperbola. [AP Mar. 16] [TS Mar. 15, 13]
Solution:
If e and e₁, are the eccentricity of a hyper bola and its conjugate hyperbola, then
\(\frac{1}{\mathrm{e}^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
Given e = \(\frac{5}{4}\) = \(\frac{16}{25}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
\(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 – \(\frac{16}{25}\) = \(\frac{9}{25}\) e_1² = \(\frac{25}{9}\)
e₁ = \(\frac{5}{3}\)

Question 3.
Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the x² – 4y² = 4 hyperbola. [A.P. Mar. 16; May 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{4}\)– \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
Centre is c (0, 0)
a²e² = a² + b² = 4 + 1 = 5
ae = \(\sqrt{5}\)
Foci are (±ae, 0) = (±\(\sqrt{5}\), 0)
Eccentricity = \(\frac{\mathrm{ae}}{\mathrm{a}}\) = \(\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ± \(\frac{\mathrm{ae}}{\mathrm{a}}\)
= ± 2 . \(\frac{2}{\sqrt{5}}\)
⇒ \(\sqrt{5}\) x = ± 4
⇒ \(\sqrt{5}\) x ± 4 = 0
Length of the latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2.1}{2}\) = 1

Question 4.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and
ii) Perpendicular to the line y = x – 7. [AP Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3.
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) The tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ± 1.

Question 5.
If 3x – 4y – k = 0 is a tangent to x² – 4y² = 5, find value of k. [T.S. Mar. 17]
Solution:
Equation of the hyperbola x² – 4y² = 5
\(\frac{x^{2}}{5}\) – \(\frac{y^{2}}{\left(\frac{5}{4}\right)}\) = 1
a² = 5, b² = \(\frac{5}{4}\)
Equation of the given line is 3x — 4y + k = 0
4y = 3x + k
y = \(\frac{3}{4}\) x + \(\frac{k}{4}\)
m = \(\frac{3}{4}\), c = \(\frac{k}{4}\),
Condition for tangency is c² = a²m² – b²
\(\frac{\mathrm{k}^{2}}{16}\) = 5 . \(\frac{9}{16}\) – \(\frac{5}{4}\)
k² = 45 – 20 = 25
k = ± 5.

Question 6.
Find the equations of the tangents to the hyperbola x² – 4y² = 4 Which are
i) Parallel
ii) Perpendicular to the line x + 2y = 0. [T.S. Mar. 15, Mar. 14, 11; May 06]
Solution:
Equation of the hyperbola is
x² – 4y² = 4
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
i) The tangent is parallel to x + 2y = 0
m = –\(\frac{1}{2}\)
c² = a²m² – b² = 4 . \(\frac{1}{4}\) = 1 – 1 = 0
c = 0
Equation of the parallel tangent is
y = mx + c
= –\(\frac{1}{2}\) x
2y = -x
x + 2y = 0.

ii) The tangent is perpendicular to x + 2y = 0
Slope of the tangent = m = \(\frac{-1}{\left(-\frac{1}{2}\right)}\) = 2
c² = a²m² – b² = 4 . 4 – 1 = 153
c = ±\(\sqrt{15}\)
Equation of the perpendicular tangent is
y = 2x ± \(\sqrt{15}\)

Question 7.
Prove that the point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b². [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m²(x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₁ which are the slopes of the tangents
passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = – 1 ⇒ y₁² + b²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 8.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 [Mar. 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
∴ b² = a²(e² – 1) ⇒ e² – 1 = \(\frac{b^{2}}{a^{2}}\)
e² = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{a^{2}}\)
∴ \(\frac{1}{\mathrm{e}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\) …………… (1)
Equation of the conjugate hyperbola is
Inter 2nd Year Maths 2B Hyperbola Important Questions 1

Question 9.
Find the centre eccentricity, foci, directrices and length of the latus rectum of the following hyperbolas.
i) 4x² – 9y² – 8x – 32 = 0
ii) 4 (y + 3)² – 9(x – 2)² = 1.
Solution:
i) 4x² – 9y² – 8x – 32 = 0
4(x² – 2x) – 9y² = 32
4(x² – 2x + 1) – 9y² = 36
\(\frac{(x-1)^{2}}{9}\) – \(\frac{(y)^{2}}{4}\) = 1
Centre of the hyperbola is C (1, 0)
a² = 9, b² = 4 ⇒ a = 3, b = 2
e = \(\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{9+4}{9}}=\frac{\sqrt{13}}{3}\)
Foci are (1±3. \(\frac{\sqrt{13}}{3}\), 0) = (1±\(\sqrt{13}\), 0)
Equations of differences are x = 1 ± \(\frac{3.3}{\sqrt{13}}\)
⇒ x = 1 ± \(\frac{9}{\sqrt{13}}\)
Length of the latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2.4}{3}\) = \(\frac{8}{3}\)

ii) The equation of the hyperbola is
4 (y + 3)² – 9 (x – 2)² = 1
\(\frac{y-(-3)^{2}}{1 / 4}\) = \(\frac{(x-2)^{2}}{1 / 9}\) = 1
Centre is C (2, -3)
Semi transverse axis = b = \(\frac{1}{2}\)
Semi conjugate axis = a = \(\frac{1}{3}\)
Inter 2nd Year Maths 2B Hyperbola Important Questions 2

Question 10.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{e_{1}^{2}}\) = 1. [Mar. 11]
Solution:
Inter 2nd Year Maths 2B Hyperbola Important Questions 3
Equation of the conjugate hyperbola is

Question 11.
i) If the line lx + my = 0 is a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that a²l² – b²m² = n².

ii) If the lx + my = 1 is a normal to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that \(\frac{a^{2}}{l^{2}}\) – \(\frac{b^{2}}{m^{2}}\) = (a² + b²)².
Solution:
i) Equation of the given tangent ¡s
lx + my + n = 0 ……………. (1)
Equation of the tangent P(θ) is
\(\frac{x}{a}\) sec θ – \(\frac{y}{b}\) tan θ – 1 = 0 …………….. (2)
Comparing (1) and (2)
\(\frac{\sec \theta}{a l}\) = \(\frac{\tan \theta}{-\mathrm{bm}}\) = \(\frac{-1}{n}\)
sec θ = –\(\frac{\mathrm{a} l}{\mathrm{n}}\), tan θ = \(\frac{\mathrm{bm}}{\mathrm{n}}\)
sec² θ – tan²θ = 1
= \(\frac{a^{2} l^{2}}{n^{2}}\) – \(\frac{b^{2} m^{2}}{n^{2}}\) = 1 ⇒ a²l² – b²m² = n².

ii) Equation of the given line is lx + my = 1 ……………. (1)
Equation of the normal at P(θ) is
\(\frac{a x}{\sec \theta}\) + \(\frac{b y}{\tan \theta}\) = a² + b² ……….. (2)
Comparing (1) and (2)
Inter 2nd Year Maths 2B Hyperbola Important Questions 5

Question 12.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and ii) Perpendicular to the line y = x – 7. [A.P. Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) Th tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ±1.

Question 13.
Prove that the point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b². [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m² (x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₂ which are the slopes of the tangents passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = -1 ⇒ y₁² + b² = -x₁² + a²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 14.
A circle cuts the rectangular hyperbola xy = 1 in the points (x_r, y_r), r = 1, 2, 3, 4. Prove that x₁x₂x₃x₄ = y₁y₂y₃y₄ = 1.
Solution:
Let the circle be x² + y² = a².
Since (t, \(\frac{1}{t}\)) (t ≠ 0) lies on xy = 1, the points of intersection of the circle and the hyperbola are given by
t² + \(\frac{1}{t^{2}}\) = a²
⇒ t⁴ – a²t² + 1 = 0
⇒ t⁴ + 0 . t³ – a²t² + 0 . t + 1 = 0.
If t₁, t₂, t₃ and t₄ are the roots of the above biquadratic, then t₁t₂t₃t₄ = 1.
If (x_r, y_r) = (t_r; \(\frac{1}{t_{r}}\)), r = 1, 2, 3, 4
then x₁x₂x₃x₄ = t₁t₂t₃t₄ = 1,
and y₁y₂y₃y₄ = \(\frac{1}{t_{1} t_{2} t_{3} t_{4}}\) = 1

Question 15.
If four points be taken on a rectangular hyperbola such that the chords joining any two points is perpendicular to the chord joining the other two, and if α, β, γ and δ be the inclinations to either asymptote of the straight lines joining these points to the centre, prove that
tan α tan β tan γ tan δ = 1.
Solution:
Let the equation of the rectangular hyperbola be x² – y² = a². By rotating the X-axis and the Y-axis about the origin through an angle \(\frac{\pi}{4}\) in the clockwise sense, the equation x² – y² = a² can be transformed to the form xy = c².
Let (ct_r, \(\frac{c}{t_{r}}\)), r = 1, 2, 3, 4 (t_r ≠ 0) be four point on the curve. Let the chord joining
A = (ct₁, \(\frac{c}{t_{1}}\)), B = (ct₂, \(\frac{c}{t_{2}}\)) be perpendicular to the chord joining C = (ct₃, \(\frac{c}{t_{3}}\)) and D = (ct₄, \(\frac{c}{t_{4}}\)).
The slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{\frac{c}{t_{1}}-\frac{c}{t_{2}}}{c t_{1}-c t_{2}}=\frac{-1}{t_{1} t_{2}}\)
[No chord of the hyperbola can be vertical]
Similarly slope of \(\stackrel{\leftrightarrow}{C D}\) is –\(\frac{1}{t_{3} t_{4}}\), Since \(\stackrel{\leftrightarrow}{A B}\) ⊥ \(\stackrel{\leftrightarrow}{C D}\).
\(\left(-\frac{1}{t_{1} t_{2}}\right)\left(-\frac{1}{t_{3} t_{4}}\right)\) = -1 ⇒ t₁t₂t₃t₄ = -1 ………………… (1)
We know the coordinate axes are the asymptotes of the curve, If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the positive direction of the X-axis, then tan α, tan β, tan γ and tan δ are their respective slopes. [O, the origin is the centre, None of \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) is vertical]
Inter 2nd Year Maths 2B Hyperbola Important Questions 6
If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the other asymptote the Y-axis then cot α, cot β, cot γ and cot δ are their respective inclinations so that
cot α cot β cot γ cot δ = tan α tan β tan γ tan δ = 1.

Inter 2nd Year Maths 2B Ellipse Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Ellipse Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Ellipse Important Questions

Question 1.
Find the value of k if 4x + y + k = 0 is a tangent to the ellipse x² + 3y² = 3. [AP Mar. 16, 15]
Solution:
The equation of the ellipse is
x² + 3y² = 3
\(\frac{x^{2}}{3}\) + \(\frac{y^{2}}{1}\) = 1
a² =, 3, b² = 1
Equation of the line is 4x + y + k = 0
y = -4x – k .
m = -4c = -k.
Condition for tangency is c² = a²m² + b²
(-k)² = 3 (4)² + 1 ,
k² = 48 + 1 = 49
k = ±7.

Question 2.
Find the equation of tangents to the ellipse 2x² + y² = 8 which are
i) Parallel to x – 2y – 4 = 0 [May. 05, Mar. 06] [T.S. Mar. 17]
Solution:
Slope will be : \(\frac{1}{2}\)
Equation of tangent y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = \(\frac{1}{2}\)x ± \(\sqrt{a^{2}\left(\frac{1}{2}\right)^{2}+b^{2}}\)
\(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{8}\) = 1
y = \(\frac{1}{2}\)x ± \(\sqrt{4 \times \frac{1}{4}+8}\)
y = \(\frac{1}{2}\)x ± 3
2y – x ± 6 = 0 required equation of tangents.
x – 2y ± 6 = 0.

Question 3.
Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is \(\frac{1}{2}\). [T.S. Mar. 15]
Solution:
Latus rectum = \(\frac{15}{2}\)
distance between foci = 2
\(\frac{2 b^{2}}{a}\) = \(\frac{15}{2}\) ; 2ae = 2
ae = 1
⇒ b² = a² – a² e²
⇒ b² = a² – 1
⇒ \(\frac{15}{4}\)a = a² – 1 .
⇒ 4a² – 15a – 4 = 0
a = 4 or a = –\(\frac{1}{4}\)
b² = a² – 1
= 16 – 1
Equation of the ellipse is \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{15}\) = 1

Question 4.
Find the equation of the ellipse in the standard form such that distance between foci is 8 and distance between directrices is 32. [Mar. 06, May. 07]
Solution:
Distance between foci = 8.
Distance between directrices = 32
2ae = 8
ae = 4
\(\frac{2 a}{e}\) = 32
\(\frac{a}{e}\) = 16
(ae) (\(\frac{a}{e}\)) = 64
a² = 64
b² = a² – a² e²
= 64 – 16 = 48
Equation of the ellipse is
∴ \(\frac{x^{2}}{64}\) + \(\frac{y^{2}}{48}\) = 1

Question 5.
Find the condition for the line x cos α + y sin α = p to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. [Mar. 14]
Solution:
Equation of the ellipse is
Inter 2nd Year Maths 2B Ellipse Important Questions 1

Question 6.
Find the equation of the ellipse with focus at (1, -1) e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0. [Mar. 05] [T.S. Mar. 19]
Solution:
P(x₁, y₁) is any point on the ellipse. Equation of the directrix is
x + y + 2 = 0
Draw PM perpendicular to ZM, Join SP
By Definition of ellipse SP = e. PM
SP² = e² . PM²
(x₁ – 1)² + (y₁ + 1)² (\(\frac{2}{3}\))²[latex]\frac{x_{1}+y_{1}+2}{\sqrt{1+1}}[/latex]²
(x₁ – 1)² + (y₁+ 1)² = \(\frac{4}{9} \frac{\left(x_{1}+y_{1}+2\right)^{2}}{2}\)
9[(x₁ – 1)² + (y₁ + 1)²] = 2 (x₁ + y₁ + 2]²
9[x₁² – 2x₁ + 1 + y₁² + 2y₁ + 1] = 2[x₁² + y₁² + 4 + 2x₁y₁ + 4x₁ + 4y₁]
9x₁² + 9y₁² – 18x₁ + 18y₁ + 18 = 2x₁² + 2y₁² + 4x₁y₁ + 8x₁ + 8y₁ + 8
7x₁² – 4x₁y₁ + 7y₁² – 26x₁ + 10y₁ + 10 = 0
focus of P (x₁, y₁) is 7x² – 4xy + 7y² – 26x + 10y + 10 = 0
This is the equation of the required Ellipse.
Inter 2nd Year Maths 2B Ellipse Important Questions 2

Question 7.
L Find tle length of major axis, minor axis, latus rectum, eccentricity, co-ordinates of centre, foci and the equations of directrices of the following ellipse. [TS Mar. 16; Mar. 14]
i) 9x² + 16y² = 144
ii) 4x² + y² – 8x + 2y + 1 = 0
iii) x² + 2y² – 4x + 12y + 14 = 0 [Mar. 11, May 07]
Solution:
Given equation is 9x² + 16y² = 144
\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1
∴ a = 4, b = 3
Length of major axis = 2a = 2 . 4 = 8
Length of minor axis = 2b = 2 . 3 = 6
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2.9}{4}\) = \(\frac{9}{2}\)
Eccentricity = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\)
Centre is C (0, 0)
Foci are (± ae, 0) = ( ± \(\sqrt{7}\), 0)
Equations of the directrices are x = ± \(\frac{a}{e}\)
x = ± 4 . \(\frac{4}{\sqrt{7}}\) = ± \(\frac{16}{\sqrt{7}}\)
\(\sqrt{7}\)x = ± 16

(ii) Given equation is 4x² + y² – 8x + 2y + 1 = 0
4(x² – 2x) + (y² + 2y) = – 1
4(x – 1)² + (y + 1)² = 4 + 1 – 1 = 4
\(\frac{(x-1)^{2}}{1}\) + \(\frac{(y+1)^{2}}{4}\) = 1
Hence a < b ⇒ y – axis is major axis
a = 1, b = 2
Length of major axis = 2b = 4
Length of minor axis = 2a = 2
Length of lattis rectum = \(\frac{2 a^{2}}{b}\) = \(\frac{2}{2}\) = 1
Eccentricity = \(\sqrt{\frac{b^{2}-a^{2}}{b^{2}}}=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}\)
Centre is C (-1, 1)
be = 2 . \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\)
Foci are (-1, 1 ± \(\sqrt{3}\))
Equations of the directrices are y + 1 = ± \(\frac{b}{e}\)
= ± \(\frac{4}{\sqrt{3}}\)
\(\sqrt{3}\) y + \(\sqrt{3}\) = ± 4
\(\sqrt{3}\) y + \(\sqrt{3}\) ± 4 = 0

iii) Given equation is x² + 2y² – 4x + 12y + 14 = 0
x² – 4x + 2 (y² + 6y) = 14
⇒ (x² – 4x + 4) + 2(y² + 6y + 9) = 4 + 18 – 14
⇒ (x – 2)² + 2(y + 3)² = 8
⇒ \(\frac{(x-2)^{2}}{8}\) + \(\frac{(y+3)^{2}}{4}\) = 1
⇒ \(\frac{(x-2)^{2}}{(2 \sqrt{2})^{2}}\) + \(\frac{(y+3)^{2}}{2^{2}}\) = 1
a = 2\(\sqrt{2}\), b = 2, h = 2, k = -3
Length of major axis = 2a = 2(2 \(\sqrt{2}\)) = 4 \(\sqrt{2}\)
Length of minor axis = 2b = 2(2) = 4
Length of latus rectum
= \(\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}\) = \(\frac{2(4)}{2 \sqrt{2}}\) = 2\(\sqrt{2}\)
Eccentricity = \(\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{4}{8}}=\frac{1}{\sqrt{2}}\)
Centre = (h, k) = (2,-3)
Foci = (h ± ae, k) = (2 ± 2, -3)
= (4, -3), (0, -3)
Equations of the directrices are x – h = ± \(\frac{a}{e}\)
x – 2 = \(\frac{2 \sqrt{2}}{\left(\frac{1}{\sqrt{2}}\right)}\)
x – 2 = ± 4
i.e., x = 6, x = -2.

Question 8.
Find the equations of tangent and normal to the ellipse 2x² + 3y² = 11 at the point whose ordinate is 1. [T.S. Mar. 16]
Solution:
Equation of the ellipse is 2x² + 3y² = 11
Given y = 1
2x² + 3 = 11
⇒ 2x² = 8
x² = 4
x = ±2
Points on the ellipse are P (2, 1) and Q(-2, 1)
Case i) P (2, 1)
Equation of the tangent is 2x . 2 + 3y . 1 = 11
4x + 3y = 11
The normal is perpendicular to the tangent Equation of the normal at P can be taken as
3x – 4y = k
The normal passes through p (2, 1)
6 – 4 = k ⇒ k = 2
Equation of the normal at P is 3x – 4y = 2.
Case ii) Q (-2, 1)
Equation of the tangent at Q is
2x(-2) + 3y . 1 = 11
-4x + 3y =1 1
4x – 3y + 11 = 0
Equation of the normal can be taken as
3x + 4y = k
The normal passes through Q (-2, 1)
-6 + 4 = k ⇒ k = -2
Equation of the normal at Q is 3x + 4y = -2
or 3x + 4y + 2 = 0.

Question 9.
Find the eccentricity, co-ordinates of foci. Length of latus rectum and equations of directrices of the following ellipses.
i) 9x² + 16y² – 36x + 32y – 92 = 0,
ii) 3x² + y² – 6x – 2y – 5 = 0 [T.S. Mar. 15]
Solution:
i) Given ellipse is
9x² + 16y² – 36x + 32y – 92 = 0
9(x² – 4x + 4) + 16 (y² + 2y + 1)
= 92 + 36 + 16
9(x – 2)² + 16(y + 1)² = 144
comparing with \(\frac{(x-2)^{2}}{16}\) + \(\frac{(y+1)^{2}}{9}\) = 1,
we get
a² = 16, b² = 9 ⇒ a = 4, b = 3.
Inter 2nd Year Maths 2B Ellipse Important Questions 3
Equations of the directrices are x = h ± \(\frac{\mathrm{a}}{\mathrm{e}}\)
x = 2 ± \(\frac{4 \times 4}{\sqrt{7}}\)
\(\sqrt{7x}\) = 2\(\sqrt{7}\) ± 16

ii) 3x² + y² – 6x – 2y – 5 = 0
Solution:
3(x² – 2x) + (y² – 2y) = 5
⇒ 3(x² – 2x + 1) + (y² – 2y + 1) = 9
⇒ 3(x – 1)² + (y – 1)² = 9
comparing with ⇒ \(\frac{(x-1)^{2}}{3}\) + \(\frac{(y-1)^{2}}{9}\) = 1,
we get
a < b ⇒ Y – axis is the major axis
a² = 3, b² = 9
a = \(\sqrt{3}\), b = 3, h = 1, k = 1
Length of major axis = 2b = 2(3) = 6
Length of minor axis = 2a = 2\(\sqrt{3}\)
Inter 2nd Year Maths 2B Ellipse Important Questions 4

Question 10.
Find the equation of the tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant. [A.P. Mar. 15, Mar. 07]
Solution:
Given ellipse is 9x² + 16y² = 144
Inter 2nd Year Maths 2B Ellipse Important Questions 5
Equation of the normal at P is

Question 11.
Show that the points of intersection of the perpendicular tangents to an ellipse lie on a circle. [A.P. Mar. 16]
Solution:
Let the equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
(a > b). Any tangent to it in the slope-intercept form is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) …………………. (1)
Let the perpendicular tangents intersect at P(x₁, y₁).
∴ P lies on (1) for some real m, i.e.,
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
∴ (y₁ – mx₁)² = a²m² + b².
or .
(x₁² – a²) m² – 2x₁y₁m + (y₁² – b²) = 0 being a
quadratic equation in ‘m’ has two roots say m₁ and m₂ then m₁, m₂ are the slopes of tangents from P to the ellipse
∴ m₁m₂ = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
∴ -1 = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
[∵ The tangents are perpendicular to each
other so that m₁m₂ = -1]
i.e., x₁² + y₁² = a² + b²,
If, however, one of the perpendicular tangents is vertical, then such pair of perpendicular tangents intersect at one of the points (± a, ± b) and any of these points satisfies x² + y² = a² + b².
∴ The point of intersection of perpendicular tangents to the ellipse S = O lies on the circle x² + y² = a² + b².

Question 12.
Find the eccentricity, co-ordinates of foci, Length of latus reEtum and equations of directrices of the following ellipses.
i) 9x² + 16y² – 36x + 32y – 92 = 0
ii) 3x² + y² – 6x – 2y – 5 = 0 [T.S. Mar. 15]
Solution:
Given ellipse is :
9x² + 16y² – 36x + 32y – 92 = 0
9(x² -4x + 4) + 16 (y² + 2y + 1)
= 92 + 36 + 16
9 (x – 2)² + 16 (y + 1)2² = 144
comparing with \(\frac{(x-2)^{2}}{16}\) + \(\frac{(y+1)^{2}}{9}\) = 1,
we get
a² = 16, b² = 9 ⇒ a = 4, b = 3.
e = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\)
Foci (h ± ae, k) = (2 ± 4 . \(\frac{\sqrt{7}}{4}\), -1)
= (2 ± \(\sqrt{7}\), -1)
Length of the latus rectum
= \(\frac{2 \cdot b^{2}}{a}=\frac{2.9}{4}=\frac{9}{2}\)
Equations of the directrices are x = h ± \(\frac{a}{e}\)
x = 2 ± \(\frac{4 \times 4}{\sqrt{7}}\)
\(\sqrt{7x}\) = 2\(\sqrt{7}\) ± 16

ii) 3x² + y² – 6x – 2y – 5 = 0
Solution:
3(x² – 2x) + (y² – 2y) = 5
⇒ 3(x² – 2x + 1) + (y² – 2y + 1) = 9
⇒ 3(x – 1)² + (y – 1)² = 9 .
comparing with ⇒ \(\frac{(x-1)^{2}}{3}\) + \(\frac{(y-1)^{2}}{9}\) = 1
we get
a < b ⇒ Y – axis is the major axis
a² = 3, b² = 9 .
a = \(\sqrt{3}\), b = 3, h = 1, k = 1
Length of major axis = 2b = 2(3) = 6
Length of minor axis = 2a = 2\(\sqrt{3}\)
Length of latus rectum = \(\frac{2 a^{2}}{b}=\frac{2.3}{3}\) = 2
Eccentricity = \(\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{3}{9}}=\sqrt{\frac{2}{3}}\)
Centre = (h, k) = (1, 1)
Focus = (h, k ± be) = (1, 1 ± 3 \(\left.\sqrt{\frac{2}{3}}\right)\))
= (1, 1 ± \(\sqrt{6}\))
Equation of directrices are y – k = ± \(\frac{b}{e}\)
y – 1 = ± \(\frac{3 \sqrt{3}}{\sqrt{2}}\)
y = 1 ± \(\frac{3 \sqrt{3}}{\sqrt{2}}\)

Question 13.
Find the equation of the ellipse referred to its major and minor axes as the co-ordinate axes x, y respectively with latus rectum of length 4 and the distance between foci 4\(\sqrt{2}\).
Solution:
Let the equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
(a > b)
Length of the latus rectum = \(\frac{2 b^{2}}{a}\) = 4
⇒ b² = 2a.
Foci are S (ae, 0), S’ (-ae, 0)
Distance between the foci = 2ae = 4\(\sqrt{2}\)
ae = 2\(\sqrt{2}\)
b² = a² (1 – e²) = a² – (ae)²
2a = a² – 8 ⇒ a² – 2a – 8 = 0
(a – 4) (a + 2) = 0
a = 4 or – 2
a > 0 ⇒ a = 4
b² = 2a = 2 . 4 = 8
Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{8}\) = 1
x² + 2y² = 16.

Question 14.
If the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse.
Solution:
Let \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a> b) be the ellipse in its standard form.
Length of latus rectum = \(\frac{1}{2}\) (minor axis)
2 \(\frac{b^{2}}{a}\) = \(\frac{1}{2}\) (2b)
2 \(\frac{b^{2}}{a}\) = b
a = 2b
a² =4 b², ⇒ a² = 4a² (1 – e²)
∴ 1 – e² = \(\frac{1}{4}\) ⇒ e² = \(\frac{3}{4}\)
⇒ e = \(\frac{\sqrt{3}}{2}\).

Question 15.
If θ₁, θ₂ are the eccentric angles of the extremeties of a focal chord (other that the vertices) of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) and e Its eccentricity. Then show that
i) e cos \(\frac{\left(\theta_{1}+\theta_{2}\right)}{2}\) = cos \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\)
ii) \(\frac{e+1}{e-1}\) = cot \(\left(\frac{\theta_{1}}{2}\right)\) . cot \(\left(\frac{\theta_{2}}{2}\right)\)
Solution:
Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1,
(a > b)
Inter 2nd Year Maths 2B Ellipse Important Questions 7
sin θ₁ . cos θ₂ – e sin θ₁ = cos θ₁ sin θ₂ – e . sin θ₂
sin θ₁ . cos θ₂ – cos θ₁ sin θ₂ = e sin θ₁ – e sin θ₂
sin (θ₁ – θ₂) = e (sin θ₁ – sin θ₂)
2 sin \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\) . cos \(\frac{\left(\theta_{1}-\theta_{2}\right)}{2}\)
= e [2 cos \(\frac{\theta_{1}+\theta_{2}}{2}\) . sin \(\frac{\theta_{1}-\theta_{2}}{2}\)]
Inter 2nd Year Maths 2B Ellipse Important Questions 8

Question 16.
C is the centre1 AA’ and BB’ are major and minor axis of the ellipse.
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. If PN is the ordinate of a point P on the ellipse then show that
\(\frac{(\mathrm{PN})^{2}}{\left(\mathrm{~A}^{\prime} \mathrm{N}\right)(\mathrm{AN})}\) = \(\frac{(\mathrm{BC})^{2}}{(\mathrm{CA})^{2}}\)
Solution:
Inter 2nd Year Maths 2B Ellipse Important Questions 9
Equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
P(a cos θ, b sin θ) any point on the ellipse.
PN = b sin θ; AN = a – a cos θ,
AN = a + a cos θ; BC = b, CA = a
(A’N). (AN) = (a + a cos θ) (a – a cos θ)
= a² – a²cos²θ
= a² (1 – cos θ)
= a² sin²θ
\(\frac{(\mathrm{PN})^{2}}{\left(\mathrm{~A}^{\prime} N\right)(\mathrm{AN})}=\frac{\mathrm{b}^{2} \sin ^{2} \theta}{\mathrm{a}^{2} \sin ^{2} \theta}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\)
\(\frac{B C^{2}}{(C A)^{2}}=\frac{b^{2}}{a^{2}} \Rightarrow \frac{P^{2}}{\left(A^{1} N\right)(A N)}=\frac{(B C)^{2}}{(C A)^{2}}\)

Question 17.
S and Tare the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the eccentricity of the ellipse.
Solution:
Inter 2nd Year Maths 2B Ellipse Important Questions 10
Let \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) be an ellipse whose foci are S and T, B is an end of the minor axis such that STB is equilateral, then SB = ST = TB. We have S(ae, 0).
T = (-ae, 0) and B(0, b)
Consider SB = ST ⇒ (SB)² = (ST)²
⇒ (ae)² + b² = 4a²e²
∴ a²e² + a² (1 – e)² = 4a²e²
[∵ b² = a² (1 – e²)]
e² = \(\frac{1}{4}\)
∴ Eccentricity of the ellipse is \(\frac{1}{2}\).

Question 18.
Show that among the points on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b), (-a, 0) is the farthest point and (a, 0) is the nearest point from the focus (ae, 0).
Solution:
Let P = (x, y) be any point on the ellipse so that – a ≤ x ≤ a and S = (ae, 0) be the focus.
Since (x, y) is on the ellipse,
y² = \(\frac{b^{2}}{a^{2}}\) (a² – x²)
= (1 – e²)(a² – x²) ………….. (1) [∵ b² = a²(1 – e²)]
Then we know that
sp² = (x – ae)² + y²
= (x – ae)² + (1 – e²)(a² – x²)
= -2xae + a² + e²x²
= [a – ex]²
∴ SP = [a – ex]
we have – a ≤ x ≤ a
⇒ -ae ≤ xe ≤ ae
⇒ -ae – a ≤ xe – a ≤ ae – a …………………. (2)
∴ ex – a < 0
∴ SP = a – ex …………………… (3)
From (2) and (3)
ae + a ≥ SP ≥ a – ae
⇒ a – ae ≤ SP ≤ ae + a
∴ Max SP = ae + a when P = (-a, 0)
and Min SP = a – ae when P = (a, 0)
Hence the nearest point is (a, 0) and the farthest one is (-a, 0).

Question 19.
The orbit of the Earth is an ellipse with eccentricity \(\frac{1}{60}\) with the Sun at one of its foci, the major axis being approximately 186 × 10⁶ miles in length. Find the shortest and longest distance of the Earth from the Sun.
Solution:
We take the orbit of the Earth to be
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b).
Since the major axis is 186 × 10⁶ miles,
2a = 186 × 10⁶ miles
∴ a = 93 × 10⁶ miles
If e be the eccentricity of the orbit, e = \(\frac{1}{60}\)
We know, the longest and shortest distances of the Earth from the Sun are respectively
a + ae and a – ae (problem 7)
Here, the longest distance
= 93 × 10⁶ × (1 + \(\frac{1}{60}\))
= 9455 × 10⁴ miles.
and the shortest distance
= 93 × 10⁶ × (1 – \(\frac{1}{60}\)) miles
= 9145 × 10⁴ miles.

Question 20.
Find the equation of the tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant. A.P. [Mar. 15, Mar. 07]
Solution:
Given ellipse is 9x² + 16y² = 144
Inter 2nd Year Maths 2B Ellipse Important Questions 11
Equation of the normal at P is

Question 21.
If a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b) meets its major.axis and minor axis atM and N respectively, then prove that \(\frac{a^{2}}{(C M)^{2}}\) + \(\frac{b^{2}}{(C N)^{2}}\) = 1. Where C is the centre of the ellipse.
Solution:
Let P(θ) (a cos θ, b sin θ) is any point on the ellipse then Equation of the tangent at P (θ) is
Inter 2nd Year Maths 2B Ellipse Important Questions 13
a² . \(\frac{\cos ^{2} \theta}{a^{2}}\) + b² . \(\frac{\sin ^{2} \theta}{b^{2}}\)
= cos² θ + sin² θ = 1.

Question 22.
Find the condition for the line
i) lx + my + n = 0 to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
ii) lx+ my n = 0 to be a normal to the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1.
Solution:
i) Equation of the ellipse is \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
Equation .of the tangent at P(θ) is
\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 ……………………… (1)
Equation of the given line is
lx + my = -n …………………. (2)
(1), (2) represent the same line. Comparing the co-efficients
Inter 2nd Year Maths 2B Ellipse Important Questions 14
Comparing (1) and (2)

Question 23.
If the normal at one end of a latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 passes through one end of the minor axis, then show that e⁴ + e² = 1 [e is the eccentricity of the ellipse]
Solution:
Let L be the one end of the latus rectum of
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. Then the coordinates of
L = (ae, \(\frac{b^{2}}{a}\))
Hence equation of the normal at L is
Inter 2nd Year Maths 2B Ellipse Important Questions 17
is a line passes through the one end
B’ = (0, -b)
or minor axis of \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 as shown in figure.
\(\frac{\mathrm{a}(0)}{\mathrm{e}}\) – a(-b) = a² – b²
ab = a2² – a² (1 – e²)
ab = a²e² ⇒ e² = \(\frac{b}{a}\) ⇒ e⁴ = \(\frac{b^{2}}{a^{2}}\)
= \(\frac{a^{2}\left(1-e^{2}\right)}{a^{2}}\) = 1 – e² ⇒ e⁴ + e² = 1.

Question 24.
If PN is the ordinate of a point P on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 and the tangent at P meets the X – axis at T then show that (CN) (CT) = a² where C is the centre of the ellipse.
Solution:
Let P(θ) = (acosθ, bsinθ) be a point on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1. Then the equation of the tangent at P(θ) is \(\frac{x \cos \theta}{a}\) + \(\frac{y \sin \theta}{b}\) = 1 or \(\frac{x}{\frac{a}{\cos \theta}}+\frac{y}{\frac{b}{\sin \theta}}=1\) meets the X – axis at T
Inter 2nd Year Maths 2B Ellipse Important Questions 18
x – intercept (CT) = \(\frac{a}{\cos \theta}\) and the ordinate of P is PN = bsinθ
then its absicca CN = a cos θ. (see Fig.)
∴ (CN) . (CT) =, (a cos θ) (\(\frac{a}{\cos \theta}\)) = a².

Question 25.
Show that the points of intersection of the perpendicular tangents to an ellipse lie on a circle. [A.P. Mar. 16]
Solution:
Let the equation of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a > b). Any tangent to it in the slope-intercept
form is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) …………………. (1)
Let the perpendicular tangents interested at
P(x₁, y₁).
∴ p lies on (1) for some real m, i.e.,
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
∴ (y₁ – mx₁)² = a²m² + b².
or
(x₁² – a²) m² – 2x₁y₁m + (y₁² – b²) = 0 is a quadratic equation in ‘m’, has two roots say m₁ and m₂ then m₁, m₂ are the slopes of tangents from P to the ellipse
∴ m₁m₂ = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
∴ – 1 = \(\left(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\right)\)
[∵ The tangents are perpendicular to each other so that m₁ m₂ = -1]
i.e., \(x_{1}^{2}\) + \(y_{1}^{2}\) = a² + b².
If, however, one of the perpendicular tangents is vertical, then such a pair of perpendicular tangents intersect at one of the points (a, ± b) and any of these points satisfy x² + y² = a² + b².
∴ The point of intersection of perpendicular tangents to the ellipse S = 0 lies on the circle x² + y² = a²+ b².

Inter 2nd Year Maths 2B Parabola Important Questions

August 25, 2022

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Parabola Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Parabola Important Questions

Question 1.
Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2). [T.S. Mar. 15]
Solution:
Let S = (1, -7), A(1, -2)
h = 1, k = -2, a = -2 + 7 = 5
The Axis of the parabola is parallel to the y-axis
The equation of the parabola is
(x – h)² = – 4a (y – k)
(x – 1)² = – 20(y + 2)
x² – 2x + 1, = – 20y – 40
⇒ x² – 2x + 20y + 41 – 0.

Question 2.
If the normal at the point t₁ on the parabola y² =, 4ax meets it again at point t² then prove that t₁t₂ + t₁² + 2 = 0. [May 07]
Solution:
Equation of normal is
y – y₁ = \(\frac{-y_{1}}{2 a}\) (x – x₁)
y – 2at₁ = \(\frac{-2 a t_{1}}{2 a}\) (x – at₁²) ……………… (i)
Equation of the line (i) again meets parabola at (at₂², 2at₁)
∴ 2at₂ – 2at₁ = t₁ (at₂² – at₁²)
\(-\frac{2}{t_{1}}\) = t₁ + t₂ ⇒ -2 = t₁² + t₁t₂
⇒ t₁² + t₁t₂ + 2 = 0

Question 3.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the y – x + 4y + 5 = 0. [Mar 05]
Solution:
y² – x + 4y + 5 = 0 ⇒ (y – (-2))² = (x – 1), comparing with (y – k)² = 4a(x – h),we get (h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus (h + a, k) = (\(\frac{5}{4}\), -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0

Question 4.
Find the coordinates of the points on the parabola y² = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x₁, y₁) be a point on the parabola
y² = 2x whose focal distance is \(\frac{5}{2}\) then
y₁² = 2x₁ and x₁ + a = \(\frac{5}{2}\)
⇒ x₁ + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x₁ = 2
∴ y₁² = 2(2) = 4 ⇒ y₁ = ±2
∴ The required points are (2, 2) and (2, -2)

Question 5.
Find the value of k if the line 2y = 5x + k is a tangent to the parabola y² = 6x. [T.S. Mar. 16]
Solution:
Given line is 2y = 5x + k
⇒ y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\))
Comparing y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\)) with y = mx + c
We get m = \(\frac{5}{2}\), c = \(\frac{k}{2}\)
y = (\(\frac{5}{2}\))x + \(\frac{k}{2}\) is a tangent to y² = 6x
c = \(\frac{a}{m}\)
⇒ \(\frac{k}{2}=\frac{\left(\frac{3}{2}\right)}{\left(\frac{5}{2}\right)}\) ⇒ k = \(\frac{6}{5}\)

Question 6.
Show that the equations of common tangents to the circle x² + y² = 2a² and the parabola y² = 8ax are y = ± (x + 2a). [Mar. 06]
Solution:
The equation of tangent to parabola
y² = 8ax is y = mx + \(\frac{2a}{m}\)
m²x – my + 2a = 0 ……………….. (1)
If (i) touches circle x² + y² = 2a², then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a\(\sqrt{2}\) of the circle.
\(\left|\frac{2 a}{\sqrt{m^{2}+m^{4}}}\right|=a \sqrt{2}\)
or 4 = 2 (m⁴ + m²)
m⁴ + m² – 2 = 0
(m² + 2) (m² – 1) = 0 or m = ±1
Required tangents are
y = (1) x + \(\frac{2 a}{(1)}\), y = (-1) x + \(\frac{2 a}{(-1)}\)
y = ± (x + 2a)

Question 7.
Find the co-ordinates of the point on the parabola y² = 8x whose focal distance is 10. [T.S. Mar. 17] [A.P. Mar. 16; Mar. 14, 11]
Solution:
Equation of the parabola is
y² = 8x
4a = 8 ⇒ a – 2
Inter 2nd Year Maths 2B Parabola Important Questions 1
Co-ordinates of the focus S are (2, 0) Suppose P(x, y) is the point on the parabola.
Given SP = 10 ⇒ SP² = 100
(x – 2)² + y² = 100
But y² = 8x
⇒ (x – 2)² + 8x = 100
⇒ x² – 4x + 4 + 8x – 100 = 0
⇒ x² + 4x – 96 = 0 ⇒ (x + 12) (x – 8) = 0
x + 12 = 0 or x – 8 = 0
x = -12, or 8
Case (i) x = 8
y² = 8.x = 8.8 = 64
y = ±8
Co-ordinates of the required points are (8, 8) and (8, -8)
Case (ii) x = -12
y² = 8(-12) = -96 < 0
y is not real.

Question 8.
If (\(\frac{1}{2}\), 2) is one extermity of a focal chord of the parabola y² = 8x. Find the co-ordinates of the other extremity. [May 06]
Solution:
A = (\(\frac{1}{2}\), 2); S = (2, 0)
B = (x₁, y₁) ⇒ (\(\frac{y_{1}^{2}}{8}\), y₁)
ASB is a focal chord.
∴ Slopes of SA and BS are same.
Inter 2nd Year Maths 2B Parabola Important Questions 2
or 4y₁² + 24y₁ – 64 = 0
⇒ y₁² + 6y₁ – 16 = 0
⇒ (y₁ + 8) (y, – 2) = 0
y₁ = 2, 8
x₁ = \(\frac{1}{2}\), 8; So (8, -8) other extremity.

Question 9.
Show that the common tangent to the parabola y² = 4ax and x² = 4by is xa^1/3 + yb^1/3 + a^2/3b^2/3 = 0 [T.S. Mar. 17] [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y² = 4ax ………………. (1)
and x² = 4by ……………… (2)
Equation of any tangent to (1) is of the form
y = mx + \(\frac{a}{m}\) …………….. (3)
If the line (3) is a tangent to (2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2), we get x² = 4b (mx + \(\frac{a}{m}\)) is 3x² – 4bm²x – 4ab = 0 should have equal roots Therefore its discriminent must be zero. Hence
16b²m⁴ – 4m (-4ab) = 0
16b(bm⁴ + am) = 0
m(bm³ + a) = 0 But m ≠ 0
∴ m = – a^1/3/b^1/3 substituting in (3) the equation of the common tangent becomes
y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) or
a^1/3x + b^1/3y + a^2/3b^2/3 = 0 .

Question 10.
Prove that the area of the triangle formed by the .tangents at (x₁, y₁), (x₂, y₂) and (x₃, y₃) to the parabola y² = 4ax (a > 0) is \(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x₁, y₁) = (at₁², 2at₁),
E(x₂, y₂) = (at₂², 2at₂),
and F(x₃, y₃) = (at₃², 2at₃), be three points on the parabola
y² = 4ax (a > 0).
The equation of the tangents at D, E and F are
t₁y = x + at₁² ……………… (1)
t₂y = x + at₂² ………………. (2)
t₃y = x.+ at₃² ………………. (3)
(1) – (2) ⇒ (t₁ – t₂) y = a(t₁ – t₂) (t₁ + t₂)
⇒ y = a(t₁ + t₂) substituting in (1)
we get x = at₁t₂.
∴ The point of intersection of the tangents at D and E is say P(at₁t₂, a(t₁ + t₂))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at₂t₃, a(t₂ + t₃) and R(at₃t₁ a(t₃ + t₁)) respectively
Area of ∆PQR
Inter 2nd Year Maths 2B Parabola Important Questions 3
\(\frac{1}{16a}\) |2a(t₁ – t₂) 2a(t₂ – t₃) 2a(t₃ – t₁)|
\(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units.

Question 11.
Prove that the two parabolas y² = 4ax and x² = 4by intersect (other than the origin) at an angle of Tan^-1 \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y⁴ = 16a²x²
= 16a²(4by)
= 64a²by
∴ y[y³ – 64a²b] = 0
=> y³ – 64a²b = 0
=> y = (64a²b)^1/3 [∵ y > 0]
= 4a^2/3b^1/3
Inter 2nd Year Maths 2B Parabola Important Questions 4
Also from y² = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)
= 4a^1/3b^2/3
∴ P = (4a^1/3b^2/3, 4a^2/3b^1/3)
Differentiating both sides of y² 4ax w.r.t ‘x’, we get
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)
If m₁ be the slope of the tangent at P to y² = 4ax, then
m₁ = \(\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}\)
Similarly, we get m₂ = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m₂ is the slope of the tangent at P to x² = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)
so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 12.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the following parabolas.
(i) y² = 16x
(ii) x² = -4y
(iii) 3x² – 9x + 5y – 2 = 0
(iv) y² – x + 4y + 5 = 0 [Mar. 05]
Solution:
i) y² = 16x, comparing with y² = 4ax,
we get 4a = 16 ⇒ a = 4
The coordinates of the vertex = (0, 0)
The coordinates of the focus = (a, 0) = (4, 0)
Equation of the directrix: x + a = i.e., x + 4 = 0
Axis of the parabola y = 0

ii) x² = -4y, comparing with x² = -4ay,
we get 4a = 4 ⇒ a = 1
The coordinates of the vertex = (0, 0
The coordinates of the focus = (0, -a) = (0, 1)
The equation of the directrix y – a = 0
i.e., y – 1 = 0 .
Equation of the axis x = 0

iii) 3x² – 9x + 5y – 2 = 0
3(x² – 3x) = 2 – 5y
⇒ 3(x² – 2x(\(\frac{3}{2}\)) + \(\frac{9}{4}\)) = 2 – 5y + \(\frac{27}{4}\)
(x – \(\frac{3}{2}\))² = –\(\frac{5}{3}\) (y – \(\frac{7}{4}\)),
Comparing with (x – h)² = -4a (y – k) we get
a = \(\frac{5}{12}\), h = \(\frac{3}{2}\), k = \(\frac{7}{4}\)
∴ Coordinates of the vertex = (h, k)
= (\(\frac{3}{2}\), \(\frac{7}{4 }\))
Coordinates of the focus = (h, k – a)
= (\(\frac{3}{2}\), \(\frac{7}{4}\) – \(\frac{5}{12}\)) = (\(\frac{3}{2}\), \(\frac{4}{3}\))
Equation of the directrix is y – k – a = 0
i.e., 6y – 13 = 0
Equation of the axis is x – h = 0
i.e., 2x – 3 = 0

iv) y² – x + 4y + 5 = 0 ⇒ (y- (-2))² = (x – 1),
comparing with (y – k)² = 4a(x – h),we get
(h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus
(h + a, k) = (\(\frac{5}{4}\), -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0

Question 13.
Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Solution:
The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis, focus is above the vertex.
a = distance between focus and vertex = 3.
∴ Equation of the parabola
(x – 3)² = 4(3) (y + 2)
i.e., (x – 3)² = 12(y + 2).

Question 14.
Find the coordinates of the points on the parabola y² = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x₁, y₁) be a point on the parabola
y² = 2x whose focal distance is \(\frac{5}{2}\) then
y₁² = 2x₁ and x₁ + a = \(\frac{5}{2}\)
⇒ x₁ + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x₁ = 2
∴ y₁² = 2(2) = 4 ⇒ y₁ = ±2
∴ The required points are (2, 2) and (2, -2)

Question 15.
Find the equation of the parabola passing through the points (-1, 2), (1, -1) and (2, 1) and having its axis parallel to the X-axis.
Solution:
Since the axis is parallel to x-axis the equation of the parabola is in the form of
x = -ly² + my + n. .
Since the parabola passes through (-1, 2), we have
-1 = l( 2)² + m(2) + n
⇒ 4l + 2m + n = – 1 ……………………. (1)
Similarly, since the parabola passes through (1,-1) and (2, 1) we have
l – m + n = 1 ……………… (2)
l + m + n = 2 ……………… (3)
Solving (1), (2) and (3)
we get l = –\(\frac{7}{6}\), m = \(\frac{1}{2}\) and n = \(\frac{8}{3}\).
Hence the equation of the parabola is
x = –\(\frac{7}{6}\) y² + \(\frac{1}{2}\) y + \(\frac{8}{3}\) (or)
7y² – 3y + 6x- 16 = 0.

Question 16.
A double ordinate of the curve y² = 4ax is of length 8a. Prove that the line from the vertex to its ends are at right angles.
Solution:
Let P = (at², 2at) and P’ = (at², -2at) be the ends of double ordinate PP’. Then
8a = PP’ = \(\sqrt{0+(4 a t)^{2}}\) = 4at ⇒ t = 2.
∴ P = (4a, 4a), P’= (4a, -4a) .
Slope of \(\overline{\mathrm{AP}}\) × slope of \(\overline{\mathrm{AP}^{\prime}}\)
= (\(\frac{4a}{4a}\))(-\(\frac{4a}{4a}\)) = -1
∴ ∠PAP’ = \(\frac{\pi}{2}\)

Question 17.
i) If the coordinates of the ends of a focal chord of the parabola y² = 4ax are (x₁, y₁) and (x₂, y₂), then prove that x₁x₂ = a², y₁y₂ = -4a².
ii) For a focal chord PQ of the parabola y² = 4ax, if SP = l and SQ = l’ then prove that \(\frac{1}{l}\) + \(\frac{1}{l}\) = \(\frac{1}{a}\).
Solution:
i) Let P(x₁, y₁) = (at₁², 2at₁) and Q(x₂, y₂) = (at₂², 2at₁) be two end points of a focal chord.
P, S, Q are collinear
Slope of \(\overline{\mathrm{PS}}\) = Slope of \(\overline{\mathrm{QS}}\)
\(\frac{2 a t_{1}}{a t_{1}^{2}-a}=\frac{2 a t_{2}}{a t_{2}^{2}-a}\)
t₁t₂² – t₁ = t₂t₁² – t₂
t₁t₂ (t₂ – t₁) + (t₂ – t₁) = 0
1 + t₁t₂ = 0 ⇒ t₁t₂ = -1 ………………… (1)
From (1) x₁x₂ = at₁² at₂² = a²(t₂t₁)² = a²
y₁y₂ = 2at₁2at₂ = 4a²(t₂ t₁) = -4a²

ii) Let P(at₁², 2at₁) and Q(at₂², 2at₁) be the extremities of a focal chord of the parabola, then t₁t₂ = -1 (from (1))
Inter 2nd Year Maths 2B Parabola Important Questions 5

Question 18.
If Q is the foot of the perpendicular from a point P on the parabola y² = 8(x – 3) to its directrix. S is the focus of the parabola and if SPQ is an equilateral triangle then find the length of side of the triangle.
Solution:
Given parabola y² = 8(x – 3) then
its vertex A = (3, 0) and focus = (5, 0)
[4a = 8 ⇒ a = 2] since PQS is an equilateral triangle
Inter 2nd Year Maths 2B Parabola Important Questions 6
Hence length of each side of triangle is ‘8’.

Question 19.
The cable of a uniformely loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 72 mt. long is supported by vertical wires attached to the cable, the longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the road-way 18 mts. from the middle.
Inter 2nd Year Maths 2B Parabola Important Questions 7
Solution:
Let AOB be the cable [0 is its lowest point and A, B are the highest points]. Let PRQ be the bridge suspended with PR = RQ = 36 mts. (see above Fig.).

PA = QB = 30 mts (longest vertical sup-porting wires)

OR = 6 mts (shortest vertical supporting wire) [the lowest point of the cable is up¬right the mid-point R of the bridge]

Therefore, PR = RQ = 36 mts. We take the origin of coordinates at 0, X-axis along the tangent at O to the cable and the Y-axis along \(\overleftrightarrow{\mathrm{RO}}\). The equation of the cable would, therefore, be x² = 4ay for some a > 0. We get B = (36, 24) and 36² = 4a × 24.
Therefore, 4a = \(\frac{36 \times 36}{24}\) = 54 mts.
If RS = 18 mts. and SC is the vertical through S meeting the cable at C and the X-axis at D, then SC is the length of the supporting wire required. If SC = l mts, then
DC = (l – 6) mts. As such C = (18, l – 6).
Since C is on the cable, 18² = 4a (l – 6)
⇒ l – 6 = \(\frac{18^{2}}{4 a}\) = \(\frac{18 \times 18}{54}\) = 6
⇒ l = 12

Question 20.
Find the condition for the straight line lx + my + n = 0 to be a tangent to the parabola y² = 4ax and find the co-ordinates of the point of contact.
Solution:
Let the line lx + my + n = 0 be a tangent to the parabola y² = 4ax at (at², 2at). Then the equation of the tangent at P(t) is x – yt + at² = 0 then it represents the given line
lx + my + n = 0, then
Inter 2nd Year Maths 2B Parabola Important Questions 8

Question 21.
Show that the straight line 7x + 6y = 13 is a tangent to the parabola y² – 7x – 8y + 14 = 0 and find the point of contact.
Solution:
Equation of the given line is 7x + 6y = 13, equation of the given parabola is y² – 7x – 8y + 14 = 0.
By eliminating x, we get the ordinates of the points of intersection of line and parabola adding the equations y² – 2y + 1 = 0.
i.e., (y – 1 )² = 0 ⇒ y = 1, 1.
∴ The given line is tangent to the given parabola.
If y = 1 then x = 1 hence the point of contact is (1, 1).

Question 22.
Prove that the normal chord at the point other than origin whose ordinate is equal to its abscissa subtends a right angle at the focus.
Solution:
Let the equation of the parabola be
y² = 4ax and P(at², 2at) be any point …………………. (1)
On the parabola for which the abscissa is equal to the ordinate.
i.e., at² = 2at ⇒ t = 0
or t = 2. But t ≠ 0. Hence the point (4a, 4a) at which the normal is
y + 2x = 2a(2) + a(2)³ (or)
y = (12a – 2x) ………………….. (2)
Substituting the value of
y = 12a – 2x in (1) we get
(12a – 2x)² = 4ax (or)
x² – 13ax + 36a² = (x – 4a) (x – 9a) = 0
⇒ x = 4a, 9a
corresponding values of y are 4a and -6a. Hence the other points of intersection of that normal at P(4a, 4a) to the given parabola is Q(9a, -6a), we have S(a, 0).
Slope of the \(\overline{\mathrm{SP}}\) = m₁ = \(\frac{4a-0}{4a-a}\) = \(\frac{4}{3}\),
Slope of the \(\overline{\mathrm{SQ}}\) = m₂ = \(\frac{-6a-0}{9a-a}\) = \(\frac{3}{4}\)
clearly m₁m₂ = -1, so that \(\overline{\mathrm{SP}}\) ⊥ \(\overline{\mathrm{SQ}}\)

Question 23.
From an external point P, tangent are drawn to the parabola y² = 4ax and these tangent make angles θ₁, θ₂ with its axis, such that tan θ₁ + tan θ₂ is constant b. then show that P lies on the line y = bx.
Solution:
Let the coordinates of P be (x₁, y₁) and the equation of the parabola y² = 4ax. Any tangent to .the parabola is y = mx + \(\frac{a}{m}\), if this passes through (x₁, y₁) then
y₁ = mx, + \(\frac{a}{m}\)
i.e., m²x₁ – my₁ + a = 0 ………………… (1)
Let the roots of (1) be m₁, m₂₁.
Then m₁ + m₂ = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)
⇒ tan θ₁ + tan θ₂ = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)
[∵ The tangents make angles θ₁, θ₂ with its axis (x – axis) then their slopes m₁ = tan θ₁ and m₂ = tan θ₂].
∴ b = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\) ⇒ y₁ = bx₁
∴ P(x₁, y₁) lies on the line y = bx.

Question 24.
Show that the common tangent to the parabola y² = 4ax and x² = 4by is xa^1/3 + yb^1/3 + a^2/3b^2/3 = 0. [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y² = 4ax ………………. (1)
and x² = 4by …………………. (2)
Equation of any tangent to (1) is of the form
y = mx + \(\frac{a}{m}\)
If the line (3) is a tangent to'(2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2),
we get x² = 4b (mx + \(\frac{a}{m}\)) is mx² – 4bm²x – 4ab = 0 should have equal roots therefore its discriminent must be zero. Hence
16b²m⁴ – 4m (-4ab) = 0
16b(bm⁴ + am) = 0
m(bm³ + a) = 0 But m ≠ 0
∴ m = – a^1/3/b^1/3 substituting in (3) the equation of the common tangent becomes
y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) (or)
∴ a^1/3 x + b^1/3y + a^2/3b^2/3 = 0

Question 25.
Prove that the area of the triangle formed ‘ by the tangents at (x₁, y₁), (x₂, y₂) and (x₃, y₃) to the parabola y² = 4ax (a > 0) is \(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x₁, y₁) = (at₁², 2at₁)
E(x₂, y₂) = (at₃², 2at₂)
and F(x₃, y₃) = (at₂³, 2at₃)
be three points on the parabola
y² = 4ax (a > 0).
The equation of the tangents at D, E and F are
t₁y = x + at₁² ………………. (1)
t₂y = x + at₂² ………………. (2)
t₃y = x + at₃² ………………. (3)
(1) – (2) ⇒ (t₁ – t₂) y = a(t₁ – t₂) (t₁ + t₂)
⇒ y = a(t₁ + t₂) substituting in (1)
we get x = at₁t₂
∴ The point of intersection of the tangents at D and E is say P(at₁t₂, a(t₁ + t₂))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at₂t₃, a(t₂ + t₃) and R(at₃t₁, a(t₃ + t₁)) respectively
Area of ∆PQR
Inter 2nd Year Maths 2B Parabola Important Questions 9

Question 26.
Prove that the two parabolas y² = 4ax and x² = 4by intersect (other than the origin) at an angle of Tan^-1 \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y⁴ = 16a²x²
= 16a²(4by)
= 64a²by
∴ y[y³ – 64a²b] = 0
⇒ y³ – 64a²b = 0
⇒ y = (64a²b)^1/3 [∵ y > 0]
= 4a^2/3b^1/3
Inter 2nd Year Maths 2B Parabola Important Questions 4
Also from y² = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)
= 4a^1/3b^2/3
∴ P = (4a^1/3b^2/3, 4a^2/3b^1/3)
Differentiating both sides of y² 4ax w.r.t ‘x’, we get
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)
If m₁ be the slope of the tangent at P to y² = 4ax, then
m₁ = \(\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}\)
Similarly, we get m₂ = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m₂ is the slope of the tangent at P to x² = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)
so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 27.
Prove that the orthocenter of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.
Solution:
Let y² = 4ax be the parabola and
A = (at₁², 2at₁),
B = (at₂², 2at₂),
C = (at₃², 2at₃) be any three points on it.
Now we consider the triangle PQR formed by the tangents to the parabola at A, B, C
where P = (at₁t₂, a(t₁ + t₂)),
Q = (at₂t₃, a(t₂ + t₃)) and R = (at₃t₁, a(t₃ + t₁)).
Equation of \(\overleftrightarrow{\mathrm{QR}}\) (i.e., the tangent at C) is x – t₃ y + at₃² = 0.
Therefore, the attitude through P of triangle PQR is
t₃x + y = at₁t₂t₃ + a(t₁ + t₂) ………………….. (1)
Similarly, the attitude through Q is
t₁x + y = at₁t₂t₃ + a(t₂ + t₃) ………………….. (2)
Solving (1) and (2), we get (t₃ – t₁)
x’ = a(t₁ – t₃) i.e., x = – a.
Therefore, the orthocenter of the triangle PQR, with abscissa as -a, lies on the directrix of the parabola.

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