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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:

Question 2.
x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y = 2x² sec²2x [May 11]
Solution:

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

∴ 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:

= 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:

Question 14.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0 [T.S. Mar. 15]
Solution:

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:

Question 17.
Solve (1 + y²)dx = (tan^-1y – x)dy. [A.P. Mar. 15]
Solution:
Given \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\tan ^{-1} y-x}{1+y^{2}}\)

Question 18.
(x² – y²)dx – xy dy = 0 [May 06]
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy

–\(\frac{1}{4}\) [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:


= 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

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

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

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

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

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:

Question 38.
Solve \(\frac{d y}{d x}=\frac{y^{2}+2 y}{x-1}\)
Solution:

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:

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

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

Question 44.
Solve \(\sqrt{1+x^{2}} \sqrt{1+y^{2}}\) dx + xy dy = 0
Solution:
Given equation can be written as

Question 45.
Solve \(\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}\)
Solution:
Put x – 2y = t

Question 46.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\sqrt{y-x}\)
Solution:
Put y – x = t²

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:

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

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:

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

\(\log v \sqrt{2 v-3}=-3 \log \frac{x}{c}=\log \frac{c^{3}}{x^{3}}\)

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

-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

Question 59.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^{2}+y^{2}}{2 x^{2}}\)
Solution:
This is a homogeneous equation.

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


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

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

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:

Question 67.
Solve (1 + x²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2xy – 4x² = 0
Solution:
Given equation can be written as

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

Question 71.
Solve (1 + y²)dx = (tan^-1y – x)dy. [A.P. Mar. 15]
Solution:

is the solution.

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:

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:

Question 6.
\(\int_{0}^{2 \pi}\) sin²x cos⁴; x dx [T.S. Mar. 15; Mar 14]
Solution:
sin²x cos⁴x is even function.

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 = \(\frac{\pi}{2}\)

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:

Question 10.
I = \(\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\) [Mar. 08]
Solution:

= –\(\frac{1}{40} \ln \left[\frac{1 / 4}{9 / 4}\right]\) = \(\frac{1}{40}\) . 2ln . 3 = \(\frac{1}{20}\) ln 3

Question 11.
y = x³ + 3, y = 0, x = -1, x = 2 [Mar. 05]
Solution:
Required area PABQ

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

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:

= \(\frac{1}{3}\left[\ln \frac{4}{2}\right]=\frac{1}{3} \ln 2\)

Question 15.
\(\int_{0}^{\pi} \frac{x}{1+\sin x} d x\) [May 11]
Solution:

Question 16.
\(\int_{0}^{\pi} \frac{x \sin ^{3} x}{1+\cos ^{2} x} d x\) [T.S. Mar. 15; Mar. 11]
Solution:

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 = \(\frac{\pi}{4}\)

Question 18.
\(\int_{0}^{\pi / 4} \log (1+\tan x) d x\) [A.P. Mar. 16]
Solution:
I = \(\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x\)

Question 19.
y = 4x – x², y = 5 – 2x. [T.S. Mar. 16]
Solution:

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

= 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:

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

= \(\int_{\pi / 6}^{\pi / 3} d x=(x)_{\pi / 6}^{\pi / 3}\)
= \(\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)
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:


= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 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:

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

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) = \(\sqrt{4-x^{2}}\)

Question 30.
Evaluate \(\int_{0}^{16} \frac{x^{1 / 4}}{1+x^{1 / 2}} d x\)
Solution:
Put t⁴ = x ⇒ dx = 4t³ . dt

Question 31.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin |x| d x\)
Solution:

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.

Question 33.
Evaluate \(\int_{0}^{\pi / 2} \frac{\cos ^{5 / 2} x}{\sin ^{5 / 2} x+\cos ^{5 / 2} x} d x\)
Solution:

= \(\int_{0}^{\pi / 2} d x=(x)_{0}^{\pi / 2}=\frac{\pi}{2}\)
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:


= \(\frac{\pi}{4 \sqrt{2}}\) 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:

Question 36.
Find \(\int_{-a}^{a}\left(x^{2}+\sqrt{a^{2}-x^{2}}\right) d x\)
Solution:

= 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:


= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 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:

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:

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:

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.

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:

\(\int_{0}^{\pi}\) 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

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:

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 \(\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\)

= 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:

Question 54.
Find the area cut off between the line y = 0 and the parabola y = x² – 4x + 3.
Solution:

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:

= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = 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:

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:

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:

= \(\left(-\frac{y^{3}}{3}+y\right)_{0}^{1}\) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 59.
Find the area enclosed by the curves y = 3x and y = 6x – x².
Solution:

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

= 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:

= \(\left(\frac{2}{3} x \sqrt{x}-\frac{x^{3}}{3}\right)_{0}^{1}\)
= \(\frac{2}{3}\) – \(\frac{1}{3}\) = \(\frac{1}{3}\)

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|

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:

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:

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

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:

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

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:

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:

Question 22.
\(\int \sqrt{1+3 x-x^{2}} d x\) [May 11]
Solution:
\(\int \sqrt{1+3 x-x^{2}} d x=\int \sqrt{1-\left(x^{2}-3 x\right)} d x\)
= \(\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x\)

Question 23.
\(\int \frac{9 \cos x-\sin x}{4 \sin x+5 \cos x} d x\) [T.S. Mar. 17; Mar. 08]
Solution:

Question 24.
\(\int \frac{d x}{5+4 \cos 2 x}\) [Mar. 11]
Solution:

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:

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}

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 = \(-\frac{\sin ^{n-1} x \cdot \cos x}{n}+\frac{n-1}{n} \cdot I_{n-2}\)

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

Question 30
\(\int \frac{1}{1+\sin x+\cos x}\) dx [T.S. Mar. 15]
Solution:

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 + \(\frac{3}{2}\) = \(\frac{13}{2}\)

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

Question 33.
\(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^{2}}}\) [May 05]
Solution:

Question 34.
\(\int \frac{d x}{4 \cos x+3 \sin x}\) [Mar. 06]
Solution:

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


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

Question 37.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:

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:

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

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

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:

Question 56.
Evaluate \(\int \frac{1}{a^{2}-x^{2}}\) dx for x ∈ I = (-a, a)
Solution:

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:

Question 60.
Evaluate \(\int \sqrt{9 x^{2}-25} d x\) on [\(\frac{5}{3}\), ∞)
Solution:

Question 61.
Evaluate \(\int \sqrt{16-25 x^{2}}\) dx on (\(\frac{-4}{5}\), \(\frac{4}{5}\))
Solution:

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

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

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

Question 67.
Evaluate \(\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\) on I ⊂ R \ {2nπ : n ∈ z}.
Solution:

= -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

Question 70.
Evaluate \(\int \frac{d x}{4 x^{2}-4 x-7}\)
Solution:

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 – \(\left(\sqrt{\frac{7}{2}}\right)^{2}\))
\(\int \frac{d x}{5-2 x^{2}+4 x}\)
= \(-\frac{1}{2} \int \frac{1}{\left((x-1)^{2}-\sqrt{\frac{7}{2}}\right)^{2}} d x+C\)

Question 72.
Evaluate \(\int \frac{d x}{x^{2}+x+1}\)
Solution:

Question 73.
Evaluate \(\int \frac{d x}{\sqrt{x^{2}+2 x+10}}\)
Solution:

Question 74.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^{2}}}\)
Solution:

Question 75.
Evaluate \(\int \sqrt{3+8 x-3 x^{2}} d x\)
Solution:

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) – \(\frac{1}{2}\)

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 + \(\frac{3}{4}\)
= \(\frac{-5}{4}\)

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
= \(\int \frac{2 x-2}{\sqrt{x^{2}-2 x+10}} d x+7 \int \frac{d x}{\sqrt{x^{2}-2 x+10}}+C\)

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:

Question 81.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\) [Mar. 13]
Solution:
t = tan \(\frac{x}{2}\) ⇒ dx = \(\frac{2 d t}{1+t^{2}}\)

Question 82.
Find \(\int \frac{d x}{d+e \tan x}\)
Solution:

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:

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

Question 85.
Find \(\int \frac{x^{3}-2 x+3}{x^{2}+x-2}\) dx.
Solution:

Question 86.
Find \(\int \frac{d x}{x^{2}-81}\)
Solution:

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

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 . \(\frac{5}{4}\) = \(\frac{11}{2}\)

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

Question 90.
Evaluate \(\int\) x³ . e^5x dx.
Solution:
We take a = 5, use the reduction formula

Question 91.
Evaluate \(\int\) sin⁴ x dx. [Mar. 14]
Solution:

Question 92.
Evaluate \(\int\) tan⁶ x dx.
Solution:

Question 93.
\(\int\) sec⁵ x dx.
Solution:
Reduction formula is

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

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 = \(\frac{1}{3}\)

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:

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)

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]

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.

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

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.

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.

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\(\sqrt{3}\)

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

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)

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 \(\frac{\theta_{1}-\theta_{2}}{2}\)]

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:

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:

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

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

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

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

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

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.

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

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.

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

\(\frac{1}{16a}\) |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

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))

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

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.

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

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

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

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.

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

Intermediate 2nd Year Maths 2B System of Circles Important Questions

Question 1.
x² + y² + 4x + 8 = 0, x² + y² – 16y + k = 0 [A.P. & T.S. Mar. 16]
Solution:
g₁ = 2; f₁ = 0; c₁ = 8
g₂ = 0; f₂ = – 4; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(2) (0) + 2(0) (-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Question 2.
(x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b) [A.P. Mar. 15]
Solution:
(x² + y² – 2xa – 2yb – c²)
– (x² + y² – 2xb – 2ya – c²) = 0
– 2x (a – b) – 2y(b – a) = 0
(or) x – y = 0

Question 3.
Find the angle between the circles given by the equations. [T.S. Mar.17]
i) x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0.
Solution:
C₁ = (6, 3)
r₁ =(36 + 9 – 41)^1/2
r₁ = 2

C₂ = (-2, -3)
r₂ =(4 + 9 + 59)^1/2
r₂ = (72)^1/2 = 6\(\sqrt{2}\)

C₁C₂ = d = \(\sqrt{(6+2)^{2}+(3+3)^{2}}\)
= \(\sqrt{64+36}\) = 10
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
= \(\frac{100-4-72}{2 \times 2 \cdot \sqrt{72}}=\frac{24}{4 \times 6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
θ = 45°

Question 4.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 = 0 and x² + y² – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……………….. (i)
Orthogonal to circle
2g (-2) +2f(-3) = 11 + c ……………………. (ii)
2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
Subtracting it we get
-6g + 2f = 10 ……………………… (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 ……………………. (v)
Solving (iv) and (y)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 5.
Show that the angle between the circles x² + y² = a², x² + y² = ax + ay is \(\frac{3\pi}{4}\) [Mar. 14]
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0
C₁ (0, 0), C₂ (\(\frac{a}{2}\), \(\frac{a}{2}\))
∴ C₁C₂² = (0 – \(\frac{a}{2}\))² + (0 – \(\frac{a}{2}\))²

Question 6.
If x + y = 3 ¡s the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter. [A.P. Mar. 15]
Solution:
Required equation of circle passing through
intersection S = 0 and L = 0 is S + λL = 0
(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0
(x² + y² + x(-2 + λ)+ y(4 + λ) – 8 – 3λ = 0 ………………. (i)
x² + y² + 2gx + 2fy + c = 0 ……………………. (ii)
Comparing (i) and (ii) we get
g = \(\frac{(-2+\lambda)}{2}\), f = \(\frac{(4+\lambda)}{2}\)
Centre lies on x + y = 3
∴ \(-\left(\frac{-2+\lambda}{2}\right)-\left(\frac{4+\lambda}{2}\right)\) = 3
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = – 4
Required equation of circle be
(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0
x² + y² – 6x + 4 = 0

Question 7.
If two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
Solution:
C₁ = (-g, -f)
r₁ = \(\sqrt{g^{2}+f^{2}}\)
C₂ = (-g^-1, -f^-1)
r₂ = \(\sqrt{g^{\prime 2}+r^{\prime 2}}\)
C₁C₂ = r₁ + r₂
(C₁C₂)² = (r₁r₂)²
(g’ – g)² + (f’ – f)² = g² + f² + g’² + f’² + \(2 \sqrt{g^{2}+f^{2}} \sqrt{g^{2}+f^{1^{2}}}\)
-2(gg’ + ff’) = 2{g²g’² + g²f’² + f²g’²}^1/2
Squaring again
(gg’ + ff’)² = g²g’² + f²f’² + g²f’² + g’²f²
g²g’² + f²f’² + 2gg’ff’ = g²g’² + f²f’² + g²f’² + g’²f’²
2gg’ff’ = g²f’² + f²g’²
⇒ g²f’² + g’²f’² – 2gg’ff’ = 0
(or) (gf’ – fg’)² = 0 (or) gf’ = fg’

Question 8.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 …………………… (1)
and S ≡ x² + y² + 6x + 2y – 90 = 0 …………………(2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10.
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In
this case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) in the ratio 5 : 10
i.e., 1: 2 (externally)
∴ Point of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5)

Question 9.
Find the angle between the circles x² + y² + 4x – 14y + 28 = 0 and x² + y² + 4x – 5 = 0
Solution:
Equations of the given circles are .
x² + y² + 4x – 14y + 28 = 0
x² + y² + 4x – 5 = 0
Centres are C₁ (-2, 7), C₂ (-2, 0)
C₁C₂ = \(\sqrt{(-2+2)^{2}+(7-0)^{2}}\)
= \(\sqrt{0+49}\) = 7
r₁ = \(\sqrt{4+49-28}\) = \(\sqrt{25}\) = 5
r₂ = \(\sqrt{4+5}\) = \(\sqrt{9}\) = 3
If θ is the angle between the given circles,
then cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
cos θ = \(\frac{49-25-9}{2(5)(3)}\) = \(\frac{15}{2.5 .3}\) = \(\frac{1}{2}\) = cos 60°
Angle between the circles = θ = \(\frac{\pi}{3}\)

Question 10.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45° find k.
Solution:
Suppose θ is the angle between the circles
x² + y² – 12x – 6y + 41 = 0
and x² + y² + kx + 6y – 59 = 0
g₁ = -6, f₁ = -3, c₁ = 41,
g₂ = \(\frac{k}{2}\), f₂ = 3, c₂ = -59

\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
Squaring and cross – multiplying
4(\(\frac{k^{2}}{4}\) + 68) = 18k²
\(\frac{2\left[k^{2}+272\right]}{4}\) = 9k²
k² + 272 = 18 k²
17k² = 272
k² = \(\frac{272}{17}\)
k² = 16
k = ±4.

Question 11.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
x² + y² – 8x – 2y + 16 = 0 and …………….. (1)
x² + y² – 4x – 4y – 1 = 0. ………………. (2)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………. (3)
Then the circle (3) is orthogonal to (1) and (2).
∴ By applying the condition of orthogonality give in x² + y² + 2gx + 2fy + c = 0 we get
2g(-4) + 2f(-1) = c + 16 and …………………. (4)
2g (-2) + 2f(-2) = c – 1 ……………….. (5).
Given that the circle (3) is passing through (1, 1)
∴ 1² + 1² + 2g(1) + 2f(1) + c = 0
2g + 2f + c + 2 = 0 ………………….. (6)
Solving (4), (5) and (6) for g, f and c, we get
g = –\(\frac{7}{3}\), f = \(\frac{23}{6}\), c = -5
Thus the equation of the required circle is
3(x² + y²) – 14x + 23y – 15 = 0.

Question 12.
Find the equation of the circle which is orthogonal to each of the following three circles.
x² + y² + 2x + 17y + 4 = 0 ………………… (1)
x² + y² + 7x + 6y + 11 = 0 ………………. (2)
and x² + y² – x + 22y + 3 = 0 ………………… (3)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………… (4)
Since this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x² + y² + 2gx + 2fy + c = 0
we have
2(g)(1) + 2(f) (\(\frac{17}{2}\)) = c + 4 ……………….. (5)
2(g) (\(\frac{7}{2}\)) + 2(f)(3) = c + 11 ………………. (6)
and 2(g) (-\(\frac{1}{2}\)) + 2(f)(11) = c + 3 ………………….. (7)
Solving (5), (6) and (7) for g, f, c we get
g = -3, f = -2 and c = -44
Thus the equation of the required circle is
x² + y² – 6x – 4y – 44 = 0.

Question 13.
If the straight line is represented by
x cos α + y sin α = p ………………….. (1)
intersects the circle
x² + y² = a² ……………… (2)
at the points A and B, then show that the equation of the circle with \(\overline{\mathrm{AB}}\) as diameter is(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.
Solution:
The equation of the circle passing through the points A and B is S ≡ x² + y² + 2gx + 2fy + c = 0
(x² + y² – a²) + λ(x cos α + y sin α – p) = 0 ……………… (3)
The centre of this circle is
\(\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1)
∴ \(-\frac{\lambda \cos \alpha}{2}\) (cos α) – \(\frac{\lambda \sin \alpha}{2}\) (sin α) = p
i.e., –\(\frac{\lambda}{2}\) (cos² α + sin² α) p
i.e., λ = -2p
Hence the equation of the required circle is
(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.

Question 14.
Find the equation of the circle passing through the points of intersection of the circles.
x² + y² – 8x – 6y + 21 = 0 ……………….. (1)
x² + y² – 2x – 15 = 0 ……………….. (2)
and (1, 2).
Solution:
The equation of circle passing through the points of intersection of (1) arid (2) is
(x² + y² – 8x – 6y + 21) + λ (x² + y² – 2x – 15) = 0 ……………….. (3)
If it passes through (1, 2). we obtain
(1 + 4 – 8 – 12 + 21) + λ(1 + 4 – 2 – 15) = 0
i.e., 6 + λ(-12) = 0
i.e., λ = \(\frac{1}{2}\)
Hence the equation of the required circle is
(x² + y² – 8x – 6y + 21) + \(\frac{1}{2}\) (x² + y² – 2x – 15) = 0
i.e., 3(x² + y²) – 18x – 12y + 27 = 0.

Question 15.
Let us find the equation the radical axis of the circles S ≡ x² + y² – 5x + 6y + 12 = 0 and S’ ≡ x² + y² + 6x – 4y – 14 = 0
Solution:
The given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
i.e., 11x – 10y – 26 = 0

Question 16.
Let us find the equation of the radical axis of the circles
2x² + 2y² + 3x + 6y – 5 = 0 ………………….. (1)
and 3x² + 3y² – 7x + 8y – 11 = 0 ……………….. (2)
Solution:
Hence the given equations are not in general form, we get :
x² + y² + \(\frac{3}{2}\)x + 3y – \(\frac{5}{2}\) = 0 and
x² + y² – \(\frac{7}{3}\)x + \(\frac{8}{3}\) y – \(\frac{11}{3}\) = 0
Now the radical axis equation of given circles is
(\(\frac{3}{2}\) + \(\frac{7}{3}\))x + (3 – \(\frac{8}{3}\))y + (-\(\frac{5}{2}\) + \(\frac{11}{3}\)) = 0
i.e., 23x + 2y + 7 = 0

Question 17.
Let us find the radical centre of the circles
x² + y² – 2x + 6y = 0 ………………… (1)
x² + y² – 4x – 2y + 6 = 0 …………………. (2)
and x² + y² – 12x + 2y + 3 = 0 ………………. (3)
Solution:
The radical axis of (1) and (2) and (3)
x + 4y – 3 = 0 ………………… (4)
8x – 4y + 3 = 0 …………………. (5)
10x + 4y – 3 = 0 ……………… (6)
Solving (4) and (5) for the point of intersection we get (0, \(\frac{3}{4}\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.

Question 18.
Find the equation and length of the common chord of the two circles
S ≡ x² + y² + 3x + 5y + 4 = 0
and S’ ≡ x² + y² + 5x + 3y + 4 = 0
Solution:
Equations of the given circles are
S ≡ x² + y² + 3x + 5y + 4 = 0 ………………… (1)
S’ ≡ x² + y² + 5x + 3y + 4 ………………. (2)
Equations of the common chord is S – S’ = 0
-2x + 2y = 0
L ≡ x – y = 0 …………….. (3)

Question 19.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 ……………….. (1)
and S’ ≡ x² + y² + 6x + 2y – 90 = 0 ……………. (2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
S – S’ = 0
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
∴ Point, of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5).

Question 20.
Find the equation of the circle whose diameter is the common chord of the circles
S ≡ x² + y² + 2x + 3y + 1 = 0 ……………….. (1)
and S’ ≡ x² + y² + 4x + 3y + 2 = 0 ……………….. (2)
Solution:
Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
i.e., 2x + 1 = 0 …………………… (3)
The equation of any circle passing through
the points of intersection of (1) and (3) is (S + λL = 0)
(x² + y² + 2x + 3y + 1) + λ(2x + 1) = 0
x² + y² + 2(λ + 1)x + 3y + (1 + λ) = 0 ……………………. (4)
The centre of this circle is (-(λ + 1), \(\frac{3}{2}\)).
For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
∴ 2{-(λ + 1)} + 1 = 0
⇒ λ = –\(\frac{1}{2}\)
Thus equation of the circle whose diameter is the common chord (1) and (2)
(put λ = \(\frac{1}{2}\) in equation (4))
2(x² + y²) + 2x + 6y + 1 = 0

Question 21.
Let us find the equation of a circle which cuts each of the following circles orthogonally
S’ ≡ x² + y² + 3x + 2y + 1 = 0 ………………… (1)
S” ≡ x² + y² – x + 6y + 5 = 0 …………….. (2)
and S” ≡ x² + y² + 5x – 8y + 15 = 0 ……………………. (3)
Solution:
The centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
x – y = 1 ………………… (4)
and the radical axis of (2) and (3) is
3x – 7y = -5 …………………… (5)
The point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
= \(\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \sqrt{3}\)
Thus the required circle is
(x – 3)² + (y – 2)² = (3\(\sqrt{3}\))²
x² + y² – 6x – 4y – 14 = 0.

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

Intermediate 2nd Year Maths 2B Circle Important Questions

Question 1.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3), find g, f and its radius. [Mar. 11]
Solution:
Circle is x² + y² + 2gx + 2fy – 12 = 0
C = (-g, -f) C = (2, 3)
∴ g = – 2, f = – 3, c = – 12
Radius = \(\sqrt{g^{2}+f^{2}-c}\)
= \(\sqrt{4+9+12}\) = 5 units

Question 2.
Obtain the parametric equation of x² + y² = 4 [Mar. 14]
Solution:
Equation of the circle is x² + y² = 4
C (0, 0), r = 2
Parametric equations are
x = – g + r cos θ = 2 cos θ
y = – b + r sin θ = 2 sin θ, 0 < θ < 2π

Question 3.
Obtain the parametric equation of (x – 3)² + (y – 4)² = 8² [A.P. Mar. 16; Mar. 11]
Solution:
Equation of the circle is (x – 3)² + (y – 4)² =8²
Centre (3, 4), r = 8
Parametric equations are
x = 3 + 8 cos θ, y = 4 + 8 sin θ, 0 < θ < 2π

Question 4.
Find the power of the point P with respect to the circle S = 0 when
ii) P = (-1,1) and S ≡ x² + y² – 6x + 4y – 12
Solution:
Power of the point = S₁₁
= 1 + 1 + 6 + 4 – 12 = 0

iii) P = (2, 3) and S ≡ x² + y² – 2x + 8y – 23
Power of the point = S₁₁
= 4 + 9 – 4 + 24 – 23 = 10

iv) P = (2, 4) and S ≡ x² + y² – 4x – 6y – 12
Power of the point = 4 + 16 – 8 – 24 – 12
= -24.

Question 5.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1 then find k. [Mar. 15; Mar. 01]
Solution:
= \(\sqrt{S_{11}}=\sqrt{(5)^{2}+(4)^{2}+8 k}\)
But length of tangent = 1
∴ 1 = \(\sqrt{25+16+8k}\)
Squaring both sides we get 1 = 41 + 8k
k = – 5 units.

Question 6.
Find the polar of (1, -2) with respect to x² + y² – 10x – 10y + 25 = 0 [T.S. Mar. 15]
Solution:
Equation of the circle is x² + y² – 10x – 10y + 25 = 0
Equation of the polar is S₁ = 0
Polar of P(1,-2) is
x . 1 + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0
⇒ x – 2y – 5x – 5 – 5y + 10 + 25 = 0
⇒ -4x – 7y + 30 = 0
∴ 4x + 7y – 30 = 0

Question 7.
Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35. [T.S. Mar. 16]
Solution:
Equation of the circle is
x² + y² = 35
Polar of P(1, 3) is x. 1 + y. 3 = 35
x + 3y = 35
P(1, 3) and Q(2, k) are conjugate points
The polar of P passes through Q
2 + 3k = 35
3k = 33
k = 11

Question 8.
If the circle x² + y² – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is
x² + y² – 4x + 6y + a = 0
2g = – 4, 2f = 6, c = a
g = -2, f = 3, c = a
radius = 4 ⇒ \(\sqrt{g^{2}+f^{2}-c}\) = 4
\(\sqrt{4+9-a}\) = 4
13 – a = 16
a = 13 – 16 = -3

Question 9.
Find the value of ‘a’ if 2x² + ay² – 3x + 2y – 1 =0 represents a circle and also find its radius.
Solution:
General equation of second degree
ax² + 2hxy + by² + 2gx + 2fy + c = 0
Represents a circle, when
a = b, h = 0, g² + f² – c ≥ 0
In 2x² + ay² – 3x + 2y – 1 = 0
a = 2, above equation represents circle.
x² + y² – \(\frac{3}{2}\) x + y – \(\frac{1}{2}\) = 0
2g = – \(\frac{3}{2}\); 2f = 1; c = – \(\frac{1}{2}\)
c = (-g, -f) = (\(\frac{+3}{4}\), \(\frac{-1}{2}\))
Radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}\)
= \(\frac{\sqrt{21}}{4}\) units

Question 10.
If the abscissae of points A, B are the roots of the equation, x² + 2ax – b² = 0 and ordinates of A, B are root of y² + 2py – q² = 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter. [Mar. 14]
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
x² – x(x₁ + x₂) + x₁x₂ + y² – y (y₁ + y₂) + y₁y₂ = 0
x₁, x₂ are roots of x² + 2ax – b² = 0
y₁, y₂ are roots of y² + 2py – q² = 0
x₁ + x₂ = – 2a
x₁x₂ = -b²

y₁ + y₂ = -2p
y₁y₂ = -q²
Equation of circle be
x² – x (- 2a) – b² + y²– y (- 2p) – q² = 0
x² + 2xa + y² + 2py – b² – q² = 0

Question 11.
Find the equation of a circle which passes through (4, 1) (6, 5) and having the centre on 4x + 3y – 24 = 0 [A.P. Mar. 16; Mar. 14]
Solution:
Equation of circle be x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
4² + 1² + 2g(4) + 2f(1) + c = 0 ………….. (i)
6² + 5² + 2g(6) + 2f(5) + c = 0 ……………… (ii)
Centre lie on 4x + 3y – 24 = 0
∴ 4(-g) + 3 (-f) – 24 = 0
(ii) – (i) we get
44 + 4g + 8f = 0
Solving (iii) and (iv) we get
f = -4, g = -3, c = 15
∴ Required equation of circle is
x² + y² – 6x – 8y + 15 = 0

Question 12.
Find the equation of a circle which is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (- 2, 14). [Mar. 14]
Solution:
x² + y² – 6x – 4y – 12 = 0 …………… (i)
C = (- g, – f)
= (3, 2)
Equation of circle concentric with (i) be
(x – 3)² + (y – 2)² = r²
Passes through (-2, 14)
∴ (- 2 – 3)² + (14 – 2)² = r²
169 = r²
Required equation of circle be
(x – 3)² + (y – 2)² = 169
x² + y² – 6x – 4y – 156 = 0

Question 13.
Find the equation of the circle whose centre lies on the X-axis and passing through (- 2, 3) and (4, 5). [A.P. & T.S. Mar. 15]
Solution:
x² + y² + 2gx + 2fy + c = 0 ……………… (i)
(- 2, 3) and (4, 5) passes through (i)
4 + 9 – 4g + 6f + c = 0 ………………………. (ii)
16 + 25 + 8g + 10f + c = 0 …………………. (iii)
(iii) – (ii) we get
28 + 12g + 4f = 0
f + 3g = – 7
Centre lies on X – axis then f = 0
g = -, \(\frac{7}{3}\), f = 0, c = \(\frac{67}{3}\) -, we get by substituting g; f in equation (ii)
Required equation will be 3(x² + y²) – 14x – 67 = 0

Question 14.
Find the equation of circle passing through (1, 2); (3, – 4); (5, – 6) three points.
Solution:
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
1 + 4 + 2g + 4f + c = 0 …………………… (i)
9 + 16 + 6g – 8f + c = 0 …………………… (ii)
25 + 36 + 10g – 12f + c = 0 ………………… (iii)
Subtracting (ii) – (i) we get
20 + 4g – 12f = 0
(or) 5 + g – 3f = 0 ……………….. (iv)
Similarly (iii) – (ii) we get
36 + 4g – 4f = 0
(or) 9 + g – f = 0 ………………… (v)
Solving (v) and (iv) we get
f = -2, g = – 11, c = 25
Required equation of circle be x² + y² – 22x – 4y + 25 = 0

Question 15.
Find the length of the chord intercepted by the circle x² + y² – 8x – 2y – 8 = 0 on the line x + y + 1 = 0 [T.S. Mar. 16]
Solution:
Equation of the circle is x² + y² – 8x – 2y – 8 = 0
Centre is C(4, 1), r = \(\sqrt{16+1+8}\) = 5
Equation of the line is x + y + 1 = 0
P = distance from the centre = \(\frac{|4+1+1|}{\sqrt{1+1}}\)
= \(\frac{6}{\sqrt{2}}\) = 3\(\sqrt{2}\)
Length of the chord = 2\(\sqrt{r^{2}-p^{2}}\)
= 2\(\sqrt{25-18}\)
= 2\(\sqrt{7}\) units.

Question 16.
Find the pair of tangents drawn from (1, 3) to the circle x² + y² – 2x + 4y – 11 =0 and also find the angle between them. [T.S. Mar. 16]
Solution:
SS₁₁ = S₁²
(x² + y² – 2x + 4y – 11) (1 + 9 – 2 + 12 – 11) = [x + 3y – 1 (x + 1) + 2 (y + 3) – 11]²
(x² + y² – 2x + 4y – 11) 9 = [5y – 6]²
9x² + 9y² – 18x + 36y – 99
= 25y² + 36 – 60y
9x² – 16y² – 18x + 96y – 135 = 0
cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\) = \(\frac{|9-16|}{\sqrt{(25)^{2}}}\)
= \(\frac{|-7|}{25}\) = \(\frac{7}{25}\) ⇒ θ = cos^-1 (\(\frac{7}{25}\))

Question 17.
Find the value of k if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x² + y² – 5x + 8y + 6 = 0. [T.S. Mar. 17]
Solution:
Equation of the circle is x² + y² – 5x + 8y + 6 = 0
Polar of P(4, 2) is
x . 4 + y . 2 – \(\frac{5}{2}\) (x + 4) + 4 (y + 2) + 6 = 0
8x + 4y – 5x – 20 + 8y + 16 + 12 = 0
3x + 12y + 8 = 0
P(4, 2), Q(k, -3) are conjugate points
Polar of P passes through Q
∴ 3k – 36 + 8 = 0
3k = 28 ⇒ k = \(\frac{28}{3}\)

Question 18.
If (2, 0), (0,1), (4, 5) and (0, c) are concyclic, and then find c. [A.P. & T.S. Mar. 15]
Solution:
x² + y² + 2gx + 2fy + c₁ = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c₁ = 0 …………………. (i)
0 + 1 + 2g. 0 + 2f + c₁ = 0 – (ii)
16 + 25 + 8g + 10f + c₁ = -0 ……………. (iii)
(ii) – (i) we get
– 3 – 4g + 2f = 0
4g – 2f = – 3 ……………… (iv)
(ii) – (iii) we get
– 40 – 8g – 8f = 0 (or)
g + f = – 5 …………………… (v)
Solving(iv) and (v) we get
g = –\(\frac{13}{6}\), f = –\(\frac{17}{6}\)
Substituting g and f values in equation (i) we get
4 + 4 (-\(\frac{13}{6}\)) + c₁ = 0
c₁ = \(\frac{14}{3}\)
Now equation x² + y² – \(\frac{13}{3}\) x – \(\frac{17}{3}\) y + \(\frac{14}{3}\) = 0
Now circle passes through (0, c) then
c² – \(\frac{17}{3}\) c + \(\frac{14}{3}\) = 0
3c² – 17c + 14 = 0
⇒ (3c – 14) (c – 1) = 0
(or)
c = 1 or \(\frac{14}{3}\)

Question 19.
Find the length of the chord intercepted by the circle x² + y² – x + 3y – 22 = 0 on the line y = x – 3. [May 11; Mar. 13]
Solution:
Equation of the circle is
S ≡x² + y² – x + 3y – 22 = 0
Centre C(\(\frac{1}{2}\), –\(\frac{3}{2}\))

Question 20.
Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0 [Mar. 11]
Solution:
Equation of the line is 3x + 4y + 4 = 0
P = Length of the perpendicular .

Length of the chord = 2λ = 2 ⇒ λ = 1
If r is the radius of the circle then
r² = 2² + 1² – 4 + 1 = 5
Centre of the circle is (-2, 3)
Equation of the circle is (x + 2)² + (y – 3)² = 5
x² + 4x + 4 + y² – 6y + 9 – 5 = 0
i.e., x² + y² + 4x – 6y + 8 = 0

Question 21.
Find the pole of 3x + 4y – 45 = 0 with respect to x² + y² – 6x – 8y + 5 = 0. [A.P. Mar. 16]
Solution:
Equation of polar is
xx₁ + yy₁ – 3(x + x₁) – 4(y + y₁) + 5 = 0
x(x₁ – 3) + y(y₁ – 4) – 3x₁ – 4y₁ + 5 = 0 …………………. (i)
Polar equation is same 3x + 4y – 45 = 0 ……………….. (ii)
Comparing (i) and (ii) we get

Question 22.
i) Show that the circles x² + y² – 6x – 2y + 1 = 0 ; x² + y² + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact. [A.P. Mar. 16; Mar. 11]
Solution:
Equations of the circles are
S₁ ≡ x² + y² – 6x – 2y + 1 = 0
S₂ ≡x² + y² + 2x – 8y + 13 = 0
Centres are A (3, 1), B(-1, 4)
r₁ = \(\sqrt{9+1+1}\) = 3, r₁ = \(\sqrt{1+16-13}\) = 2
AB = \(\sqrt{(3+1)^{2}+(1-4)^{2}}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5
AB = 5 = 3 + 2 = r₁ + r₁
∴ The circles touch each other externally.
The point of contact P divides AB internally in the ratio r₁ : r₂ = 3 : 2
Co- ordinates of P are
\(\left(\frac{3(-1)+2.3}{5}, \frac{3.4+2.1}{5}\right)\) i.e., P\(\left(\frac{3}{5}, \frac{14}{5}\right)\)
Equation of the common tangent is S₁ – S₂ = 0
-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0

ii) Show that x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are
S₁ ≡ x² + y² – 6x – 9y + 13 = 0
S₂ ≡ x² + y² – 2x – 16y = 0
centres are A(3, \(\frac{9}{2}\)), B(1, 8)
r₁ = \(\sqrt{9+\frac{81}{4}-13}\) = \(\frac{\sqrt{65}}{2}\), r₂ = \(\sqrt{1+64}\)
= \(\sqrt{65}\)
AB = \(\sqrt{(3-1)^{2}+\left(\frac{9}{2}-8\right)^{2}}\) = \(\sqrt{4+\frac{49}{4}}\)
= \(\frac{\sqrt{65}}{2}\)
AB = |r₁ – r₂|
∴ The circles touch each other internally. The point of contact ‘P’ divides AB
externally in the ratio r₁ : r₂ = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\)
= 1 : 2 Co-ordinates of P are
\(\left(\frac{1(1)-2(3)}{1-2}, \frac{1(8)-2\left(\frac{9}{2}\right)}{1-2}=\left(\frac{-5}{-1}, \frac{-1}{-1}\right)\right.\) = (5, 1)
p = (5, 1)
∴ Equation of the common tangent is
S₁ – S₂ = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 23.
Find the direct common tangents of the circles. [T.S. Mar. 15]
x² + y² + 22x – 4y – 100 = 0 and x² + y² – 22x + 4y + 100 = 0.
Solution:
C₁ = (-11, 2)
C₂ = (11, -2)
r₁ = \(\sqrt{121+4+100}\) = 15
r₂ = \(\sqrt{121+4-100}\) = 5
Let y = mx + c be tangent
mx – y + c = 0
⊥ from (-11, 2) to tangent = 15
⊥ from (11 ,-2) to tangent = 5

Squaring and cross multiplying
25 (1 + m²) = (11 m + 2 – 22m – 4)²
96m² + 44m – 21 = 0
⇒ 96m² + 72m – 28m – 21 = 0
m = \(\frac{7}{24}\), \(\frac{-3}{4}\)
c = \(\frac{25}{2}\)
y = – \(\frac{3}{4}\)x + \(\frac{25}{2}\)
4y + 3x = 50
c = -22m – 4
= -22(\(\frac{7}{24}\)) – 4
= \(\frac{-77-48}{12}\) =\(\frac{-125}{12}\)
y = \(\frac{7}{24}\) x – \(\frac{125}{12}\)
⇒ 24y = 7x – 250
⇒ 7x – 24y – 250 = 0

Question 24.
Find the transverse common tangents of the circles x² + y² – 4x – 10y + 28 = 0 and x² + y² + 4x-6y + 4 = 0 [A.P. Mar. 15; Mar. 14]
Solution:
C₁ =(2, 5), C₂ = (-2, 3)
r₁ = \(\sqrt{4+25-28}\) = 1,
r₂ = \(\sqrt{4+9-4}\) = 3
r₁ + r₂= 4
C₁C₂ = \(\sqrt{(2+2)^{2}+(5-3)^{2}}\)
= \(\sqrt{16+4}\) = \(\sqrt{20}\)
‘C’ divides C₁C₂ in the ratio 5 : 3

Equation of the pair transverse of the common tangents is
S₁² = SS₁₁
(x . 1 + \(\frac{9}{2}\)y – 2(x + 1) – 5(y + \(\frac{9}{2}\)) + 28)² = [1 + \(\frac{81}{4}\) – 4 – 10 × \(\frac{9}{2}\) + 28]
= – (x² + y² – 4x – 10y + 28)
⇒ (-x – \(\frac{1}{2}\)y + \(\frac{7}{2}\))²
= \(\frac{1}{4}\) (x² + y² – 4x – 10y + 28)
(-2x – y + 7)² = (x² + y² – 4x – 10y + 28)
4x² + y² + 4xy – 28x – 14y + 49 = x² + y² – 4x – 10y + 28
3x² + 4xy – 24x – 4y + 21 = 0
(3x + 4y – 21); (x – 1) = 0
3x + 4y – 21 = 0; x – 1 = 0

Question 25.
Show that the circles x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 touch each other. Also find the point of contact and common tangent at this point of contact. [Mar. 13]
Solution:
Equations of the circles are
x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0
Centres are C₁(2, 3), C₂ = (-3, -9)
r₁ = \(\sqrt{4+9+12}\) = 5
r₂ = \(\sqrt{9+81-26}\) = 8
C₁C₂ = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)
= \(\sqrt{25+144}\) = 13 = r₁ + r₂
∴ Circle touch externally .
Equation of common tangent is S₁ – S₂ = 0
-10x -,24y – .38 = 0
5x + 12y + 19 = 0

Question 26.
Find the equation of circle with centre (1, 4) and radius ‘5’.
Solution:
Here (h, k) = (1, 4) and r = 5.
∴ By the equation of the circle with centre at C (h, k) and radius r is
(x – h)² + (y – k)² = r²
(x – 1)² + (y – 4)² = 5²
i.e., x² + y²– 2x – 8y – 8 = 0

Question 27.
Find the centre and radius of the circle x² + y² + 2x – 4y – 4 = 0.
Solution:
2g = 2, 2f = -4, c = -4
g = 1, f = -2, c = -4
Centre (-g, -f) = (-1, 2)
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+4-(-4)}\) = 3

Question 28.
Find the centre and radius of the circle 3x² + 3y² – 6x + 4y – 4 = 0.
Solution:
Given equation is
3x² + 3y² – 6x + 4y – 4 = 0
Dividing with 3, we have
x² + y² – 2x + \(\frac{4}{3}\) y – \(\frac{4}{3}\) = 0
2g = -2, 2f = \(\frac{4}{3}\), c = –\(\frac{4}{3}\)
g = -1, f = \(\frac{2}{3}\), c = –\(\frac{4}{3}\)
Centre (-g, -f) = (1, \(\frac{-2}{3}\))
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+\frac{4}{9}+\frac{4}{3}}\)
= \(\sqrt{\frac{9+4+12}{9}}\) = \(\sqrt{\frac{25}{9}}\) = \(\frac{5}{3}\)

Question 29.
Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6).
Solution:
Let C(-1, 2) be the centre of the circle

Since P(5,6) is a point on the circle CP = r
CP² = r² ⇒ r² = (-1 – 5)² + (2 – 6)²
= 36 + 16 = 52
Equation of the circle is (x + 1)² + (y – 2)²
= 52
x² + 2x + 1 + y² – 4y + 4 – 52 = 0
x² + y² + 2x – 4y – 47 = 0

Question 30.
Find the equation of the circle passing through (2, 3) and concentric with the circle x² + y² + 8x + 12y + 15 = 0.
Solution:
The required circle is concentric with the circle x² + y² + 8x + 12y + 15 = 0
∴ The equation of the required circle can be taken as
x² + y² + 8x + 12y + c = 0

This circle passes through P(2, 3)
∴ 4 + 9 + 16 + 36 + c = 0
c = – 65
Equation of the required circle is x² + y² + 8x + 12y – 65 = 0

Question 31.
From the point A(0, 3) on the circle x² + 4x + (y – 3)² = 0 a chord AB is drawn and extended to a point M such that AM = 2 AB. Find the equation of the locus of M.
Solution:
Let M = (x’, y’)
Given that AM = 2AB

AB + BM = AB + AB
⇒ BM = AB
B is the mid point of AM
Co- ordinates of B are \(\left(\frac{x^{\prime}}{2}-\frac{y^{\prime}+3}{2}\right)\)
B is a point on the circle
(\(\frac{x^{\prime}}{2}\))² + 4(\(\frac{x^{\prime}}{2}\)) + (\(\frac{y^{\prime}+3}{2}\) – 3)² = 0
\(\frac{x^{\prime 2}}{4}\) + 2x’ + \(\frac{y^{\prime 2}-6 y^{\prime}+9}{4}\) = 0
x’² + 8x’ + y’² – 6y’ + 9 = 0
Lotus of M(x’, y’) is x² + y² + 8x – 6y + 9 = 0, which is a circle.

Question 32.
If the circle x² + y² + ax + by – 12 = 0 has the centre at (2, 3) then find a, b and the radius of the circle.
Solution:
Equation of the circle is
x² + y² + ax + by – 12 = 0
Centre = (\(-\frac{a}{2}\), \(-\frac{b}{2}\)) = (2, 3)
\(-\frac{a}{2}\) = 2, \(-\frac{b}{2}\) = 3
a = – 4, b – -6
g = -2, f = -3, c = -12
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{4+9+12}\) = 5

Question 33.
If the circle x² + y² – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is x² + y² – 4x + 6y + a = 0
2g = – 4, 2f = 6, c = a
g =-2, f = 3, c = a
radius = 4 ⇒ \(\sqrt{g^{2}+f^{2}-c}\) = 4
\(\sqrt{4+9-a}\) = 4
13 – a = 16
a = 13 – 16 = -3

Question 34.
Find the equation of the circle passing through (4, 1), (6, 5) and having the centre on the line 4x + y – 16 = 0.
Solution:
Let the equation of the required circle be x² + y² + 2gx + 2fy + c = 0 .
This circle passes through A(4, 1)
16 + 1+ 8g + 2f + c = 0
8g + 2f + c = -17 ………………. (1)
The circle passes through B(6, 5)
36 + 25 + 12g + 10f + c = 0
12g + 10f + c = -61 ……………… (2)
The centre (-g, -f) lies on 4x + y – 16 = 0
– 4g – f – 16 = 0
4g + f + 16 = 0 ……………….. (3)
(2) – (1) gives 4g + 8f = -44 ……………….. (4)
4g + f = -16 …………….. (3)
7f = -28
f = \(\frac{-28}{7}\) = -4
From(3) 4g – 4 = -16
4g = -12 ⇒ g = -3
From(1) 8(-3) + 2(-4) + c = -17
c = -17 + 24 + 8 = 15
Equation of the required circle is
x² + y² – 6x – 8y + 15 = 0

Question 35.
Suppose a point (x₁, y₁) satisfies x² + y²+ 2gx + 2fy + c = 0 then show that it represents a circle whenever g, f and c are real.
Solution:
Comparing with the general equation of second degree co-efficient of x² = coefficient of y² and coefficient of xy = 0
The given equation represents a circle if g² + f² – c ≥ 0
(x₁, y₁) is a point on the given equation
x² + y² + 2gx + 2fy + c = 0, we have
x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0
g² + f² – c = g² + f² + x₁² + y₁² + 2gx₁ + 2fy₁ = 0
= (x₁ + g)² + (y₁ + f)² ≥ 0
g, f and c are real
∴ The given equation represents a circle.

Question 36.
Find the equation of the circle whose extremities of a diameter are (1, 2) and (4, 5).
Solution:
Here (x₁, y₁) = (1, 2) and (x₂, y₂) = (4, 5)
Equation of the required circle is
(x – 1) (x – 4) + (y – 2) ( y – 5) = 0
x² – 5x + 4 + y² – 7y + 10 = 0
x² + y² – 5x – 7y + 14 = 0

Question 37.
Find the other end of the diameter of the circle x² + y² – 8x – 8y + 27 = 0 if one end of it is (2, 3).
Solution:
A = (2, 3) and AB is the diameter of the circle x² + y² – 8x – 8y + 27 = 0

Centre of the circle is C = (4, 4)
Suppose B(x, y) is the other end
C = mid point of AB = \(\left(\frac{2+x}{2}, \frac{3+y}{2}\right)\) = (4, 4)
\(\frac{2+x}{2}\) = 4
2 + x = 8
x = 6
\(\frac{3+y}{2}\) = 4
3 + y = 8
y = 5
The other end of the diameter is B(6, 5)

Question 38.
Find the equation of the circum – circle of the traingle formed by the line ax + by + c = 0 (abc ≠ 0) and the co-ordinate axes.
Solution:
Let the line ax + by + c = 0 cut X, Y axes at A and B respectively co-ordinates of O are (0, 0) A are

Suppose the equation of the required circle is x² + y² + 2gx + 2fy + c = 0
This circle passes through 0(0, 0)
∴ c = 0
This circle passes through A(-\(\frac{c}{a}\), 0)
\(\frac{c^{2}}{a^{2}}\) + 0 – 2\(\frac{\mathrm{gc}}{\mathrm{a}}\) = 0
2g . \(\frac{c}{a}\) = \(\frac{c^{2}}{a^{2}}\) ⇒ 2g = \(\frac{c}{a}\) ⇒ g = \(\frac{c}{2a}\)
The circle passes through B (0, –\(\frac{c}{b}\))
0 + \(\frac{c^{2}}{b^{2}}\) + 0 – 2g \(\frac{c}{b}\) = 0
2f\(\frac{c}{b}\) = \(\frac{c^{2}}{b^{2}}\) ⇒ 2g = \(\frac{c}{b}\) ⇒ f = \(\frac{c}{2b}\)
Equation of the circle, through O, A, B is
x² + y² + \(\frac{c}{a}\) x + \(\frac{c}{b}\) y = 0
ab(x² + y²) + (bx + ay) = 0
This is the equation of the circum circle of ∆OAB

Question 39.
Find the equation of the circle which passes through the vertices of the triangle formed by L₁ = x + y + 1 = 0; L₂ = 3x + y- 5 = 0 and L₃ = 2x + y – 5 = 0.
Solution:
Suppose L₁, L₂,: L₂, L₃ and L₃, L₁ intersect in A, B and C respectively.
Consider a curve whose equation is
k (x + y + 1) (3x + y – 5) + l(3x + y – 5) (2x + y – 5) + m(2x + y – 5) (x + y + 1) = 0 ………………. (1)
This equation represents a circle
i) Co-efficient of x² = Co – efficient of y²
3k + 6l + 2m = k + l + m
2k + 5l + m = 0 ……………….. (2)
ii) Co-efficient of xy = 0
4k + 5l + 3m = 0 ……………….. (3).
Applying cross multiplication rule for (2) and (3) we get

Substituting in (1), equation of the required circle is
5(x + y + 1) (3x + y – 5) – 1 (3x + y – 5)
(2x + y – 5) – 5(2x + y – 5) (x + y + 1) = 0
i.e., x² + y² – 30x – 10y + 25 = 0

Question 40.
Find the centre of the circle passing through the points (0, 0), (2, 0)and (0, 2).
Solution:
Here (x₁, y₁) = (0, 0); (x₂, y₁) = (2, 0);
(x₃, y₃) = (0, 2)
c₁ = -(x₁² + y₁²) = 0
c₂ = – (x₂² + y₂²) = -(2² + 0²) = -4
c₃ = -(x₃² + y₃²) = -(0² + 2²) – 4
The centre of the circle passing through three non-collinear points P(x₁, y₁), Q(x₂, y₂) and R(x₃, y₃)

Thus the centre of the required circle is (1, 1)

Question 41.
Obtain the parametric equations of the circle x² + y² = 1.
Solution:
Equation of the circle is x² + y² = 1
Centre is (0, 0) radius = r = T

The circle having radius r is x = r cos θ,
y = sin θ where 0 < θ < 2π
The parametric equation of the circle
x² + y² = 1 and
x = 1 . cos θ = cos θ
y = 1 . sin θ = sin θ, θ < θ < 2π
Note: Every point on the circle can be expressed as (cos θ, sin θ)

Question 42.
Obtain the parametric equation of the circle represented by
x² + y² + 6x + 8y – 96 = 0.
Solution:
Centre (h, k) of the circle is (-3, -4)
radius = r = \(\sqrt{9+16+96}\) = \(\sqrt{121}\) = 11
Parametric equations are
x = h + r cos θ = -3 + 11 cos θ
y = k + r sin θ = -4 + 11 sin θ
where 0 < θ < 2π

Question 43.
Locate the position of the point (2, 4) with respect to the circle. x² + y² – 4x – 6y + 11 = 0.
Solution:
Here (x₁, y₁) = (2, 4) and
S ≡ x² + y² – 4x – 6y + 11
S₁₁ = 4 + 16 – 8 – 24 + 11
= 31 – 32 = – 1 < 0
∴ The point (2, 4) lies inside the circle S = 0

Question 44.
Find the length of the tangent from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0.
Solution:
Here (x₁, y₁) = (1, 3) and
S = x² + y² – 2x + 4y – 11 = 0
P(x₁, y₁) to S = 0 is \(\sqrt{S_{11}}\)
Length of the tangent = \(\sqrt{S_{11}}\)
= \(\sqrt{1+9-2+12-11}\) = \(\sqrt{9}\) = 3

Question 45.
If a point P is moving such that the length of tangents drawn from P to
x² + y² – 2x + 4y – 20 = 0 ……………… (1)
and x² + y² – 2x – 8y + 1 = 0 ……………….. (2)
are in the ratio 2 : 1.
Then show that the equation of the locus of P is x² + y² – 2x – 12y + 8 = 0.
Solution:
Let P(x₁, y₁) be any point on the locus and \(\overline{\mathrm{PT}_{1}}\), \(\overline{\mathrm{PT}_{2}}\) be the lengths of tangents from P to the circles (1) and (2) respectively.
x² + y² – 2x + 4y – 20 = 0 and
x² + y² – 2x – 8y + 1 = 0
\(\frac{\overline{\mathrm{PT}_{1}}}{\overline{\mathrm{PT}_{2}}}=\frac{2}{1}\)
i.e., \(\sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}+4 y_{1}-20}\)
= \(2 \sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}-8 y_{1}+1}\)
3 (x₁² + y₁²) – 6x₁ – 36y₁ + 24 = 0
Locus of P (x₁, y₁) is
x² + y² – 2x – 12y + 8 = 0

Question 46.
If S ≡ x² + y² + 2gx + 2fy + c = 0. represents a circle then show that the straight line lx + my + n = 0
i) touches the circle S = 0 if
(g² + f² – c) = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) meets the circle S = 0 in two points if
g² + f² – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) will not meet the circle if
g² + f² – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)
Solution:
Let ‘c’ be the centre and ‘r’ be the radius of the circle S = 0
Then C = (-g, -f) and r = \(\sqrt{g^{2}+f^{2}-c}\)
i) The given straight line touches the circle
if r = \(\frac{|l(-\mathrm{g})+\mathrm{m}(-\mathrm{f})-\mathrm{n}|}{\sqrt{l^{2}+\mathrm{m}^{2}}}\)
\(\sqrt{g^{2}+f^{2}-c}\) = \(\frac{|-(l g+m f-n)|}{\sqrt{l^{2}+m^{2}}}\)
squaring on both sides, we get
g² + f² – c = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) The given line lx + my + n = 0 meets the circle s = 0 in two points if
g² + f² – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) The given line lx + my + n = 0 will not meet the circle s = 0 if
g² + f² – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

Question 47.
Find the length of the chord intercepted by the circle x² + y² + 8x – 4y – 16 = 0 on the line 3x – y + 4 = 0.
Solution:
The centre of the given circle c = (-4, 2) and radius r = \(\sqrt{16+4+16}\) = 6. Let the perpendicular distance from the centre to the line 3x-y + 4 = 0 be ‘d’ then
d = \(\frac{|3(-4)-2+4|}{\sqrt{3^{2}+(-1)^{2}}}\) = \(\frac{10}{\sqrt{10}}\) = \(\sqrt{10}\)
Length of the chord + \(\sqrt{r^{2}-d^{2}}\)
= 2\(\sqrt{6^{2}-(\sqrt{10})^{2}}\) = 2\(\sqrt{26}\)

Question 48.
Find the equation of tangents to x² + y² – 4x + 6y – 12 = 0 which are parallel to x + 2y- 8 = 0.
Solution:
Here g = -2, f = 3, r = \(\sqrt{4+9+12}\) = 5
and the slope of the required tangent is \(\frac{-1}{2}\)
The equations of tangents are
y + 3 = \(\frac{-1}{2}\) (x – 2) ± 5 \(\sqrt{1+\frac{1}{4}}\)
2(y + 3) = – x + 2 ± 5\(\sqrt{5}\)
x + 2y + (4 ± 5\(\sqrt{5}\)) = 0

Question 49.
Show that the circle S ≡ x² + y² + 2gx + 2fy + c = 0 touches the
i) X – axis if g² = c
ii) Y – axis if f² = c.
Solution:
i) We know that the intercept made by S = 0 on X – axis is 2\(\sqrt{g^{2}-c}\)
If the circle touches the X – axis then
22\(\sqrt{g^{2}-c}\) 0 ⇒ g² = c

Question 50.
Find the equation of the tangent to x² + y² – 6x + 4y – 12 = 0 at (- 1, 1).
Solution:
Here (x₁, y₁) = (-1, 1) and
S ≡ x² + y² – 6x + 4y – 12 = 0
∴ The equation of the tangent is
x(-1) + y (1) – 3(x – 1) + 2(y + 1)- 12 = 0
The equation of the tangent at the point P(₁, y₁) to the circle
S ≡ x² + y² + 2gx + 2fy + c = 0 is S₁ = 0
⇒ – x + y – 3x + 3 + 2y + 2 – 12 = 0
⇒ – 4x + 3y – 7 = 0
(or) 4x – 3y + 7 = 0

Question 51.
Find the equation of the tangent to x² + y² – 2x + 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
Solution:
Here (x₁, y₁) = (3, -1) and
S ≡ x² + y² – 2x + 4y = 0
The equation of the tangent at (3, -1) is
x (3) + y (-1) – (x + 3) + 2(y – 1) = 0
3x – y – x – 3 + 2y – 2 = 0
2x + y – 5 = 0
Slope of the tangent is m = -2, for the circle
g = -1, f = 2, c = 0
r = \(\sqrt{1+4-0}\) = \(\sqrt{5}\)
Equations of the tangents are
y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x² + y² = r² and the slope of the tangent is m.
y + 2 = -2(x – 1) ± \(\sqrt{5} \sqrt{1+4}\)
y + 2 = -2x + 2 ± 5
2x 4- y = ± 5
The tangents are
2x + y + 5 = 0 and 2x + y – 5 = 0
The tangent parallel to the given tangent is 2x + y ± 5 = 0

Question 52.
If 4x – 3y + 7 = 0 is a tangent to the circle represented by x² + y² – 6x + 4y – 12 = 0, then find its point of contact.
Solution:
Let (x₁, y₁) be the point of contact
Equation of the tangent is
(x₁ + g) x + (y₁ + f) y + (gx₁ + fy₁ + c) = 0
We have \(\frac{x_{1}-3}{4}\) = \(\frac{y_{1}+2}{-3}\)
= \(\left(\frac{-3 x_{1}+2 y_{1}-12}{7}\right)\) ……………… (1)
From first and second equalities of (1), we get
3x₁ + 4y₁ = 1 ………………. (2)
Now by taking first and third equalities of (1), we get
19x₁ – 8y₁ = -27 ………………. (3)
Solving (2) and (3) we obtain
x₁ = -1;, y₁ = 1
Hence the point of contact is (-1, 1).

Question 53.
Find the equations of circles which touch 2x – 3y + 1 = 0 at (1, 1) and having radius \(\sqrt{13}\).
Solution:
The centres of the required circle lies on a line perpendicular to 2x – 3y + 1 =0 and passing through (1, 1)

The equation of the line of centre can be taken as
3x + 2y + k = 0
This line passes through (1, 1)
3 + 2 + k = 0 ⇒ k = -5
Equation of AB is 3x + 2y – 5 = 0
The centres A and B are situated on
3x + 2y – 5 = 0 at a distance \(\sqrt{13}\) from (1, 1).
The centres are given by
(x₁ ± r cos θ, y₁ ± r sin θ)
\(\left(1+\sqrt{13}\left(-\frac{2}{\sqrt{13}}\right) 1+\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\) and
\(\left(1-\sqrt{13} \frac{(-2)}{\sqrt{13}}, 1-\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\)
i.e., (1 -2, 1 +3) and (1 + 2, 1 – 3)
(-1, 4) and (3, -2)
Centre (3, -2), r = \(\sqrt{13}\)
Equation of the required circles are
(x + 1 )² + (y – 4)² = 13 and
(x – 3)² + (y + 2)² = 13
i.e., x² + y² + 2x – 8y + 4 = 0
and x² + y² – 6x + 4y = 0

Question 54.
Show that the line 5x + 12y – 4 = 0 touches the circle x² + y² – 6x + 4y + 12 = 0.
Solution:
Equation of the circle is
x² + y² – 6x + 4y + 12 = 0
Centre (3, -2), r = \(\sqrt{9+4-12}\) = 1 ……………….. (1)
The given line touches the circle if the perpendicular distance from the centre on the given line is equal to radius of the circle, d = perpendicular distance from (3, -2)
= \(\frac{|5(3)+12(-2)-4|}{\sqrt{25+144}}\)
= \(\frac{13}{13}\) = 1 = radius of the circle ………………… (2)
∴ The given line 5x + 12y – 4 = 0 touches the circle.

Question 55.
If the parametric values of two points A and B lying on the circle x² + y² – 6x + 4y – 12 = 0 are 30° and 60° respectively, then find the equation of the chord joining A and B.
Solution:
Here g = -3, f = -2; r = \(\sqrt{9+4-12}\) = 5
∴ The equation of the chord joining the points θ₁ = 30°, θ₂ = 60°
Equation of chord joining the point; (-g+ r cos θ₁(-f + r sin θ₁) where r is the radius of the circle; θ₂ and (-g + r cos θ₂, -f + r sin θ₂) is (x + g) cos (\(\frac{\theta_{1}+\theta_{2}}{2}\)) + (y + f)
sin (\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos (\(\frac{\theta_{1}+\theta_{2}}{2}\))
(\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos \(\frac{\theta_{1}+\theta_{2}}{2}\)
(x – 3) cos [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]
(y + 2) sin [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]
= 5 cos [latex]\frac{60^{\circ}-30^{\circ}}{2}[/latex]
i.e., (x – 3) cos 45 ° + (y + 2) sin 45°
= 5 cos 15°
= \(\frac{(x-3)+(y+2)}{\sqrt{2}}=5\left[\frac{(\sqrt{3}+1)}{2 \sqrt{2}}\right]\)
i.e., 2x + 2y – (7 + 5\(\sqrt{3}\)) = 0.