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Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e^2x dx, x ∈ R.
Solution:
∫e^2x dx = \(\frac{e^{2x}}{2}\) + C

Question 2.
∫sin 7x dx, x ∈ R.
Solution:
∫sin 7x dx = \(\frac{\cos 7x}{7}\) + C

Question 3.
∫\(\frac{x}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{xdx}{1+x^2}\) dx = \(\frac{1}{2}\)∫\(\frac{2xdx}{1+x^2}\) = \(\frac{1}{2}\) log(1+x²) + C

Question 4.
∫2xsin(x²+1) dx, x ∈ R.
Solution:
∫2x.sin(x²+1) dx, x ∈ R.
t = x² + 1 ⇒ dt = 2x dx
∫2x. sin(x²+1) dx = ∫ sin t dt = -cos t + C
= -cos (x²+1) + C

Question 5.
∫\(\frac{(logx)^2}{x}\) dx on I ⊂ (0, ∞).
Solution:
∫\(\frac{(logx)^2}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{(logx)^2}{x}\) dx = ∫t² dt
= \(\frac{(t^3}{3}\) + C = \(\frac{(logx)^3}{3}\) + C

Question 6.
dx on I ⊂ (1, ∞).
Solution:

Question 7.
∫\(\frac{\sin(Tan^{-1}x)}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\) dx
t = tan^-1 x ⇒ dt = \(\frac{dx}{1+x^2}\)
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\)dx = ∫ sin t
= -cos t + t
= -cos (tan^-1 x) + C

Question 8.
∫\(\frac{1}{8+2x^2}\) dx on R.
Solution:

Question 9.
∫\(\frac{3x^2}{1+x^6}\)dx, on R.
Solution:

Question 10.
∫\(\frac{2}{\sqrt{25+9x^2}}\)dx on R.
Solution:

Question 11.
∫\(\frac{3}{\sqrt{9x^2-1}}\)dx on (\(\frac{1}{3}\), ∞).
Solution:

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx cos nx = \(\frac{1}{2}\)∫ 2 sin mx. cos nx dx
= \(\frac{1}{2}\)∫sin(m+n)x + sin(m-n)x) dx
= –\(\frac{1}{2}\)(\(\frac{\cos (m+n) x}{m+n}+ cos\frac{(m-n)x}{m-n}\)) + C

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx. sin nx dx = \(\frac{1}{2}\) ∫2 sin mx. sin nx dx
= \(\frac{1}{2}\)∫cos (m – n)x – cos (m + n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m-n) x}{m-n}+\frac{\sin (m+n)x}{m+n}\)) + C

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫cos mx. cos nx dx = \(\frac{1}{2}\) ∫2 cos mx.cos nx dx
= \(\frac{1}{2}\)∫cos (m + n)x – cos (m – n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n)x}{m-n}\)) + C

Question 15.
∫sin x sin 2x. sin 3x dx on R.
Solution:
sin 2x . sin 3x = \(\frac{1}{2}\)(2 sin 3x. sin 2x)
= \(\frac{1}{2}\)(cos x – cos 5x)
sin x sin 2x sin 3x
= \(\frac{1}{2}\)(sin x . cos x – cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1.2}{2}\)cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1}{2}\) – (sin 6x – sin 4x)

Question 16.
∫\(\frac{\sin x}{\sin(a+x)}\) dx on I ⊂ R\{nπ – a : n ∈ Z}.
Solution:
sin x = sin (a + x – a)
= sin (a + x) . cos a – cos (a + x) sin a
∫\(\frac{\sin x}{\sin(a+x)}\)dx
= cos a ∫ dx – sin a ∫\(\frac{\cos (a+x)}{\sin(a+x)}\) dx
= x . cos a – sin a . log |sin (a + x)| + c

II. Evaluate the following integrals.

Question 1.
∫(3x – 2)^½ dx on (\(\frac{2}{3}\), ∞)
Solution:
t = 3x – 2 ⇒ dt = 3 dx
∫(3x – 2)^½ dx = \(\frac{1}{3}\) ∫t^½ dt = \(\frac{1}{3}\) \(\frac{t^{3/2}}{3/2}\) + C
= \(\frac{2}{9}\)(3x – 2)^3/2 + C

Question 2.
∫\(\frac{1}{7x+3}\) dx on I ⊂ R\{-\(\frac{3}{7}\)}.
Solution:
∫\(\frac{1}{7x+3}\) dx
t = 7x + 3
⇒ dt = 7 dx
= ∫\(\frac{1}{7x+3}\) dx = \(\frac{1}{7}\) ∫\(\frac{dt}{t}\)
= \(\frac{1}{7}\) log |t| + C = \(\frac{1}{7}\) log|7x + 3| + C

Question 3.
∫\(\frac{log(1+x)}{1+x}\) dx on (-1, ∞).
Solution:
∫\(\frac{log(1+x)}{1+x}\) dx
t = 1 + x ⇒ dt = dx

Question 4.
∫(3x² – 4)x dx on R.
Solution:
∫(3x² – 4)x dx
t = 3x² – 4 ⇒ dt = 6x dx
∫(3x² – 4)x dx = \(\frac{1}{6}\)∫t dt = \(\frac{1}{6}\).\(\frac{t^2}{2}\) + C
= \(\frac{(3x^2-4)^2}{12}\) + C

Question 5.
∫\(\frac{dx}{\sqrt{1+5x}}\) dx on (-\(\frac{1}{5}\), ∞).
Solution:

Question 6.
∫(1 – 2x³)x² dx on R.
Solution:
∫(1 – 2x³)x² dx
t = 1 – 2x³ ⇒ dt = -6x² dx
∫(1 – 2x³)x² dx = –\(\frac{1}{6}\)∫t dt
= –\(\frac{1}{6}\) . \(\frac{t^2}{2}\) + C
= \(\frac{-(1-2x^3)^2}{12}\) + C

Question 7.
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx on I ⊂ R \ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx
t = 1 + tan x ⇒ dt = sec² x dx

Question 8.
∫x³ sinx⁴ dx on R.
Solution:
∫x³ . sinx⁴ dx
t = x⁴ ⇒ dt = 4x³ dx
∫x³ . sinx⁴ dx = \(\frac{1}{4}\)∫sin t. dt
= –\(\frac{1}{4}\)cos t + C
= –\(\frac{1}{4}\). cos x⁴ + C

Question 9.
∫\(\frac{\cos x}{(1+sin x)^2}\) dx on I ⊂ R\{2nπ + \(\frac{3 \pi}{2}\) : n ∈ Z}.
Solution:

Question 10.
∫\(\sqrt[3]{sin x}\) cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:

Question 11.
∫ 2x e^x² dx on R.
Solution:
∫ 2x. e^x² dx
t = x² ⇒ dt = 2x dx
∫ 2xe^x² dt = ∫ e^t dt = e^t + C
= e^x² + C

Question 12.
∫\(\frac{e^{log x}}{x}\) dx on (0, ∞).
Solution:
∫\(\frac{e^{log x}}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\).dx
∫\(\frac{e^{log x}}{x}\) dx = ∫e^t. dt
= e^t + C
= e^{log x} + C
= x + C

Question 13.
∫\(\frac{x^2}{\sqrt{1-x^6}}\) dx on I = (-1, 1).
Solution:

Question 14.
∫\(\frac{2x^3}{1+x^8}\) dx on R.
Solution:
t = x⁴ ⇒ dt = 4x³ dx
∫\(\frac{2x^3}{1+x^8}\) = \(\frac{1}{2}\)∫\(\frac{dt}{1+t^2}\)
= \(\frac{1}{2}\) tan^-1 t + C
= \(\frac{1}{2}\) tan^-1(x⁴) + C

Question 15.
∫\(\frac{x^8}{1+x^18}\) dx on R.
Solution:
t = x⁹ ⇒ dt = 9x⁸ dx

Question 16.
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx on I ⊂ R/{x ∈ R : cos (xe^x) = 0}.
Solution:
t = x . e^x
dt = (x . e^x + e^x)dx = e^x(1 + x)dx
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx = ∫\(\frac{dt}{\cos^2 t}\) dx
= ∫sin² t dt
= tan t + C
= tan (x. e^x) = C

Question 17.
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx on I ⊂ R\ {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
Let t = a + b cot x
dt = -b cosec² x dx
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx = –\(\frac{1}{b}\) ∫\(\frac{bt}{t^5}\)
= –\(\frac{1}{b}\) ∫t^-5 dt
= –\(\frac{1}{b}\) \(\frac{t^{-4}}{-4}\) + C
= \(\frac{1}{4b t^{4}}\) + C
= \(\frac{1}{4b(a+b cotx)^{4}}\) + C

Question 18.
∫e^x sin e^x dx on R.
Solution:
t = e^x ⇒ dt = e^x dx
∫e^x . sin e^x dx = ∫ sin t dt
= -cos t + C
= -cos (e^x) + C

Question 19.
∫\(\frac{\sin(log x)}{x}\) dx on (-1, ∞).
Solution:
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{\sin(log x)}{x}\)dx = ∫sin t dt
= -cos t + C
= -cos (log x) + C

Question 20.
∫\(\frac{1}{x log x}\) dx on (-1, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) . dx
∫\(\frac{1}{x log x}\) dx = ∫\(\frac{1}{t}\)dt = log t + C = log(log x + C)

Question 21.
∫\(\frac{(1+log x)^n}{x}\) dx on (e^-1, ∞). n ≠ -1.
Solution:
t = 1 + log x
dt = \(\frac{1}{x}\) dx

Question 22.
∫\(\frac{\cos(log x)}{x}\) dx on (0, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) dx
∫\(\frac{\cos(log x)dx}{x}\) = ∫cos t dt
= sin t + C
= sin (log x) + C

Question 23.
∫\(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx on (0, ∞).
Solution:

Question 24.
∫\(\frac{2x+1}{x^2+x+1}\) dx on R.
Solution:

Question 25.
∫\(\frac{ax^{n-1}}{bx^n+c}\) dx, where n ∈ N, a, b, c are real nembers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xⁿ ≠ –\(\frac{c}{b}\)}.
Solution:
t = bxⁿ + C
dt = nbx^n-1dx

Question 26.
∫\(\frac{1}{x log x[log(log x)]}\) dx on (1, ∞).
Solution:

Question 27.
∫cot hx dx on R.
Solution:
t = sinh x ⇒ dt = cosh x dx
∫cot hx dx = ∫\(\frac{dt}{t}\)
= log |t| +C
= log |log(log x)| + C

Question 28.
∫\(\frac{1}{\sqrt{1-4x^2}}\) dx on (-\(\frac{1}{2}\), \(\frac{1}{2}\)).
Solution:

Question 29.
∫\(\frac{dx}{\sqrt{25+x^2}}\) dx on R.
Solution:

Question 30.
∫\(\frac{1}{(x+3\sqrt{x+2}}\) dx on I ⊂ (-2, ∞).
Solution:
x + 2 = t²
dx = 2t dt

Question 31.
∫\(\frac{1}{1+\sin 2x}\) dx on I ⊂ R\{\(\frac{n \pi}{2}\) + (-1)ⁿ \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:

Question 32.
∫\(\frac{x^2+1}{x^4+1}\) dx on R.
Solution:
∫\(\frac{x^2+1}{x^4+1}\) dx
Dividing Nr and Dr by x²

Question 33.
∫\(\frac{dx}{\cos^2+\sin 2x}\) on I ⊂ R\({(2n+1)\(\frac{\pi}{2}\) : n ∈ Z}∪{2nπ + tan^-1 \(\frac{1}{2}\) : n ∈ Z})
Solution:
∫\(\frac{dx}{\cos^2+\sin 2x}\)
Dividing Nr and Dr by cos² x

Question 34.
∫\(\sqrt{1-sin 2x}\) dx on I ⊂ {2nπ – \(\frac{3 \pi}{4}\), 2nπ + \(\frac{\pi}{4}\)}, n ∈ Z.
Solution:

Question 35.
∫\(\sqrt{1+cos 2x}\) dx I ⊂ {2nπ – \(\frac{\pi}{2}\), 2nπ + \(\frac{\pi}{2}\)}, n ∈ Z.
Solution:
∫\(\sqrt{1+cos 2x}\) dx = ∫\(\sqrt{2cos^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

Question 36.
∫\(\frac{\cos x + \sin x}{\sqrt{1+\sin 2x}}\) dx on I ⊂ {2nπ – \(\frac{\pi}{4}\), 2nπ + \(\frac{3\pi}{4}\)}, n ∈ Z.
Solution:

Question 37.
∫\(\frac{\sin 2x}{(a+b\cos x)^2}\) dx on {R, if |a| > |b| I ⊂ {x ∈ R : a + b cos x ≠ x}, if |a| < |b|.
Solution:
Put a + b cos x = t ⇒ cos x = \(\frac{t-a}{b}\)
Then b(-sin x)dx = dt
⇒ sin dx = \(\frac{-1}{b}\) dt

Question 38.
∫\(\frac{\sec x}{(\sec x+\tan x)^2}\) dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
Put sec x + tan x = t
Then (sec x tan x + sec² x) dx = dt
sec x (sec x + tan x) dx = dt

Question 39.
∫\(\frac{dx}{a^2\sec^2 x+b^2\cos^2 x}\) on R, a ≠ 0, b ≠ 0.
Solution:

Question 40.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\({a + nπ : n ∈ Z} ∪ {b + nπ n ∈ Z}).
Solution:

Question 41.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\ ({a + \(\frac{(2n+1)\pi}{2}\) : n ∈ Z} ∪ {b + \(\frac{(2n+1)}{2}\)π : n ∈ Z}).
Solution:

III. Evaluate the following integers.

Question 1.
∫\(\frac{\sin 2x}{a \cos^2 x + b \sin^2 x}\) dx on I ⊂ R\ {x ∈ R |a cos² x + b sin² x = 0}.
Solution:
t = a cos² x + b sin² x
dt = (a(2cos x)(-sin x) + b(2sin x cos x))dx

Question 2.
∫\(\frac{1-\tan x}{1+\tan x}\) dx for x ∈ I ⊂ R\ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:

Question 3.
∫\(\frac{\cot(log x)}{x}\) dx, x ∈ I ⊂ (0, ∞)\{e^nπ : n ∈ Z}.
Solution:
t = log x ⇒ dt = \(\frac{dx}{x}\)
∫\(\frac{1-\tan x}{1+\tan x}\) dx = ∫cot t dt
= log(sin t) + C
= log |sin (log x)| + C

Question 4.
∫e^x . cot e^x dx, x ∈ I ⊂ R\ {log n π : n ∈ Z}.
Solution:
t = e^x ⇒ dt = e^x dx
∫e^x . cot e^x dx = ∫cot t dt
= log (sin t) + C
= log |sin(log x)| + C

Question 5.
∫sec (tan x) sec² x dx, on I ⊂ {x ∈ E : tan x ≠ \(\frac{(2k+1)\pi}{2}\) for any k ∈ Z}. where E = R / {\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫sec(tan x) sec² x dx = ∫sec t. dt
= log tan (\(\frac{\pi}{4}+\frac{t}{2}\) + C
= log (tan(\(\frac{\pi}{4}+\frac{\tan x}{2}\))) + C

Question 6.
∫\(\sqrt{sin x}\)cos dx on {2nπ, (2n + 1)π}, (n ∈ Z).
Solution:
t = sin x ⇒ dt = cos x dx
∫\(\sqrt{sin x}\)cos dx = ∫√t dt
= \(\frac{2}{3}\) = t^3/2 + C
= \(\frac{2}{3}\) = (sin x)^3/2 + C

Question 7.
∫tan⁴ x sec² x dx, x ∈ I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫tan⁴ x sec² x dx = ∫t⁴dt
= \(\frac{t^5}{5}\) + C = \(\frac{(\tan x)^2}{5}\) + C

Question 8.
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) dx, x ∈ T ⊂ \{-4, 1}
Solution:
t = x² + 3x – 4
dt = (2x + 3)dx
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) = ∫\(\frac{dt}{\sqrt{t}}\)
= 2√t + C
= \(\sqrt{x^2+3x-4}\) + C

Question 9.
∫cosec² x\(\sqrt{\cot x}\) dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = cot x ⇒ dt = -cosec² x dx
∫cosec² x\(\sqrt{\cot x}\) dx = -∫√t dt
= –\(\frac{2}{3}\) t√t + C
= –\(\frac{2}{3}\) cot(x)^3/2 + C

Question 10.
∫sec x log(sec x + tan x)dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = log(sec x + tanx)
dt = \(\frac{(\sec x.\tan x+\sec^2 x)}{(\sec x +\tan x}\) = sec x dx
∫sec x .log(sec x + tan x)dx = ∫t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{(log(\sec x+\tan x)^2}{2}\) + C

Question 11.
∫sin³ x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin³ x
sin³ x = \(\frac{1}{4}\)(3 sin x – sin 3x)
∫sin³ x dx = \(\frac{3}{4}\)∫sin x – \(\frac{1}{4}\)∫sin 3x dx
= –\(\frac{3}{4}\) cos x + \(\frac{1}{12}\) cos 3x + C
= \(\frac{1}{12}\)(cos 3x – 9 cos x) + C

Question 12.
∫cos³ x dx on R.
Solution:
cos 3x = 4 cos³ x – 3 cos x
∫cos³ x dx = \(\frac{3}{4}\)∫cos x + \(\frac{1}{4}\)∫cos 3x dx
= –\(\frac{3}{4}\) sin x + \(\frac{1}{12}\) sin 3x + C
= \(\frac{1}{12}\)(9 sin x + 9 sin 3x) + C

Question 13.
∫cos x cos 2x dx on R.
Solution:
cos 2x cos x = \(\frac{1}{2}\)(2cos 2x cos x)
∫cos x cos 2x dx = \(\frac{1}{2}\)∫(cos 3x + cos x) dx
= \(\frac{1}{2}\)∫cos 3x + \(\frac{1}{2}\)∫cos x
= \(\frac{1}{2}\)(\(\frac{\sin 3x}{3}\) + sin x) + C
= \(\frac{\sin 3x+3\sin x}{6}\) + C

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos 3x cos x = \(\frac{1}{2}\)(2cos 3x . cos x)
= \(\frac{1}{2}\)(cos 4x + cos 2x)
∫cos x cos 3x dx = \(\frac{1}{2}\)∫cos 4x dx + \(\frac{1}{2}\)∫cos 2x dx
= \(\frac{1}{2}\)(\(\frac{\sin 4x}{4}+\frac{\sin 2x}{2}\) + C
= \(\frac{1}{8}\)(sin 4x + 2sin 2x) + C

Question 15.
∫cos⁴ x dx on R.
Solution:
cos⁴ x = (cos² x)² = (\(\frac{1+\cos 2x}{2}\))²
= \(\frac{1}{4}\) (1 + 2 cos 2x+ cos² 2x)
= \(\frac{1}{4}\) (1 + 2 cos 2x + = \(\frac{1+\cos 4x}{2}\))
= \(\frac{1}{8}\) (2 + 4 cos 2x + 1 + cos 4x)
= \(\frac{1}{8}\) (3 + 4 cos 2x + cos 4x)
= \(\frac{1}{8}\)(3∫dx + 4∫cos 2x dx + ∫cos 4x dx)
= \(\frac{1}{8}\)(3x + 4\(\frac{\sin 2x}{2}+\frac{\sin 4x}{4}\)) + C
= \(\frac{1}{32}\)(12x + 8 sin 2x + sin 4x) + C

Question 16.
∫x \(\sqrt{4x+3}\) dx on (-\(\frac{3}{4}\), ∞).
Solution:

Question 17.
∫\(\frac{dx}{\sqrt{a^2-(b+cx)^2}}\) on {x ∈ R : |b + cx| < a}. where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:

Question 18.
∫\(\frac{dx}{a^2+(b+cx)^2}\) on R, a, b, c are real numbers, c ≠ 0 and a > 0.
Solution:

Question 19.
∫\(\frac{dx}{1+e^x}\), x ∈ R
Solution:

Question 20.
∫\(\frac{x^2}{(a+bx)^x}\) dx, x ∈ I ⊂ R\{-\(\frac{a}{b}\)}, where a, b are real nembers, b ≠ 0.
Solution:
Put t = a + bx
dt = b dx ⇒ dx = \(\frac{1}{b}\). dt

Question 21.
∫\(\frac{x^2}{\sqrt{1-x}}\) dx, x ∈ (-∞, 1).
Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(a)

I. Evaluate the following integrals.

Question 1.
∫(x³ – 2x² + 3) dx on R.
Solution:
∫(x³ – 2x² + 3) dx = \(\frac{x^4}{4}-\frac{2}{3}\)x³ + 3x + c

Question 2.
∫2x√x dx on (0, ∞).
Solution:
∫2x√x dx = 2 ∫ x^3/2 dx = \(\frac{2x^{5/2}}{5/2}\)
= \(\frac{4}{5}\)x^5/2 + c

Question 3.
∫\(\sqrt[3]{2 x^2}\) dx’on (0, ∞).
Solution:
∫\(\sqrt[3]{2 x^2}\) dx = ∫ 2^1/3. x^2/3 dx
= 2^1/3. \(\frac{x^{5/3}}{5/3}\) + c
= \(\sqrt[3]{2}\).\(\frac{3}{5}\)x^5/3 + c

Question 4.
∫\(\frac{x^2+3x-1}{2x}\)dx, x ∈ I ⊂ R\{0}.
Solution:

Question 5.
∫\(\frac{1-\sqrt{x}}{x}\)dx on (0, ∞).
Solution:

Question 6.
∫(\(1+\frac{2}{x}-\frac{3}{x^2}\)) dx on I⊂R\{0}
Solution:

Question 7.
∫(x + \(\frac{4}{1+x^2}\))dx on R.
Solution:

Question 8.
∫(e^x \(\frac{1}{x}-\frac{2}{\sqrt{x^2+1}}\))dx on I⊂R\[-1, 1].
Solution:

Question 9.
∫(\(\frac{1}{1-x^2}+\frac{1}{1+x^2}\))dx on (-1, 1).
Solution:

Question 10.
∫(\(\frac{1}{1-x^2}+\frac{2}{1+x^2}\))dx on (-1, 1).
Solution:

Question 11.
∫e^{log(1+tan²x)} dx on I ⊂ R \{\(\frac{(2n+1)\pi}{2}\):n ∈ Z}
Solution:
∫e^{log(1+tan²x)} dx = ∫e^log(sec²x) dx
= ∫sec²x dx = tan x + c

Question 12.
∫\(\frac{\sin^{2}x}{1+\cos2x}\) dx on I ⊂ R \{(2n ± 1)π : n ∈ Z}
Solution:

II. Evaluate the following intergrals.

Question 1.
∫(1 – x²)³ dx on (-1, 1).
Solution:
∫(1 – x²)³ dx = ∫(1 – 3x² + 3x⁴ – x⁶)dx
= x – x³ + \(\frac{3}{5}\)x⁵ – \(\frac{x^7}{7}\) + c

Question 2.
∫(\(\frac{3}{\sqrt{x}}-\frac{2}{x}+\frac{1}{3x^2}\)) dx on (0, ∞).
Solution:

Question 3.
∫(\(\frac{\sqrt{x}+1}{x}\))² dx on (0, ∞).
Solution:

Question 4.
∫(\(\frac{(3x+1)^2}{2x}\)) dx, x ∈ I ⊂ R\ {0}.
Solution:

Question 5.
∫(\(\frac{2x-1}{3\sqrt{x}}\))² dx on (0, ∞).
Solution:

Question 6.
∫(\(\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x^2-1}}-\frac{3}{2x^2}\))² dx on (0, ∞).
Solution:

Question 7.
∫(sec² x – cos x + x²) dx, x ∈ I ⊂ R/{\(\frac{n \pi}{2}\) : n is an odd integer}.
Solution:
∫(sec² x – cos x + x²) dx
= ∫sec² x dx – ∫cos x + ∫x² dx
= tan x – sin x + \(\frac{x^3}{3}\) + C

Question 8.
∫(sec x tan x + \(\frac{3}{x}\) – 4) dx, x ∈ I ⊂ R\ ({\(\frac{n \pi}{2}\) : n is an odd integer} ∪ {0}).
Solution:
∫(sec x tan x + \(\frac{3}{x}\) – 4) dx
= sec x tan x dx + 3∫\(\frac{dx}{x}\) – 4 ∫dx
= sec x + 3 log |x| – 4x + c

Question 9.
∫(√x – \(\frac{2}{1-x^2}\)) dx on (0, 1).
Solution:

Question 10.
∫(x³ – cos x + \(\frac{4}{\sqrt{x^2+1}}\)) dx
Solution:
∫(x³ – cos x + \(\frac{4}{\sqrt{x^2+1}}\)) dx
= ∫ x³ dx – ∫cos x dx + 4 ∫\(\frac{dx}{\sqrt{x^2+1}}\)
= \(\frac{x^4}{4}\) – sin x + 4 sinh^-1 x + C

Question 11.
∫(cosh x + \(\frac{1}{\sqrt{x^2+1}}\))dx, x ∈ R.
Solution:
∫(cosh x + \(\frac{1}{\sqrt{x^2+1}}\))dx
= ∫cosh x dx + ∫\(\frac{dx}{\sqrt{x^2+1}}\)
= sinh x + sinh^-1 x + c

Question 12.
∫(sinh x + \(\frac{1}{(x^2-1)^{1/2}}\)) dx, x ∈ I ⊂ (-∞, -1) ∪ (1, ∞).
Solution:
∫(sinh x + \(\frac{1}{(x^2 – 1)^{1/2}}\)) dx
= ∫sinh x dx + ∫\(\frac{dx}{\sqrt{x^2-1}}\)
= cosh x + log(x + \(\sqrt{x^2-1}\)) + C

Question 13.
∫\(\frac{a^{x}-b^{x}}{a^{x}b^{x}}\) dx (a > 0, a ≠ 1 and b > 0, b ≠ 1) on R.
Solution:

Question 14.
∫sec² x cosec² x dx on I ⊂ R\ (nπ : n ∈ Z} ∪ { (2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}).
Solution:
∫sec² x cosec² x dx

= ∫\(\frac{1}{\cos^{2}x}\)dx + ∫\(\frac{1}{(\sin^{2}x}\)dx
= ∫sec² x dx + ∫cosec² x dx
= tan x – cot x + C

Question 15.
∫\(\frac{1+\cos^{2}x}{1+\cos2x}\) dx on I ⊂ R\{nπ :n ∈ Z}
Solution:

Question 16.
∫\(\sqrt{1-cos2x}\)dx on I ⊂ [2nπ, (2n + 1)π], n ∈ Z.
Solution:

Question 17.
∫\(\frac{1}{\cosh x+\sinh x}\) dx on R.
Solution:
∫\(\frac{1}{\cosh x+\sinh x}\) dx
= ∫\(\frac{\cosh x-\sinh x}{\cosh^{2}x-\sinh^{2}x}\) dx
= ∫(cosh x – sinh x) dx
= sinh x – cosh x + C

Question 18.
∫\(\frac{1}{1+\cos x}\) dx on I ⊂ R \{(2n + 1)π : n ∈ Z}.
Solution:

= ∫cosec² (x) dx – ∫cosec x cot x dx
= -cot x + cosec x + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a)

I.

Question 1.
One focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is \(\frac{3}{2}\).
Solution:
S(1 – 3) is the focus, equation of the directrix is y – 2 = 0

P(x₁, y₁) is any point on the hyperbola join SP and draw PM perpendicular to the directrix S.P = e. PM ⇒ SP² = e². PM²
(x₁ – 1)² + (y₁ + 3)² = \(\frac{9}{4}\)|\(\frac{(y_1-2}{\sqrt{1+0}}\)|²
x²₁ + 1 – 2x₁ + y²₁ + 9 + 6y₁ = \(\frac{9}{4}\) (y₁ – 2)²
4x²₁ + 4y²₁ – 8x₁ + 24y₁ + 40 = 9(y²₁ + 4 – 4y₁) = 9y²₁ – 36y₁ + 36
4x²₁ – 5y²₁ – 8x₁ + 60y₁ + 4 = 0
Focus of P(x², y²) is
4x² – 5y² – 8x + 60y + 4 = 0
This is the equation of the required hyperbola.

Question 2.
If the lines 3x – 4y = 12 and 3x + 4y = 12 meets on a hyperbola S = 0 then find the eccentricity of the hyperbola S = 0.
Solution:
The given lines as
3x – 4y = 12
3x + 4y = 12
The combined equation of the lines
(3x – 4y) (3x + 4y) = 144
9x² – 16y² = 144

Question 3.
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8.
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)
Equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1
9x² – 16y² = 144

Question 4.
Find the equation of the hyperbola, whose asymptotes are the straight lines (x + 2y + 3) = 0, (3x + 4y + 5) = 0 and which passes through the point (1, -1).
Solution:
Combined equation of the asymptotes as
(x + 2y + 3) (3x + 4y + 5) = 0
∴ Equation of the hyperbola can be taken as
(x + 2y + 3) (3x + 4y + 5) + k = 0
The hyperbola passes through P(1, -1)
(1 – 2 + 3) (3-4 + 5) + k = 0
8 + k = 0 ⇒ k = -8
Equation of the hyperbola is
(x + 2y + 3) (3x + 4y + 5) – 8 = 0
3x² + 6xy + 9x + 4xy + 8y² + 12y + 5x +10y + 15 – 8 = 0
3x² + 10xy + 8y² + 14x + 22y + 7 = 0

Question 5.
If 3x – 4y + k = 0 is a tangent to x² – 4y² = 5, find value of k.
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
\(\frac{k^2}{16}\) = 5. \(\frac{9}{16}-\frac{5}{4}\)
k² = 45 – 20 = 25
k = ±5.

Question 6.
Find the product of lengths of the perpen-diculars from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1 to its asymptotes.
Solution:
Equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1
a² = 16, b² = 9
Product of the perpendiculars from any point as the hyperbola to its asymptotes

Question 7.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate-hyperbola.
Solution:
If e and er, are the eccentricity of a hyperbola and its conjugate hyperbola, then

Question 8.
Find the equation of the hyperbola whose asymptotes are 3x = ± 5y and the vertices are (± 5, 0).
Solution:
The equation of the asymptotes are
3x = ±5y
3x – 5y = 0, 3x + 5y = 0
Combined equation of the asymptotes is
(3x – 5y) (3x + 5y) = 0
9x² – 25y² = 0
Equation of the hyperbola is 9x² – 25y² = k
The hyperbola passes through the vertex (5, 0)
9(5)² – 0 = k ⇒ k = 225
Equation of the hyperbola is
9x² – 25y² = 225

Question 9.
Find the equation of the normal at θ = \(\frac{\pi}{3}\) to the hyperbola 3x² – 4y² = 12.
Solution:
Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^2}{4}+\frac{y^2}{3}\) = 1
Equation of the normal is

Question 10.
If the angle between the asymptotes is 30° then find its eccentricity.
Solution:
Angle between the asymptotes = 2θ = 30°
θ = 15°

Eccentricity = e = √6 – √2

II.

Question 1.
Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the following hyperbola.
i) 16y² – 9x² = 144
Solution:
Equation of the hyperbola is
16y² – 9x² = 144
\(\frac{y^2}{9}-\frac{x^2}{16}\) = 1
a² = 16, b² = 9
Centre C(0, 0)
b²e² = a² + b² = 16 + 9
= 25 ⇒ be = 5
Foci are 5(0, ±ae) = (0, ±5)
Eccentricity = \(\frac{be}{b}=\frac{5}{3}\)
Equation of the directrices are y = ±b/e
± 3. \(\frac{5}{3}\)
5y = ± 9
Length of the latus return = 2. \(\frac{a^2}{b}\)
= 2. \(\frac{16}{3}\) = \(\frac{32}{3}\)

ii) x² – 4y² = 4
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 = √5
Foci are (±ae, 0) = (± √5, 0)
Eccentricity = \(\frac{ae}{a}=\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ±\(\frac{a}{e}\)
= ± 2. \(\frac{2}{\sqrt{5}}\)
⇒ √5 = ± 4
⇒ √5x ± 4 = 0
Length of the latus rectum = \(\frac{2b^2}{a}=\frac{2.1}{2}\) = 1

iii) 5x² – 4y² + 20x + 8y = 4
Solution:
5(x² + 4x + 4) -4 (y² – 2y + 1) = 4 + 20 – 4
5(x + 2)² – 4(y – 1)² = 20
\(\frac{(x+2)^2}{4}+\frac{(y-1)^2}{5}\) = 1
a² = 4, b² = 5 => a < b
Centre C(-2, +1)
a²e² = a² + b² = 4 + 5 = 9
ae = 3
Eccentricity = \(\frac{ae}{a}=\frac{3}{2}\)
Foci are (h ± ae, k) = (-2 ± 3, 1)
= (-5, 1) and (1, 1)
Equations of the directrices are x – h = ±\(\frac{a}{e}\)
x + 2 = ± 2. \(\frac{2}{3}\)
3x + 6 = ± 4
3x+ 10 = 0 or 3x + 2 = 0
Length of the latus rectum = \(\frac{2b^2}{a}=\frac{2.5}{2}\) = 5

iv) 9x² – 16y² + 72x – 32y – 16 = 0
Solution:
Equation of the hyperbola is
9x² – 16y² + 72x – 32y -16 = 0
⇒ 9(x² +8x) – 16(y² + 2y) = 16
⇒ 9(x² + 8x +16) – 16 (y² + 2y + 1)
= 16 + 144 – 16
⇒ 9(x + 4)² – 16(y + 1)² = 144
\(\frac{(x+4)^2}{16}+\frac{(y+1)^2}{9}\) = 1
Comparing with \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}\) = 1
a² = 16, b² = 9, h = -4, k = -1
Centre (h, k) = (-4, -1)

Foci = (h ± ae, k) = (- 4±4.\(\frac{5}{4}\), 1)
= (-4 ± 5, -1)
= (1, -1) and (-9, -1)
Equation of the directrices are
x + 4 = ± 4. \(\frac{4}{5}\)
= ± \(\frac{16}{5}\)
5x + 20 = ± 16
Equation of the directices are 5x + 4 = 0
and 5x + 36 = 0
Length of Latus rectum = \(2\frac{b^2}{a}=\frac{2.9}{4}=\frac{9}{2}\)

Question 2.
Find the equation to the hyperbola whose foci are (4, 2) and (8, 2) and eccentricity is 2.
Solution:
Foci are (4, 2) and (8, 2) Centre C is the mid point of the foci
∴ Centre is (\(\frac{4+8}{2}\), \(\frac{2+2}{2}\)) = (6, 2)
ae = 6 – 4 = 2
Given e = 2 ⇒ a = \(\frac{ae}{e}=\frac{2}{2}\) = 1
b² = a² (e² – 1) = 1(4 – 1) = 3
Equation of the hyperbola is

Multiplying with 3
3(x – 6)² – (y – 2)² = 3
⇒ 3(x² – 12x + 36) – (y² – 4y + 4) = 3
⇒ 3x² – 36x + 108 – y² + 4y – 4 -3 = 0
3x² – y² – 36x + 4y + 101 = 0

Question 3.
Find the equation of the hyperbola of given length of transverse axis 6 whose vertex bisects the distance between the centre and the focus.
Solution:
Given CA = AS

a = ae – a
2a = ae ⇒ e = 2
Length of transverse axis = 2a = 6 ⇒ a = 3
b² = a²(e² – 1) = 9(4 – 1) = 27
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
\(\frac{x^2}{9}-\frac{y^2}{27}\) = 1
3x² – y² = 27

Question 4.
Find the equations of the tangents to the hyperbola x² – 4y² = 4 Which are i) Parallel ii) Perpendicular to the line x + 2y = 0.
Solution:
Equation of the hyperbola is x² – 4y² = 4
\(\frac{x^2}{4}-\frac{y^2}{1}\) = 1
a² = 4, b² = 1

i) The+anget is parallel to x + 2y = 0
m = –\(\frac{1}{2}\)
c² = a²m² – b² = 4.\(\frac{1}{4}\) – 1 = 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 = 16
c = ±√15
Equation of the perpendicular tangent is
y = 2x ± √15

Question 5.
Find the equations of tangents drawn to the hyperbola 2x² – 3y² = 6 through (-2, 1).
Solution:
Equation of the hyperbola is 2x² – 3y² = 6
\(\frac{x^2}{3}-\frac{y^2}{2}\) = 1
Suppose the slope of the tangent is m¹
The tangent passes through P (-2, 1)
Equation of the tangent is
y – 1 = m(x + 2) = mx + 2m
y = mx + (2m + 1) ………. (1)
Condition for tangency is c² = a²m² – b²
(2m + 1)² = 3m² – 2
4m² + 4m + 1 = 3m² – 2
m² + 4 m + 3 = 0
(m + 1) (m + 3) = 0
m = -1 or – 3

Case (i): m = -1
Substituting in (1) equation of the tangent is
y = -x – 1
x + y + 1 =0

Case (ii) : m = – 3
Equation of the tangent is
y = – 3x – 5
3x + y + 5 = 0

Question 6.
Prove that the product of the perpendicular distance from any point on a hyperbola to its asymptotes is constant.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Any point on the hyperbola is P(a sec θ, b tan θ)
Equations of the asymptotes are \(\frac{x}{a}\) = ± \(\frac{y}{b}\)
i.e., \(\frac{x}{a}-\frac{y}{b}\) = 0 and \(\frac{x}{a}+\frac{y}{b}\) = 0
PM = Perpendicular distance from P on

PN = Perpendicular distance from P on

III.

Question 1.
Tangents to the hyperbola make angles \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 make angles θ₁, θ₂ with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x² – a²) when tan θ₁ + tan θ₂ = k.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Equation of the tangent to the hyperbola can be taken as
y = mx ± \(\sqrt{a^2m^2-b^2}\)
Suppose p(x₁, y₁) is the point of intersection of the tangents
y₁ = mx₁ ± \(\sqrt{a^2m^2-b^2}\)
y₁ – mx₁ = ±\(\sqrt{a^2m^2-b^2}\)
Squaring both sides
(y₁ – mx₁)² = a²m² – b²
y²₁ + m²x²₁ – 2mx₁y₁ – a²m² + b² = 0
m² (x²₁ – a²) – 2mx₁y₁ + (y²₁ + b²) = 0
This is a quadratic equation in m giving the values for m say m₁, m₂

i.e.,k = \(\frac{2x_1y_1}{x^2_1-a^2}\)
or 2x₁y₁ = k(x²₁ – a²₁)
p(x₁, y₁) lies on the curve 2xy = k(x² – a²)

Question 2.
Show that the locus of feet of the perpen-diculars drawn from foci to any tangent of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 is the auxiliary circle of the hyperbola.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Equation of the tangent to the hyperbola is

y = mx ± \(\sqrt{a^2m^2-b^2}\)
y – mx = ± \(\sqrt{a^2m^2-b^2}\) ………… (1)
Equation to the perpendicular from foci (±ae, 0) on this tangent is 1
y = –\(\frac{1}{m}\)(x ± ae)
⇒ my = – (x ± ae)
x + my = ±ae …………. (2)
Squaring and adding (1) and (2)
(y – mx)² + (x + my)² = a²m² – b² + a²e²
⇒ y² + m²x² – 2mxy + x² + m²y² + 2mxy
= a²m² – a²(e² – 1) + a²e² .
⇒ (x² + y²) (1 + m²) = a²m² – a²e² + a² + a²e²
= a²(1 + m²)
or x² + y² = a² which is the auxiliary circle.
Locus of the feet of the perpendiculars drawn from the foci to any tangent to the hyperbola lies on the auxiliary circle.

Question 3.
Show that the equation \(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1 represents.
i) An ellipse if ‘c’ is a real constant less than 5.
ii) A hyperbola if ‘c’ is any real constant between 5 and 9.
iii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (±2, 0) independent of the value of ‘c’.
i) An ellipse if ‘c’ is a real constant less than 5.
Solution:
Given equation is
\(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1
This equation represents an ellipse if 9 – c >0, 5 – c > 0
∴ c < 9, c < 5
⇒ c < 5

ii) A hyperbola if ‘c’ is any real constant between 5 and 9.
Solution:
Given equation is \(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1
This equation represents a hyperbola if
9 – c > 0 and 5 – c < 0
9 > c and 5 < c
ic. 5 < c < 9

iii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (±2, 0), independent of the value ‘c’.
Solution:
In both cases
Case (i): a² = 9 – c, b² = 5 – c
a² – b² = (9 – c) – (5 – c)
= 9 – c – 5 + c = 4
a²e² = 4 ⇒ ae = 2
Foci are (±ae, 0) (±2, 0)

Case (ii) : a² = 9 – c, b² = c – 5
a² + b² = 9 – c + c – 5 = 4
a²e² = 4 ⇒ ae = 2
Foci are (± ae, 0) = (± 2, 0)

Question 4.
Show that the angle between two asymptotes of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 is 2 Tan^-1(\(\frac{b}{a}\)) Or 2 Sec^-1(e).
Solution:
Equations of the asymptotes are \(\frac{x}{a}-\frac{y}{b}\) = 0
and \(\frac{x}{a}+\frac{y}{b}\) = 0

If 2 θ is the angle between the asymptotes then tan θ = \(\frac{b}{a}\) = slope of the asymptotes of θ = tan^-1\(\frac{b}{a}\)
Angle between the asymptotes

Angle between the asymptotes
= 2tan^-1\(\frac{b}{a}\) or 2 Sec^-1(e)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(b)

I.

Question 1.
Find the equation of tangent and normal to the ellipse x + 8y = 33 at (-1, 2).
Solution:
Equation of the tangent is
\(\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\)
x(-1) + 8y(2) = 33
⇒ -x +16y = 33
⇒ x – 16y + 33 = 0
Equation of the normal is
16x + y + k = 0
It passes through P(-1, 2)
-16 + 2 + k = 0 ⇒ k =14
Equation of the normal is 16x + y + 14 = 0.

Question 2.
Find the equation of tangent and normal to the ellipse
x² + 2y² – 4x + 12y + 14 = 0 at (2, – 1).
Solution:
xx₁ + 2yy₁ – 2(x + x₁) + 6(y + y₁) + 14 = 0
⇒ 2x – 2y – 2(x + 2) + 6(y – 1) + 14 = 0
⇒ 4y + 4 = 0
y = – 1 required equation of tangent.
Slope of tangent is ‘0’
Equation of normal be
y + 1 = \(\frac{-1}{0}\) (x – 2)
x = 2 equation of normal.

Question 3.
Find the equation of the tangents to 9x² + 16y² = 144, which makes equal intercepts on the co-ordianate axis.
Solution:
Equation of the ellipse is 9x² + 16y² = 144
⇒ \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
Equation of the tangent is
\(\frac{x}{a}\). cos θ + \(\frac{y}{b}\) sin θ = 1
Slope of the tangent = –\(\frac{b \cos \theta}{a \sin \theta}\) = -1
cot θ = \(\frac{a}{b}=\frac{4}{3}\)
cos θ = ± \(\frac{4}{5}\), sin 0 = ± \(\frac{3}{5}\)
Equation of the tangent is
\(\frac{x}{4}\)(±\(\frac{4}{5}\)) + \(\frac{y}{3}\)(±\(\frac{3}{5}\)) = 1
x ± y ± 5 = 0.

Question 4.
Find the co-ordinates for the points on the ellipse x² + 3y² = 37 at which the normal is parallel to the line 6x – 5y = 2.
Solution:
Equation of the ellipse is x² + 3y² = 37
⇒ \(\frac{x^2}{37}+\frac{y^2}{\left(\frac{37}{3}\right)}\) = 1
a² = 37, b² = \(\frac{37}{3}\)
Slope of the normal =

The normal is parallel to 6x – 5y = 2

Case i) The co-ordinates of P are
(a cos θ, b sin θ)

Case ii) The co-ordinates of P are (a cos θ, b sin θ)

Question 5.
Find the value of k if 4x+y + k = 0isa tangent to the ellipse x² + 3y² = 3.
Solution:
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 = -4, c = -k.
Condition for tangency is c² = a² m² + b²
(-k)² = 3(-4)² + 1
k²= 48 + 1 = 49
k = ±7.

Question 6.
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.
Solution:
Equation of the ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) ……….. (1)
Equation of the line is x cos α + y sin α = p
y sin α = – x cos α + p cos α p

Condition for tangency is c² = a² m² + b²

II.

Question 1.
Find the equations of tangent and normal to the ellipse 2x² + 3y² = 11 at the point whose ordinate is 1.
Solution:
Equation of the ellipse is 2x² + 3y² = 11
Given y = 1
2×2 + 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 = 11
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 2.
Find the equations to the tangents to the ellipse, x² + 2y² = 3 drawn from the point (1, 2) and also find the angle between these tangents.
Solution:
Equations of the ellipse is x² + 2y² = 3
⇒ \(\frac{x^2}{3}+\frac{y^2}{\left(\frac{3}{2}\right)}\) = 1
a² = 3, b² = \(\frac{3}{2}\)
Suppose m is the slope of the tangent. It passes through P (1, 2)
Equation of the tangent is

y – 2 = m(x – 1)
= mx – m
y = mx + (2 – m)
Condition for tangency is c² = a²m² + b²
(2 – m)² = 3(m²) + \(\frac{3}{2}\)
4 + m² – 4m = 3m² + \(\frac{3}{2}\)
2m² + 4m – \(\frac{5}{2}\)
4m² + 8m – 5 = 0
(2m – 1) (2m + 5) = 0
m = \(\frac{1}{2}\) or –\(\frac{5}{2}\)

Case i) m = \(\frac{1}{2}\)
Equation of the tangent is y = \(\frac{1}{2}\) x + 2 – \(\frac{1}{2}\)
= \(\frac{x}{2}+\frac{3}{2}\)
2y = x + 3
x- 2y + 3 = 0

Case ii) m = –\(\frac{5}{2}\)
Equation of the tangent is
y = – \(\frac{5}{2}\)x + (2 + \(\frac{5}{2}\))
= –\(\frac{5}{2}\)x + \(\frac{9}{2}\)
2y = – 5x + 9
or 5x + 2y – 9 = 0
Angle between the tangents is given by

Question 3.
Fincbthe equation of tangents to the ellipse 2x² + y² = 8 which are
i) Parallel to x – 2y – 4 = 0
Solution:
Slope will be : \(\frac{1}{2}\)
Equation of tangent y = mx ± \(\sqrt{a^2m^2+b^2}\)

2y – x + 6 = 0 required equation of tangents,
x – 2y ± 6 = 0.

ii) Perpendicular to x + y + 2 = 0
Solution:
Slope of tangent be ‘1’ as it is perpendicular to above line
y = mx ± \(\sqrt{a^2m^2+b^2}\)
y = x ± \(\sqrt{4+8}\)
y = x ± 2√3
⇒ x – y ± 2√3 = 0.

iii) Which makes an angle \(\frac{\pi}{4}\) with x-axis.
Solution:
Equation of tangent y = x ± 2√3.
⇒ x – y ± 2√3 = 0.

Question 4.
A circle of radius 4, is concentric with the ellipse 3x² + 13y² = 78. Prove that a common tangent is inclined to the major axis at an angle \(\frac{\pi}{4}\).
Solution:
Equation of the ellipse is 3x² + 13y² = 78
\(\frac{x^2}{26}+\frac{y^2}{6}\) = 1 ……… (1)
Centre of the ellipse is (0, 0)
Equation of the circle is x² + y² = 16 ………. (2)
Equation of tangent at P(θ) to the circle is
x cos θ + y sin θ = 16 ……. (3)

(3) is a tangent to the ellipse.

16 = 26 cos² θ + 6 sin² θ
= 26 cos² θ + 6(1 – cos² θ)
= 26 cos² θ + 6-6 cos² θ
20 cos² θ = 10
cos² θ = \(\frac{10}{20}=\frac{1}{2}\)
cos θ = ± \(\frac{1}{\sqrt{2}}\)
θ = \(\frac{\pi}{4}\)

III.

Question 1.
Show that the foot of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve
(x² + y²)² = a²x² + b²y².
Solution:
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

CN is the perpendicular from C on the tangent slope of CN = \(\frac{y_1}{x_1}\)
∴ Slope of PT × slope of CN = -1


Locus of N (xv y,) is (x² + y²)² = a²x² + b²y²

Question 2.
Show that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.

Solution:
Hint : Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)
Equation of the tangent to the ellipse is
y = mx ± \(\sqrt{a^2m^2+b^2}\)
⇒ y – mx = ± \(\sqrt{a^2m^2+b^2}\) ………… (1)
Equation to the perpendicular from either focus (+ae, 0) on this tangent is
y = – \(\frac{1}{m}\) (x ± ae)
my = -(x ± ae)
my + x = ± ae ……….. (2)
Squaring and adding (1) and (2)
(y – mx)² + (my + x)² = a²m² + b² + a²e²
y² + m²x² – 2mxy + m²y² + x² + 2mxy = a²m² + a² – a²e² + a²e²
(x² + y²) (1 + m²) = a² (1 + m²)
⇒ x² + y² = a²
The locus is the auxiliary circle concentric with the ellipse.

Question 3.
The tangent and normal to the ellipse x² + 4y² = 4 at a point P(θ) on it meets the major axis in Q and R respectively. If 0 < θ < \(\frac{x}{2}\) and QR = 2, then show that θ = cos^–\(\frac{2}{3}\)
Solution:

Equation of the ellipse is x² + 4y² = 4
\(\frac{x^2}{4}+\frac{y^2}{1}\) = 1
Equation of the tangent at P (θ) is
\(\frac{x}{2}\).cos θ + \(\frac{y}{1}\) sin θ = 1
Equation of x-axis is y = 0
\(\frac{x}{2}\) cos θ = 1 ⇒ x = \(\frac{2}{\cos\theta}\)
Co-ordinates of Q are (\(\frac{2}{\cos\theta}\), θ)
Equation of the normal at
P(θ) is \(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = a² – b²
\(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = 3
Substituting y = 0 we get \(\frac{2x}{\cos\theta}\) = 3
x = \(\frac{3}{2}\). cos θ
Co-ordinates of R are (\(\frac{3}{2}\). cos θ, o)

Given QR = 2
\(\frac{-3 \cos ^2 \theta+4}{2 \cos \theta}\) = 2
-3 cos² θ+ 4 = 4 cos θ
3 cos² θ + 4 cos θ – 4 = 0
(3 cos θ – 2) ( cos θ + 2) = 0
3 cos θ – 2 = 0 or cos θ + 2 = 0
cos θ = \(\frac{2}{3}\) or cos θ = – 2
cos θ always lies between -1 and 1
∴ cos θ = \(\frac{2}{3}\)
i.e., θ = cos^-1(\(\frac{2}{3}\))

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(a)

I.

Question 1.
Find the equation of the ellipse with focus at (1, -1) e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0.
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

9[(x₁ – 1)² + (y₁ + 1)²] = 2[x₁ + y₁ + 2]²
9[x²₁ – 2x₁ + 1 + y²₁ + 2y₁ + 1] = 2[x²₁ + y²₁ + 4 + 2x₁y₁ + 4x₁ + 4y₁]
9x²₁ +9y²₁ – 18x₁ 18y₁ +18 = 2x²₁ +2y²₁ + 4x₁y₁ + 8x₁ + 8y₁ + 8
7x²₁ – 4x₁y₁ + 7y²₁ – 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 2.
Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is \(\frac{15}{2}\).
Solution:
Latus rectum = \(\frac{15}{2}\) ;
distance between foci = 2
\(\frac{2b^2}{2}\) = \(\frac{15}{2}\) ; 2ae = 2
ae = 1
⇒ b² = a² – a²e²
⇒ b² = a² – 1
⇒ \(\frac{15}{2}\)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 3.
Find the equation of the ellipse in the standard form such that distance between foci is 8 and distance between directrices is 32.
Solution:
Distance between foci = 8.
Distance between directrices = 32.
2ae = 8, \(\frac{2a}{e}\) = 32
ae = 4, \(\frac{a}{e}\) = 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 4.
Find the eccentricity of the ellipse, (in standard form), if its length Solution:
Latus rectum = \(\frac{2b^2}{a}\)
Major axis = 2a
Given \(\frac{2b^2}{a}\) = \(\frac{1}{2}\) .2a.
2b² = a²
But b² = a² (1 – e²)
2a²(1 – e²) = a²
1 – e² = \(\frac{1}{2}\)
e² = \(\frac{1}{2}\) ⇒ e = \(\frac{1}{\sqrt{2}}\)

Question 5.
The distance of a point on the ellipse x² + 3y² = 6 from its centre is equal to 2. Find the eccentric angles.
Solution:
Equation of the ellipse is x² + 3y² = 6
\(\frac{x^{2}}{6}+\frac{y^{2}}{2}\)
a = √6 , b = √2
Any point on the ellipse is
P (√6 cos θ, √2 sin θ)
Given CP = 2 ⇒ CP² = 4
6 cos² θ + 2 sin² θ = 4
6(1 – sin² θ + 2 sin² θ = 4
6 – 6 sin² θ + 2 sin² θ = 4

Question 6.
Find the equation of ellipse in the standard form, if it passes through the points (-2, 2) and (3, -1).
Solution:
Equation of the ellipse in standard form is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
passes through-(-2, 2); (3, -1)
\(\frac{4}{a^{2}}+\frac{4}{b^{2}}\) = 1 ……… (i)
\(\frac{9}{a^{2}}+\frac{1}{b^{2}}\) = 1 ……….(ii)
Solving (i) and (ii) we get

3x² + 5y² = 32.

Question 7.
If the ends of major axis of an ellipse are (5, 0) and (-5, 0). Find the equation of the ellipse in the standard form if its focus lies on the line 3x – 5y – 9 = 0.
Solution:
(a, 0): (5, 0), (-a, 0) : (-5, 0)
a = 5, b² = a² (1 – e²)
Focus lies on the line 3x – 5y – 9 = 0
3(ae) – 5 (0) – 9 = 0
3(5e) – 9 = 0

Equation of the ellipse is
\(\frac{x^{2}}{25}+\frac{y^{2}}{16}\) ⇒ 16x² + 25y² = 400

Question 8.
If the length of the major axis of an ellipse is three times the length of its minor axis then find the eccentricity of the ellipse.
Solution:
Major axis = 3 minor axes
2a = 3(2b) ⇒ a = 3b
a² = 9b² ⇒ a² = 9a² (1 – e²)
(1 – e²) = \(\frac{1}{9}\) ⇒ e² = 1 – \(\frac{1}{9}\) = \(\frac{8}{9}\)
e = \(\frac{2\sqrt{2}}{3}\)
Eccentricity of the ellipse = \(\frac{2\sqrt{2}}{3}\)

II.

Question 1.
Find the length of major axis, minor axis, latus rectum, eccentricity, co-ordinates of centre, foci and the equations of directrices of the following ellipse.
i) 9x² + 16y² = 144
ii) 4x² + y² – 8x + 2y + 1 =0
iii) x² + 2y² – 4x + 12y + 14 = 0
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 = \(\frac{2b^{2}}{a}=\frac{2.9}{4}=\frac{9}{2}\)
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) = (±√7, 0)
Equations of the directrices are x = ± \(\frac{a}{e}\)
x = ± 4.\(\frac{4}{\sqrt{7}}\) = ±\(\frac{16}{\sqrt{7}}\)
√7x = +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
\(\frac{(x-1)^{2}}{2}+\frac{(y+1)^{2}}{2}\)= 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 latus rectum \(\frac{2a^{2}}{a}=\frac{2}{2}=1\)
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.\(\frac{\sqrt{3}}{2}\) = √3
Foci are (-1, 1 ±√3)
Equations of the directrices are y + 1 = ± \(\frac{b}{e}\)
= ± \(\frac{4}{\sqrt{3}}\)
√3y + √3 = ±4
√3y + √3 ± 4 = o

(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

a = 2√2, b = 2, h = 2, k =-3
Length of major axis = 2a = 2(2 √2) = 4 √2
Length of minor axis = 2b = 2(2) = 4
Length of latus rectum

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 2.
Find the equation of the ellipse in the form \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}\) = 1, given the following data.
i) Centre (2, -1), one end of major axis (2, – 5), e = \(\frac{1}{3}\).
Solution:
Centre (2, -1) ⇒ h = 2, k = -1
end of major axis (2, -5)

Equation of the ellipse is

9(x – 2)² + 8(y + 1)² = 128
i.e., 9 (x – 2)² + 8 (y + 1)² = 128.

ii) Centre (4, -1), one end of major axis is (-1, -1) and passing through (8, 0).
Solution:
a = \(\sqrt{(4+1)^2+(-1+1)^2}\)
a = 5
passes through (8, 0)

Required equation be
\(\frac{(x-4)^2}{25}\) + \(\frac{9}{25}\)(y + 1)² = 1
⇒ (x – 4)² + 9(y + 1)² = 25

iii) Centre (0, -3), e = \(\frac{2}{3}\), semi-minor axis = 5.
Solution:
b = 5
⇒ b² = a² – a²e²
⇒ 25 = a² – a². \(\frac{4}{9}\) = a²\(\frac{5}{9}\)
⇒ 45 = a²

iv) Centre (2, -1); e = \(\frac{1}{2}\); length of latus rectum 4.
Solution:
Equation of the ellipse is
\(\frac{9(x-2)^2}{64}+\frac{3(y+1)^2}{16}\) = 1
9(x – 2)² + 12(y + 1)² = 64.

Question 3.
Find the radius of the circle passing through the foci of an ellipse 9x² + 16y² = 144 and having least radius.
Solution:
Equation of the ellipse 9x² + 16y² = 144.

\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
a² = 16, b² = 9
a = 4, b = 3
The circle passes through S and S¹ and has least radius.
If S S¹ is diameter.
a²e² = a² – a² (1 – e²) = a² – b² = 16 – 9 = 7
Equation of the required circle is x² + y² = 7.

Question 4.
A man running on a race course notices that the sum of the distances of the two flag posts from him is always 10m and the distance between the flag posts is 8m. Find the equation of the race course traced by the man.
Solution:
S and S¹ are the flags and P is the position of the man.

Given SP + S¹P = 10 and SS¹ = 8
The path traced by the man is an ellipse whose foci are S and S¹.
2a = 10 ⇒ a = 5
SS¹ = 8 ⇒ 2ae = 8 ⇒ ae = 4
b² = a² (1 – e²) = 25 (1 – \(\frac{16}{25}\)) = 9
Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
\(\frac{x^2}{25}+\frac{y^2}{9}\) = 1.

III.

Question 1.
A line of fixed length (a + b) moves so that its ends are always on two perpendicular straight lines fixed. Prove that a marked point on the line, which divides this line into portions of lengths ‘a’ and ‘b’ describes an ellipse and also find the eccentricity of the ellipse when a = 8, b = 12.
Solution:
Take the perpendicular lines as co-ordinates axes. AB is the fixed line. Let OA = α, and OB = β so that equation of AB is \(\frac{x}{\alpha}+\frac{y}{\beta}\) = 1
Where α² + β² = (a + b)² ………….. (1)
P(x, y) divides AB in the ratio a = b
Co-ordinates of P are (\(\frac{b\alpha}{a+b}+\frac{a\beta}{a+b}\)) = (x, y)

Substituing the values of α, β in (1) we get,

P describes an ellipse.
Given a = 8, b = 12, so that b > a

Question 2.
Prove that the equation of the chord joining the points α, β on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)

Solution:
The given points on the ellipse are
P(a cos α, b sin α) and Q (a cos β, b sin β).

Equation of the chord PQ is

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Parabola Solutions Exercise 3(b)

I.

Question 1.
Find the equation of the tangent and normal to the parabola y² = 6x at the positive end of the latus rectum.
Solution:
(a, 2a) Here 4a = 6 ⇒ a = \(\frac{3}{2}\)
(\(\frac{3}{2}\), 3)
equation of tangent yy₁ = 2a (x + x₁)
yy₁ = 3(x + x₁)
3y = 3(x + \(\frac{3}{2}\))
2y – 2x – 3 = O’ is the equation of tangent
Slope of tangent is 1
Slope of normal is -1
Equation of normal is y – 3 = -1(x – \(\frac{3}{2}\))
2x + 2y – 9 = 0

Question 2.
Find the equation of the tangent and normal to the parabola x² – 4x – 8y + 12 = 0 at (4, \(\frac{3}{2}\))
Solution:
(x – 2)² – 4 – 8y + 12 = 0
⇒ (x – 2)² – 8y + 8 = 0
⇒ (x – 2)² = 8(y – 1); 4a = 8 ⇒ a = 2
Equation of tangents at (x₁, y₁) is
(x – 2) (x₁ – 2) = 2a (y- 1 + y₁ – 1)
⇒ (x – 2) (4 – 2) = 2(2) (y – 1 + \(\frac{3}{2}\) – 1)
⇒ 2(x – 2) = 4(\(\frac{2y-1}{2}\)
x – 2y – 1 = 0
Equation of normal will be
y – y₁ = m(x – x₁)
m – Slope of normal
Slope of tangent is \(\frac{1}{2}\)
Slope of normal is – 2
y – \(\frac{3}{2}\) = -2(x – 4)
2y – 3 = -4x + 16
4x + 2y – 19 = 0

Question 3.
Find the value of k if the line 2y = 5x + k is a tangent to the parabola y² = 6x.
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

Question 4.
Find the equation of the normal to the parabola y² = 4x which is parellel to y – 2x + 5 = 0.
Solution:
Equation of the parabola is y² = 4x
∴ a = 1
The equation of the given line y – 2x + 5 = 0
Slope m = 2
The normal is parallel to the line y – 2x +5 = 0
Slope of the normal = 2
Equation of the normal at ‘t’ is
y + tx = 2at + at³
∴ Slope = -t = 2
⇒ t = -2
Equation of the normal is
y – 2x = 2.1 (-2) + 1(-2)³
= -4 – 8 = – 12
2x-y- 12 = 0.

Question 5.
Show that the line 2x – y + 2 = 0isa tangent to the parabola y² = 16x. Find the point of contact also. .
Solution:
Given line is 2x – y + 2 = 0
⇒ y = 2x + 2
Comparing with y = mx + c we get m = 2, c = 2
Comparing y² = 16x with y² = 4ax
We get 4a = 16 ⇒ a = 4
\(\frac{a}{m}=\frac{4}{2}\) = 2 = c
Point of contact = (\(\frac{a}{m^{2}},\frac{2a}{m}\)) = (\(\frac{4}{2^{2}},\frac{2(4)}{2}\))
= (1.4)

Question 6.
Find the equation of tangent to the parabola y2 = 16x inclined at an angle 60° with its axis and also find the point of contact.
Solution:
θ = 60°; m = tan 60° = √3

II.

Question 1.
Find the equations of tangents to the parabola y² = 16x which are parallel and perpendicular respectively to the line 2x – y + 5 = 0, also find the co-ordinates of the points of contact also.
Solution:
Equation of the parabola is y² = 16x
The tangent is parallel to 2x – y + 5 = 0
Equation of the tangent can be taken as y = 2x + c
Condition for tangency is c = \(\frac{a}{m}=\frac{4}{2}\) = 2
Equation of the parallel tangent is y = 2x + 2
2x – y + 2 = 0
Point of contact is (\(\frac{a}{m^{2}},\frac{2a}{m}\)) = (\(\frac{4}{4},\frac{8}{2}\)) = (1, 4)
Slope of the perpendicular tangept is
m’ = –\(\frac{1}{m}=-\frac{1}{2}\)
Equation of the perpendicular tangent is
y = m’x + c’
= (-\(\frac{1}{2}\)) x + c’
where c’ = \(\frac{a}{m’}\) = \(\frac{4}{\left(-\frac{1}{2}\right)}\) = -8
Equation of the perpendicular tangent is
y = –\(\frac{1}{2}\)x – 8
2y = – x – 16
x + 2y + 16 = 0

Question 2.
If lx + my + n = 0 is a normal to the parabola y² = 4ax, then show that al³ + 2alm² + nm² = 0.
Solution:
Equation of the parabola is y² = 4ax
Equation of the normal is y + tx = 2at + at³
tx + y – (2at + at³) = 0 ……….. (1)
Equation of the given line is
lx + my + n = 0 …………. (2)
(1), (2) represent the same line.
Comparing the co-efficients

Multiplying with m³, we get
-nm² = 2alm² + al³
⇒ al³ + 2alm² + nm² = 0

Question 3.
Show that the equations of common tangents to the circle x² + y² = 2a² and the parabola y² = 8ax are y = ± (x + 2a).
Solution:
The equation of tangent to parabola
y² = 8ax is y = mx + \(\frac{2a}{m}\) ……….. (i)
m²x – my + 2a = 0
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√2 of the circle.
\(\frac{2a}{\sqrt{m^{2} + m^{4}}}\) = a√2
or 4 = 2 (m⁴ + m²)
m⁴ + m² – 2 = 0
(m² + 2) (m² – 1) = 0 or m = ±1
Required tangents are
y = (1)x + \(\frac{2a}{(1)}\) , y = (-1) x + \(\frac{2a}{(-1)}\)
⇒ y = ±(x + 2a)

Question 4.
Prove that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Solution:
Equation of the parabola is y² = 4ax
Equation of the tangent at Q(₁) is
t₁y = x + at²₁
Equation of the tangent at R(t₂) is t₂y = x + at²₂
Solving, point of intersection is T [at₁t₂, a(t₁ + t₂)]
Equation of the chord QR is
(t₁ + t₂)y = 2x + 2at₁t₂

Question 5.
Find the condition for the line y = mx + c to be a tangent to the parabola x² = 4ay.
Solution:
Equation of tangent to x² = 4ay in terms of ‘m₁‘ [slope of tangent] is

Comparing (i) and (ij), we get

c = – am² is required condition.

Question 6.
Three normals are drawn (k, 0) to the parabola y² = 8x one of the normal is the axis and the axis and the remaining two normals are perpendicular to each other, then find the value of k.
Solution:
Equation of any normal to the parabola is
y + xt = 2at + at³
This normal passes through (k, 0)
∴ kt = 2at + at³
at³ + (2a – k)t = 0
at² + (2a – k) = 0
Given m₁ = 0, m₂m₃ = -1
(-t₂) (-t₃) = -1 ⇒ t₂t₃
\(\frac{2a-k}{a}\) = -1
2a – k = -a
k = 2a + a = 3a
Equation of the parabola is
y² = 8x
4a = 8
⇒ a = 2
k = 3a = 3(2) = 6.

Question 7.
Show that the locus of point of intersection of perpendicular tangents to the parabola y² = 4ax is the directrix x + a = 0.
Solution:
Equation of any tangent to the parabola can be taken as y = mx + \(\frac{a}{m}\)
This tangent passes through P(x₁, y₁)
y₁ = mx₁ + \(\frac{a}{m}\)
my₁ = m²x₁ + a
m²x₁ – my₁ + a = 0

The tangents are perpendicular
⇒ m₁m₂ = -1
\(\frac{a}{x_1}\) = -1
x₁ = -a
Locus of P(x₁, y₁) is x = -a, the directrix of the parabola.

Question 8.
Two Parabolas have the same vertex and equal length of latus rectum such that their axes are at right angle. Prove that the common tangents touch each at the end of latus rectum.
Solution:

Equations of the parabolas can be taken as
y² = 4ax
and x² = 4ay
Equation of the tangent at (2at, at²) to
x² = 4ay is
2atx = 2a(y + at²)
y = tx – at²
This is a tangent to y² = 4ax
∴ The condition is c = \(\frac{a}{m}\)
-at² = \(\frac{a}{t}\)
t³ = -1 ⇒ t = -1
Equation of the tangent is y = -x – a
x + y + a = 0
Equation of the tangent at L’ (a, – 2a) is
y(-2a) = 2a(x + a)
x + y + a = 0
∴ Common tangent to the parabolas touches the parabola y² = 4ax at L (a, -2a) Equation of the tangent at L (-2a, a) to
x² = 4ay
x(-2a) = 2a(y + a)
x + y + a = 0
Common tangent to the parabolas touch the parabola at L’ (-2a, a)

Question 9.
Show that the foot of the perpendicular from focus to the tangent of the parabola y² = 4ax lies on the tangent at vertex.
Solution:
Equation of any tangent to the parabola is
y = mx + \(\frac{a}{m}\)
Q(x₁, y₁) is the foot of the perpendicular
∴ y₁ = mx₁ + \(\frac{a}{m}\) ……….. (1)
Slope of SQ = \(\frac{y_{1}}{x_{1}-a}\)

SQ and PQ are perpendicular

Substituting in (1) we get

Locus of Q (x₁, y₁) is x = 0 i.e., the tangent at the vertex of the parabola.

Question 10.
Show that the tangent at one extremity of a focal chord of a parabola is parallel to the normal at the other extremity.
Solution:

P(t₁), Q(t₂) are the ends of a focai chord.
Slope of PS = Slope of PQ

Equation of the tangent at P(t₁) is
t₁y = x + at²₁
Slope of the tangent at P = \(\frac{1}{t_{1}}\) ………… (2)
Equation of the normal at Q(t₂) is
y + xt₂ = 2at₂ + at³₂
Slope of the normal at Q = -t₂ ………… (3)
Frorn (1), (2)„ (3) we get slope of the tangent at P = slope of normal at Q
Slope of the tangent at P is parallel to the normal at Q.

III.

Question 1.
If the normal at the point t, on the parabola y² = 4ax meets it again at point t² then prove that t₁t₂ + t₁² + 2 = 0.
Solution:
Equation of normal is

Equation of the line (i) again meets parabola at (at²₂, 2at₂)
∴ 2at₂ – 2at₁ = t₁(at²₂ – at²₁)
–\(\frac{2}{t_{1}}\) = t₁ + t₂ ⇒ -2 = t²₁ + t₁t₂
⇒ t²₁ + t₁t₂ + 2 = 0

Question 2.
From an external point P tangents are drawn to the parabola y² = 4ax and these tangents make angles θ₁, θ₂ with its axis such that cot θ₁ + cot θ₂ is a constant ‘a’ show that P lies on a horizontal line.
Solution:
Equation of any tangent to the parabola is
y = mx + \(\frac{a}{m}\)
This tangent passes through P(x₁, y₁)
y₁ = mx₁ + \(\frac{a}{m}\) ⇒ my₁ = m²x₁ + a
m²x₁ – my₁ + a = 0
Suppose m₁, m₂ are the roots of the equation
m₁ + m₂ = \(\frac{y_{1}}{x_{1}}\), m₁m₂ = \(\frac{a}{x_{1}}\)
Given cot θ₁ + cot θ₂ = a

Focus of P(x₁; y₁) is y = a² which is a horizontal line.

Question 3.
Show that the common tangent to the circle 2x² + 2y² = a² and the parabola 4ax intersect at the focus of the parabola y² = -4ax.
Solution:
Given circle is 2x² + 2y² = a²
with centre = (0, 0); Radius = \(\frac{a}{\sqrt{2}}\)
Given parabola is y² = 4ax a
Let y = mx + \(\frac{a}{m}\) be required tangent
But it touches 2x² + 2y² = a²
⇒ Perpendicular distance from (0, 0) = radius


Hence the focus of tangent of the parabola is y² = – 4ax.

Question 4.
The sum of the ordinates of two points on y² = 4ax is equal to the sum of the ordinates of two other points on the same curve. Show that the chord joining the first two points is parallel to the chord joining the other two points.
Solution:

Equation of the parabola is y² = 4ax
Equation of the chord joining P(t₁) and Q(t₂) is (t₁ + t₂)y = 2x + 2 at₁t₂
Slope of PQ = \(\frac{2}{t_{1}+t_{2}}\) ………… (1)
Equation of the chord joining R(t₃) and S(t₄) is (t₃ + t₄) y = 2x + 2at₃t₄
Slope of RS = \(\frac{2}{t_{3}+t_{4}}\) ………… (2)
Given 2at₁ + 2at₂ = 2at₃ + 2at₄
i.e., 2a(t₁ + t₂) = 2a(t₃ + t₄)
t₁ + t₂ = t₃ + t₄
From (1), (2), (3) we get slope of PQ = Slope of Rs
i.e., PQ and RS are parallel.

Question 5.
If a normal chord a point ‘t’ on the parabola y² = 4ax subtends a right angle at vertex, then prove that t = ± √2
Solution:
Equation of the parabola is y² = 4ax ………….. (1)
Equation of the normal at’t’ is
tx + y = 2at + at³

Homogenising (1) with the help of (2) combined equation of AQ, AR is
y² = \(\frac{4ax.(tx+y}{a(2t+t^{3})}\)
y²(2t + t³) = 4tx² + 4xy
4tx² + 4xy – (2t + t³)y² = 0
AQ, AR are perpendicular
Co-eff. of x² + Co.eff. of y² = 0
4t – 2t – t³ = 0
2t – t³ = 0
-t(t² – 2) = 0
t² – 2 = 0 ⇒ t² = 2
t = ± √2

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Parabola Solutions Exercise 3(a)

I.

Question 1.
Find the vertex and focus of 4y² + 12x – 20y + 67 = 0.
Solution:
Given equation 4y² + 12x- 20y + 67 = 0
4y² – 20y = -12x – 67
y² – 5y = – 3x – \(\frac{67}{4}\)
Adding \(\frac{25}{4}\) on both sides

Question 2.
Find the vertex and focus of x2 -16x- 6y + 6 = 0.
Solution:
Given equation is
x² – 6x – 6y + 6 = 0
x² – 6x = 6y – 6
Adding 9 on both sides
x² – 6x + 9 = 6y + 3

Question 3.
Find the equations of axis and directrix of the parabola y² + 6y – 2x + 5 = 0.
Solution:
y² + 6y = 2x – 5
Adding ‘9’ on both sides we get,
y² + 6y + 9 = 2x – 5 + 9
[y – (-3)]² = 2x + 4
[y – (-3)]² = 2[x – (-2)]
Comparing with (y – k)² = 4a (x- h) we get,
(h, k) – (-2, -3), a = \(\frac{1}{2}\)
Equation of the axis y – k = 0 i.e. y + 3 = 0
Equation of the directrix x – h + a = 0
i.e., x – (-2) + \(\frac{1}{2}\) = 0
2x + 5 = 0.

Question 4.
Find the equation of axis and directrix of the parabola 4x² + 12x – 20y + 67 = 0.
Solution:
4x² + 12x = 20y – 67
x² + 3x = 5y – \(\frac{67}{4}\)
Adding \(\frac{9}{4}\) on both sides we get

Comparing with (x – h)² = 4a(y – k)
(h, k) = (\(\frac{-3}{2},\frac{29}{10}\)) ; a = \(\frac{5}{4}\)
Equation of the axis x – h = 0, i.e., x + \(\frac{3}{2}\) = 0
2x + 3 = 0
Equation of the directrix, y – k + a = 0
y – \(\frac{29}{10}\) + \(\frac{5}{4}\) = 0
⇒ 20y – 33 = 0.

Question 5.
Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2).
Solution:
Let S = (1, -7), A(1, -2)
h = 1, k = -2, a = -2 + 7 = 5
Axis of the parabola is parallel to y-axis 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 6.
Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).
Solution:
Equation of the axis y – 3 = \(\frac{3-5}{1-3}\) (x – 1)
= x – 1
x – y + 2 = 0
The directrix is perpendicular to the axis. Equation of the directrix is x + y + k = 0
Co-ordinates of Z be (x, y)
A is the midpoint of SZ
Co-ordinates of A are \(\frac{3+x}{2},\frac{5+y}{2}\)) = (1, 3)

Co-ordinates of Z are (-1, 1)
The directrix passes through Z (-1, 1)
-1 + 1 + k = 0 ⇒ k = 0
Equation of the directrix is x – y = 0
Equation of the parabola is

⇒ 2(x² – 6x + 9 + y2 – 10y + 25) = (x + y)²
⇒ 2x² + 2y² – 12x – 20y + 68 = x² + 2xy + y²
i.e., x² – 2xy + y² – 12x – 20y + 68 = 0.

Question 7.
Find the equation of the parabola whose latus rectum is the line segment of joining the points (-3, 2) and (-3, 1).

Solution:
L (-3, 2) and L(-3, 1) are the ends of the latus rectum.
S is the midpoint of LL’
Co-ordinates of S are (-3, \(\frac{3}{2}\))
LL’ = \(\sqrt{(-3+3)^{2}+(2 – 1)^{2}}=\sqrt{0+1}\) = 1
4|a| = 1, ⇒ |a| = \(\frac{1}{4}\) ⇒ a = ± \(\frac{1}{4}\)

Case (i) a = –\(\frac{1}{4}\)
Co-ordinates of A are [-3 + \(\frac{1}{4}\), \(\frac{3}{2}\)]
Equation of the parabola is

(2y – 3)² = -(4x + 11)

Case (ii) a = –\(\frac{1}{4}\)
Co-ordinates of A are [-3 – \(\frac{1}{4}\), \(\frac{3}{2}\)]
Equation of the parabola is

i.e., (2y – 3)² = 4x + 13.

Question 8.
Find the position (interior or exterior or on) of the following points with respect to the parabola y² = 6x,
i) (6, -6)
Solution:
Equation of the parabola is
y² = 6x
i.e., S ≡ y² – 6x
S₁₁ = (-6)² – 6.6 = 36 – 36 = 0
∴ (6, – 6) lies on the parabola.

ii) (0,1)
S₁₁ = 1² – 6.0 = 1 > 0
∴ (0, 1) lies outside the parabola.

iii) (2, 3)
S₁₁ = 32 – 6.2
= 9 – 12
= – 3 < 0
∴ (2, 3) lies inside the parabola.

Question 9.
Find the co-ordinates of the point on the parabola y² = 8x whose focal distance is 10.
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 10.
If (\(\frac{1}{2}\), 2) j is one extermity of a focal chord of the parabola y² = 8x. Find the co-ordinates of the other extremity.
Solution:
A = \(\frac{1}{2}\), 2); S = (2, 0)
B = (x₁, y₁) ⇒ (\(\frac{y^{2}_{2}}{8}\), y₁)
ASB is a focal chord.
∴ Slopes of SA and BS are same.

24y₁ = -4y²₁ + 64
or 4y²₁ + 24y₁ – 64 = 0
⇒ y²₁ + 6y₁ – 16 = 0 ⇒ (y₁ + 8) (y₁ – 2) = 0
y₁ = 2, -8
x₁ = \(\frac{1}{2}\), 8; So (8, -8) other extremity.

Question 11.
Prove that the parabola y2 – 4ax, (a > O) Nearest to the focus is its vertex.
Solution:
Let P(at², 2at) be the point on the parabola y² = 4ax, which is nearest to the

focus S(a, 0) then
fSp²= (at² – a)² + (2at – 0)²
f(t) = a²2(t² – 1)(2t) + 4a²(2t)
= 4a²t(t² – 1 + 2) = 4a²t(t² + 1)
For minimum value of f(t) = 0 ⇒ t = 0
f”(L) = 4a²(3t² +1) s
f(0) = 4a² > 0
∴ At t-= 0, f(t) is minimum Then P = (0, 0)
∴ The point on the parabola y² = 4ax, which is nearest to the focus is its vertex A(0, 0).

Question 12.
A comet moves in a Parabolic orbit with the sun as focus when the comet is 2 × 10⁷ K.M from the sun, the line from the sun to it makes an fmgle \(\frac{\pi}{2}\) with the axis of the orbit. Find how near the comet comes to the sun.

Solution:
Suppose the equation of the parabolic orbit of the comet is y² = 4ax
P is the position of the comet.
Given ∠XSP = \(\frac{\pi}{2}\)
SP is perpendicular to the axis of the parabola.
SP is the semi-latus rectum
2a = 2 × 10⁷
⇒ a = 10⁷ km
A is the nearest point on the parabola from focus.
AS = a = 10⁷ km
∴ The nearest point on the parabola is 10⁷ km from the sun.

II.

Question 1.
Find the locus of the points of trisection of double ordinate of a parabola y² = 4ax (a > 0).
Solution:
Equation of the parabola is y² = 4ax
P(x, y) and Q(x, -y) are the ends of the double ordinate.

T divides PQ in the ratio 1 : 2
Co-ordinates of T are (x, \(\frac{-y+2y}{3}\))
= (x, \(\frac{y}{3}\))
T divides PQ in the ratio 2 : 1
Co-ordinates of T are (x, \(\frac{-2y+y}{3}\))
= (x, –\(\frac{y}{3}\))
Suppose Co-ordinates of the required points
L and L’ be (x₁, y₁)
y₁ = ± \(\frac{y}{3}\) ⇒ y²₁ = \(\frac{y^{2}}{9}\)
y² = 9y²₁
4ax₁ = 9y²₁
Locus of (x₁, y₁) is 9y² = 4ax.

Question 2.
Find the equation of the parabola whose vertex and focus are on the positive X – axis at a distance of a and a’ from the origin respectively.
Solution:
Given Co-ordinates of A are (a, 0) and S are (a’, 0)
AS = a’ – a

Equation of the parabola is
y² = 4(a’ – a) (x – a).

Question 3.
If L and L‘ are the ends of the latus rectum of the parabola x² = 6y, find the equations of OL and OL’ where ‘O’ is the origin. Also find the angle between them.
Solution:
x² = 6y
Curve is symmetric about Y – axis
Extremities of latus rectum are
(2a, a), (-2a, a)
4a = 6 ⇒ a = \(\frac{3}{2}\)

Question 4.
Find the equation of the parabola whose axis is parallel to X-axis and which passes through these points. (-2, 1) (1, 2), and (-1, 3)
Solution:
Axis is parallel to X – axis
General equation be
x = ay² + by + c
Passes through (-2, 1) (1, 2) (-1, 3)
-2 = a + b + c ………. (i)
1 = 4a + 2b + c ……….. (ii)
-1 = 9a + 3b + c …………. (iii)

⇒ –\(\frac{5{2}\) = a
i.e., \(\frac{21}{2}\) = b
-10 = c
x = –\(\frac{5}{2}\)y² + \(\frac{21}{2}\)y – 10
5y² + 2x – 21y + 20 = 0

Question 5.
Find the equation of the parabola whose axis is parallel to Y – axis and which passes .through the points (4, 5), (-2, 11) and (-4, 21).
Solution:
General equation be y = ax² + bx + c passes through (4, 5), (-2, 11), (-4, 21) we get
5 = 16a + 4b + c ……… (i)
11 = 4a – 2b + c ……….. (ii)
21 = 16a – 4b + c ……….. (iii)
(ii) – (i) we get
6 = -12a – 6b
(iii) – (ii) 10 = 12a – 2b
Solving we get
b = -2, a = 1/2; c = 5
y = \(\frac{1}{2}\)x² – 2x + 5
x² – 2y – 4x + 10 = 0

III.

Question 1.
Find the equation of the parabola whose focus is (-2, 3) and directrix is the line 2x + 3y – 4 = 0. Also find the length of the latus rectum and the equation of the axis of the parabola.
Solution:

Suppose P(x₁, y₁) is any point on the parabola.
S(-2, 3) is the focus.
SP² = (x₁ + 2)² + (y₁ – 3)²
Equation of the directrix is 2x + 3y – 4 = 0
PM is the perpendicular from P on the directrix
PM = \(\frac{2x_{1}+3y_{1}-4}{\sqrt{4+9}}\)
From Def. of parabola SP = PM ⇒ SP² = PM²
(x₁ + 2)² + (y₁ – 3)² = \(\frac{(2x_{1}+3y_{1}-4)^{2}}{13}\)
⇒ 13(x²₁ + 4x₁ + 4 + y²₁ – 6y₁ + 9) = (2x₁ + 3y₁ – 4)²
⇒ 13x²₁ + 13y²₁ + 52x₁ – 78y₁ + 169
= 4x²₁ + 9y²₁ + 16 + 12x₁y₁ – 16x₁ – 24y₁
⇒ 9×2₁ – 12x₁y₁ + 4y²₁ + 68x₁ – 54y₁ + 153 = 0
Locus of P(x₁, y₁) is
9x² – 12xy + 4y² + 68x- 54y + 153 = 0
Length of the latus rectum = 4a
Perpendicular distance from S on directrix

Length of the latus rectum = 4a = \(\frac{2}{\sqrt{3}}\)
The axis is perpendicular to the .directrix Equation of the directrix can be taken as
3x – 2y + k = 0
This line passes through S (-2, 3)
– 6 – 6 + k = 0 ⇒ k = 12
Equation of the axis is 3x – 2y + 12 = 0

Question 2.
Prove that the area of the triangle inscribed in the parabola y² = 4ax is \(\frac{1}{8a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units where y₁, y₂, y₃ are the ordinates of its vertices.
Solution:
Suppose P(at²₁, 2at₁), Q(at²₂, 2at₂),
R(at²₃, 2at₃) are the vertices of ∆PQR.
Area of ∆PQR = \(\frac{1}{2}\) | at²₁ (2at₂ – 2at₃) + at²₂ (2at₃ – 2at₁) + at²₃(2at₁ – 2at₂)|
= \(\frac{1}{2}\) . 2a² |t²₁ (t₂ – t₃) + t²₂(t₃ – t₁) + t²₃(t₁ – t₂)|
= a² |(t₁ – t₂) (t₂ – t₃) (t₃ – t₁)|
= \(\frac{1}{8a}\) |(2at₁ – 2at₂) (2at₂ – 2at₃) (2at₃ – 2at₁) |
= \(\frac{1}{8a}\) |(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)|
Where P(x₁, y₁), Q(x₂, y₂), R(x₃, y₃) are the vertices of ∆PQR.

Question 3.
Find the co-ordinates of the vertex and focus, the equation of the directrix and axis of the -following parabolas.
(i) y² + 4x + 4y – 3 = 0
(ii) x² – 2x + 4y – 3 = 0
Solution:
i) y² + 4x + 4y – 3 = 0
⇒ y² + 4y = – 4x + 3
⇒ y² + 4y + 4 = – 4x +. 3 + 4
⇒ (y + 2)² = – 4x + 7
⇒ [y – (-2)]² = -4[x – \(\frac{7}{4}\)]
h = \(\frac{7}{4}\), k = -2, a = 1
Vertex A(h, k) = (\(\frac{7}{4}\), -2)
Focus (h-a, k) = (\(\frac{7}{4}\) – 1, -2)
= (\(\frac{3}{4}\), -2)
Equation of the directrix x – h – a = 0
x – \(\frac{7}{4}\) – 1 = 0
4x – 11 = 0
Equation of the axis is y – k = 0
y + 2 = 0

ii) x² – 2x + 4y – 3 = 0
⇒ x² – 2x = – 4y + 3
⇒ x – 2x + 1 = – 4y + 3 + 1
(x – 1)² = – 4y + 4
= – 4[y – 1 ]
(x – 1)² = – 4[y – 1]
h = 1; k = 1; a = 1
Vertex A(h, k) = (1, 1)
Focus (h, k – a) =(1, 1 -1)
= (1, 0)
Equation of the directrix
y – k – a = 0
y – 1 – 1 = 0
y – 2 = 0
Equation of the axis is,
x – h = 0
x – 1 = 0

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(b)

I.

Question 1.
Find the equation of the radical axis of the following circles,
i) x² + y² – 3x – 4y + 5 = 0, 3(x² + y²) – 7x + 8y – 11 = 0
Solution:
S ≡ x² + y² – 3x – 4y + 5 = 0
S ≡ 3x² + 3y² – 7x + 8y + 11 = 0
S – S’ = 0 is radical axis.
(x² + y² – 3x – 4y + 5)

ii) x² + y² + 2x + 4y + 1 = 0, x² + y² + 4x + y = 0.
Solution:
S – S’ = 0 is radical axis.
(x² + y² + 2x + 4y + 1) – (x² +y² + 4x + y) = 0
⇒ -2x + 3y + 1 = 0
(or) 2x – 3y – 1 =0 required radical axis.

iii) x² + y² +4x + 6y – 7 = 0,
4(x² + y²) + 8x + 12y – 9 = 0.
Solution:
S – S’ = 0 is radical axis.
(x² + y² + 4x + 6y – 7) – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = o
⇒ 2x + 3y – \(\frac{19}{4}\) = 0 ⇒ 8x + 12y – 19 = 0

iv) x² + y² – 2x – 4y -1=0, x² + y² – 4x – 6y + 5 = 0.
Solution:
S – S’ = 0 radical axis
(x² + y² – 2x – 4y – 1) – (x² + y² – 4x – 6y + 5) = 0
2x + 2y – 6 = 0 (or) x + y – 3 = 0

Question 2.
Find the equation of the common chord of the following pair of circles.
i) x² + y² – 4x – 4y + 3 = 0, x² + y² – 5x – 6y + 4 = 0.
Solution:
(x² + y² – 4x – 4y + 3) – (x² + y² – 5x – 6y + 4) = 0
x + 2y – 1 = 0 Equation of common chord.

ii) x² + y² + 2x + 3y + 1 = 0, x² + y² + 4x + 3y + 2 = 0.
Solution:
(x² + y² +2x + 3y + 1) – (x² + y² + 4x + 3y + 2) = 0
-2x – 1 = 0 equation of common chord is

iii) (x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b)
Solution:
(x² + y² – 2xa – 2yb – c²) – (x² + y² – 2xb – 2ya – c²) = 0
-2x(a – b) – 2y(b – a) = 0
(or) x – y = 0

II.

Question 1.
Find the equation of the common tangent of the following.circles at their point of contact.
i) x² + y² + 10x – 2y + 22 = 0, x² + y² + 2x – 8y + 8 = 0.
Solution:
x² + y² + 10x – 2y + 22 = 0
x² + y² + 2x – 8y + 8 = 0
When circles touch each other then
S – S’ = 0 is required tangent (common)
∴ (x² + y² + 10x-2y + 22) – (x² + y² + 2x – 8y + 8) = 0
8x + 6y + 14 = 0 (or)
4x + 3y + 7 = 0

ii) x² + y² – 8y – 4 = 0; x² + y² – 2x – 4y = 0.
Solution:
When circles touch each other then
S – S’ = 0 is required common tangent.
(x² + y² – 8y – 4) – (x² + y² – 2x – 4y) = 0
2x – 4y – 4 = 0 (or) x – 2y – 2 = 0

Question 2.
Show that the circles x² + y² – 8x – 2y + 8 = 0 and x² + y² – 2x +, 6y + 6 = 0 touch each other and find the point of contact.
Solution:
C₁ = (4, 1) C₂ = (1, -3)
r₁ = \(\sqrt{16+1-8}\) = 3 ; r₂ = \(\sqrt{1+9-6}\) = 2
C₁C₂ = \(\sqrt{(4-1)^{2}+(1+3^{2})}\) = 5
r₁ + r₂ = C₁ + C₂ they touch each other externally

∴ Point of contact is (\(\frac{11}{5} , \frac{-7}{5}\))

Question 3.
If the two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’.
Solution:

Question 4.
Find the radical centre of the following circles.
i) x² + y² – 4x – 6y + 5 = 0 ………… (i)
x² + y² – 2x – 4y – 1 = 0 ………… (ii)
x² + y² – 6x – 2y = 0 ………… (iii)
Solution:
(i) – (ii) gives
– 2x – 2y + 6 = 0
x + y – 3 = 0 ………. (1)
(ii) – (iii) gives

Point of intersection of (1) and (2) is radical centre will (7/6,11/6) we get by solving these two equations.

ii) x² + y² + 4x – 7 = 0,
2x² + 2y² + 3x + 5y – 9 = 0,
x² + y² + y = 0.
Solution:
S = x² + y² + 4x – 7 = 0 ……….. (i)
S₁ = 2x² + 2y² + 3x + 5y – 9 = 0
= x² + y² + \(\frac{3}{2}\)x + \(\frac{5}{2}\)y – \(\frac{9}{2}\) = 0 ………. (ii)
S₁₁ = x² + y² + y = 0 ………. (iii)
Radical axis of S = 0, S₁ = 0 is S – S₁ = 0

5x – 5y – 5 = 0
x – y – 1 = 0 ………… (iv)
Radical axis of S = 0, S₁₁ = 0 is S – S₁₁ = 0
4x – y – 7 = 0 ……… (v)
x – y – 1 = 0………….. (vi)
Subtracting 3x – 6 = 0 ⇒ 3x = 6
x = \(\frac{6}{3}\) = 2
Substituting in (iv), 2 – y – 1 = 0
y = 1
Radical centre is P(2, 1)

III.

Question 1.
Show that the common chord of the circles x² + y² – 6x – 4y + 9 = 0 and x² + y² – 8x – 6y + 23 = 0 is the diameter of the second circle and also find its length.
Solution:
Common chord be
(x² + y² – 6x – 4y + 9) – (x² + y² – 8x – 6y + 23) = 0
2x + 2y – 14 = 0
x + y – 7 = 0 ………… (i)
Centre of circle (-4, -3)
(-4, -3) lies on line x + y – 7
Radius is {4² + 3² – 23}^½ = √2
Diameter = 2√2

Question 2.
Find the equation and length of the common chord of the following circles.
i) x² + y² + 2x + 2y + 1 =0,
x² + y² + 4x + 3y + 2 = 0.
Solution:
x² + y² + 2x + 2y + 1 = 0
x² + y² + 4x + 3y + 2 = 0
Equation of common chord is
S – S’ = 0 (x² + y² + 2x + 2y + 1) – (x² + y² + 4x + 3y + 2) = 0
-2x – y – 1 = 0
2x + y + 1 = 0
Centre of circle is (-1, -1)
Radius = \(\sqrt{1+1-1}\) = 1
Length of ⊥ from centre (-1, -1) to the chord is
d = \(|\frac{2(-1)+(-1)+1}{\sqrt{2^{2}+1^{2}}}|=\frac{2}{\sqrt{5}}\)

ii) x² + y² – 5x – 6y + 4 = 0, x² + y² – 2x – 2 = 0
Solution:
Common chord equation
(x² + y² – 5x- 6y + 4) – (x²+ y² – 2x – 2) = 0
-3x – 6y + 6 = 0
x + 2y – 2 = 0

Question 3.
Prove that the radical axis of the circles x² + y² + 2gx + 2fy + c = 0 and x² +y² + 2g’x + 2f’y + c‘ = 0 is the diameter of the latter circle (or the former bisects the circumference of the latter) if 2g'(g – g’) + 2f (f – f’) – c – c’.
Solution:
Radical axis is
(x² + y² + 2gx + 2fy + c) – (x² + y² + 2g’x + 2fy + c’) = 0
2(g – g’) x + 2(f – f’)y + c – c’ = 0 …………. (i)
Centre of second circle is (-g1, -f1)
Radius = \(\sqrt{g’^{2}+f’^{2}+c’}\)
Now (-g’, -f’) should lie on (i)
– 2g (g – g’) – 2f'(f – f’) + c – c’ = 0
(or) 2g (g – g’) + 2f'(f – f’) = c – c’

Question 4.
Show that the circles x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 touch each other if 1/a² + 1/b² = 1/c.
Solution:
The centres of the circles C₁ (-0, 0) and C₂ (0, -b) respectively
Radius of 1st circle be \(\sqrt{a^{2}-c}\) = r₁
Radius of 2nd circle be \(\sqrt{b^{2}-c}\) = r₂
C₁C₂ = r₁ + r₂
(C₁ C₂)² = (r₁ + r₂)²
(a² + b²) = a² – c + b² – c + 2\(\sqrt{a^{2}-c}.\sqrt{a^{2}-c}\)
c = \(\sqrt{a^{2}-c}.\sqrt{a^{2}-c}\)
c² = (a² – c) (b² – c)
c² = -c (a² + b²) + a²b² + c²
(or) c(a² + b²) = a²b² (or) \(\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)

Question 5.
Show that the circles x² + y² – 2x = 0 and x² + y² + 6x – 6y + 2 = 0 touch each other. Find the coordinates of the point of contact. Is the point of contact external or internal?
Solution:
For the circle S = x² + y² – 2x = 0
Centre C₁ = (1, 0) and Radius r₁ .= \(\sqrt{1+0}\) = 1
For the circle S’ = x² + y² + 6x – 6y + 2 = 0
Centre C₂ = (-3, 3) and

As C₁C₂ = r₁ + r₂ the two circles touch each other externally, the point of contact P divides line of centres internally in the ratio
r₁ : r₂ = 1 : 4
Hence point of contact

The contact of the circle is external.

Question 6.
Find the equation of the circle which cuts the following circles orthogonally.
i) x² + y² + 4x – 7 = 0,
2x² + 2y² + 3x + 5y – 9 = 0,
x² + y² + y = 0.
Solution:
S = x² + y² + 4x – 7 = 0 ………… (i)
S₁ = 2x² + 2y² + 3x + 5y – 9 = 0

x – y – 1 = 0 ……. (iv)
Radical axis of S = 0, S₁₁ = 0 is S – S₁₁ = 0
4x – y – 7 = 0 ……….. (v)
Solving (iv) and (v)
We get 3x – 6 = 0
x = -2
Substute x value in (iv), 2 – y – 1 =0
y = 1
Radical centre is P(2, 1)
PT =’ Length of the tangent from P to S = 0
= \(\sqrt{4+1+8-7}\)
= √6
Equation of the circles cutting the given circles orthogonally
(x – 2)² + (y – 1)² = (√6)²
x² + 4 – 4x + y² + 1 – 2y = 6
x² + y² – 4x – 2y – 1 = 0

ii) x² + y² + 2x + 4y + 1 = 0,
2x² + 2y² + 6x + 8y – 3 = 0,
x² + y² – 2x + 6y – 3 = 0.
Solution:
Equations of the required circles are
S ≡ x² + y² + 2x + 4y + 1 = 0
S₁ ≡ x² + y² + 3x + 4y – \(\frac{3}{2}\) = 0
S₁₁ ≡ x² + y² – 2x + 6y – 3 = 0
Radical axis of S = 0, S₁ = 0 is S – S₁ =0
-x + \(\frac{5}{2}\) = 0 ⇒ x = \(\frac{5}{2}\)
Radical axis of S = 0, S₁₁ = 0 is S – S₁₁ = 0
4x – 2y + 4 = 0
⇒ 2x – y + 2 = 0
x = \(\frac{5}{2}\) ⇒ 5 – y + 2 = 0
⇒ y = 7
Radical centre is P (\(\frac{5}{2}\) , 7)
PT = Length of the tangent from P to S = 0

Equation of the circles cutting the given circles orthogonally

iii) x² + y² + 2x + 17y + 4 = 0,
x² + y² + 7x + 6y + 11 = 0,
x² + y² – x + 22y + 3 = 0
Solution:
Equations of the required circles are
S ≡ x² + y² + 2x + 17y + 4 = 0 ……… (i)
S₁ ≡ x² + y² + 7x + 6y + 11 = 0 ……… (ii)
S₁₁ ≡ x² + y² – x + 22y + 3 = 0 ……… (iii)
Radical axis of S = 0, S₁ = 0 is S – S₁ = 0
-5x + 11 y – 7 = 0
5x – 11y + 7 = 0 ……… (iv)
Radical axis of S = 0, S₁₁ = 0 is S – S₁₁ =0
3x – 5y + 1 = 0 ………… (v)
Solving (iv) and (v)

Radical centre is P(3, 2)
PT = Length of the tangeqt from P to S = 0
= \(\sqrt{9+4+6+34+4}\) = √57
Equation of the circle cutting the given circles cutting orthogonally
(x – 3)² + (y – 2)² = 57
x² – 6x + 9 + y² – 4y + 4 – 57 = 0
x² + y² – 6x – 4y – 44 = 0

iv) x² + y² + 4x + 2y + 1 = 0,
2(x² + y²) + 8x + 6y – 3 = 0,
x² + y² + 6x – 2y – 3 = 0.
Solution:
Equations of the required circles are
S ≡ x² + y² + 4x + 2y + 1 = 0 ……….. (i)
S₁ ≡ x² + y² + 4x + 3y – \(\frac{3}{2}\) = 0 ………… (ii)
S₁₁ ≡ x² + y² + 6x – 2y – 3 = 0 ………. (iii)
(i) – (ii) gives radical axis of S = 0, S₁ = 0 is
S – S₁ = 0 ⇒ -y + \(\frac{5}{2}\) = 0 ⇒ y = \(\frac{5}{2}\)
Radical axis of S = 0, S₁₁ = 0 is S – S₁₁ = 0
– 2x + 4y + 4 = 0
x – 2y – 2 = 0
y = \(\frac{5}{2}\) ⇒ x – 5 – 2 = 0
x = 5 + 2 = 7
Radical centre is P (7, \(\frac{5}{2}\))
PT = Length of the tangent P to S = 0

Equation of the required circle is

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(a)

I.

Question 1.
Find k if the following pairs of circles are orthogonal.
i) x² + y² + 2by – k = 0, x² + y² + 2ax + 8=0.
Solution:
g₁ = 0; f₁ = b; c₁ = -k
g₂ = a; f₂ = 0; c₂ = 8
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(0)(a) +2(b)(0)= -k + 8
0 = – k + 8
k = 8

ii) x² + y² – 6x – 8y + 12 = 0;
x² + y² – 4x + 6y + k = 0
Solution:
g₁ = -3; f₁ = -4; c₁ = 12
g₂ = -2; f₂ = 3; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(-3)(-2) + 2(3)(-4) = 12 + k
+ 12 – 24 = 12+ k ⇒ k = -24

iii) x² + y² – 5x – 14y – 34 = 0,
x² + y² + 2x + 4y + k = 0
Solution:
g₁ = \(\frac{-5}{2}\) ; f₁ = -7 ; c₁ = -34
g₂ = 1; f₂ = 2; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(\(\frac{-5}{2}\)(1) + 2(-7)(2) = -34 + k
-5 – 28 = -34 + k
– 33 = – 34 + k
k = 34 – 33
⇒ k = 1

iv) x² + y² + 4x + 8 = 0, x² + y² – 16y +k = 0
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.
Find the angle between the circles given by the equations.
i) x² + y² – 12x – 6y + 41 = 0,
x² + y² + 4x + 6y – 59= 0.
Solution:

ii) x² + y²+ 6x – 10y – 135 = 0, x² + y² – 4x + 14y – 116 = 0.
Solution:

Question 3.
Show that the angle between the circles x + y = a , x + y = ax + ay is \(\frac{3 \pi}{4}\).
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0

Question 4.
Show that the circles given by the following equations intersect each other orthogonally.
i) x² + y² – 2x – 2y – 7 = 0,
3x² + 3y² – 8x + 29y = 0.
Solution:
C₁ = (1, 1)
g = -1, f = -1, c =-7
g’ = \(\frac{-4}{3}\), f’ = \(\frac{29}{6}\) ; c’ = o
Condition that two circles are orthogonal is

∴ -7 = -7
Hence both circles cut orthogonally.

ii) x² + y² + 4x – 2y – 11 = 0,
x² + y² – 4x – 8y + 11 =0.
Solution:
g₁ = 2, f₁ = -1, c₁ = -11
g₂ = -2, f₂ = -4, c₂ = 11
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(2)(-2) + 2(-1)(-4) = -11 + 11
-8 + 8 = 0
∴ Two circles are orthogonal.

iii) x² + y² – 2x + 4y + 4 = 0, .
x² + y² + 3x + 4y + 1 = 0.
Solution:
g = -1, f = 2, c = 4
g’ = \(\frac{3}{2}\), f’= 2, c’= 1
2gg’ + 2ff’ = c + c’
2(-1). \(\frac{3}{2}\) + 2 × 2 × 2 = 4 + 1
-3 + 8 = 5
5 = 5
Hence circles cut orthogonally.

iv) x² + y² – 2lx + g = 0, x² + y² + 2my – g = 0.
Solution:
g₁ = -l ; f₁ = 0, c₁ = g, g₂ = 0, f₂ = m, c₂= -g
2g₁g₂ + 2f₁f₂ = c₁ + c₂ is condition for two circles be orthogonal
2(-1)(0) + 2(0)(m) = g – g
0 = 0
∴ Two circles are orthogonal.

II.

Question 1.
Find the equation of the circle which passes through the origin and intersects the circles below, orthogonally.
i) x² + y² – 4x + 6y + 10 = 0, x² + y² + 12y + 6 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0 ……… (i)
Above circle passes through origin
∴ c = 0
Circle (i) is orthogonal to
x² + y² – 4x + 6y + 10 = 0 then
2g(-2) + 2f(3) = 0 + 10
-4g + 6f = 10 ………… (ii)
Circle (i) is orthogonal to
x² + y² + 12y + 6 = 0
2g(0) + 2f(6) = 6 + 0
12f = 6
f = \(\frac{1}{2}\) …………… (iii)
Solving (ii) and (iii) we get
– 4g + 6 × \(\frac{1}{2}\) = 10
-4g = 10 – 3
g = –\(\frac{7}{4}\)
∴ Equation of circle be
x² + y² – \(\frac{7}{2}\)x + y = 0
2x² + 2y² – 7x + 2y = 0.

ii) x² + y² – 4x – 6y – 3 = 0, x² + y² – 8y + 12 = 0.
Solution:
Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0
It cuts
x² + y² – 4x – 6y – 3 = 0; x² + y² – 8y + 12 = 0
g₁ = -2, f₁ = -3, c₁ = -3
g₂ = 0 ; f₂ = -4, c₂ = 12
Let g,f, c be constants of required circle.
Required circle passes through origin
∴ c = 0
Requited circle is orthogonal to both circles.
∴ 2g(-2) +2f(-3) = -3 + 0 …………… (i)
2g(0) + 2f(-4) = 12 + 0 …………….. (ii)
Solving (i) and (ii) we get
2g₁g₂ + 2f₁f₂ = c₁ + c₂
Condition of orthogonality
f = –\(\frac{3}{2}\), g = \(\frac{6}{2}\)
Required circle be x² + y² + 6x – 3y = 0

Question 2.
Find the equation of the circle which passes through the point (0, -3) and intersects the circles given by the equations x² + y² – 6x + 3y + 5 = 0 and x² + y² – x – 7y = 0 orthogonally.
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……… (i)
(i) is orthogonal with x² + y² – 6x + 3y + 5 = 0
then 2g(-3) + 2f\(\frac{+3}{2}\) = c + 5
-6g + 3f = c + 5 ……… (ii)
(i) is orthogonal with x² + y² – x – 7y = 0
-g – 7f = c ……… (iii)
Circle passes through (0, -3)
0 + 9 – 6f + c = 0 ……… (iv)
(iii) – (ii)
5g – 10f = -5
g – 2f = -1
(iii) + (iv)

x² + y² + \(\frac{4}{3}\)y + \(\frac{2}{3}\)x – 5 = 0
3x² + 3y² + 4y + 2x – 15 = 0
(or) 3x² + 3y² + 2x + 4y – 15 = 0

Question 3.
Find the equation of the circle passing through the origin, having its centre on the line x + y = 4 and intersecting the circle x² + y² – 4x + 2y + 4 = 0 orthogonally.
Solution:
Let circle be x² + y²+ 2gx + 2fy + c = 0
Equation is passing through (0, 0)
0 + 0 + 2g.0 + 2f.0 + c = 0 ⇒ c = 0
x² + y² + 2gx + 2fy = 0
Centre passes through x + y = 4
∴ -g – f = 4 ………. (i)
Circle is orthogonal to
x² + y² – 4x + 2y + 4 = 0
-4g + 2f = 4 + 0
f – 2g = 2 ……… (ii)
Solving (i) and (ii) we get
– 3g = 6
9 = -2
f = -2
Equation of circle be x² + y² – 4x – 4y = 0

Question 4.
Find the equation of the circle which passes through the points (2, 0), (0, 2) and orthogonal to the circle 2x² + 2y² + 5x – 6y + 4 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0
Passes through (2, 0), (0, 2) then

Orthogonal to x + y + \(\frac{5}{2}\)x – \(\frac{6}{2}\)y + 2 = 0
2g\(\frac{5}{4}\) + 2f(-\(\frac{2}{2}\)) = 2 + c
\(\frac{5}{2}\)g – 3f = 2 + c
But g = f
\(\frac{5}{2}\)g – 3g = 2 + c
⇒ -g = 4 + 2C
Putting value of g in equation (i)
-16 – 8c + c = -4
c = –\(\frac{12}{7}\)
– g = 4 – \(\frac{24}{7}\) = +\(\frac{4}{7}\)
g = \(\frac{-4}{7}\) = f
Equation of the circle is
x + y – \(\frac{8x}{7}\) – \(\frac{8y}{7}\) – \(\frac{12}{7}\) = 0
⇒ 7(x² + y²) – 8x – 8y – 12 = 0

Question 5.
Find the equation of the circle which cuts orthogonally the circle x² + y² – 4x + 2y- 7 = 0 and having a centre at (2, 3).
Solution:
The given circle is
x² + y² – 4x + 2y – 7 = 0 ……. (1)
Let the required circle which, cuts orthogonally the circle (1) is
x² + y² + 2gx + 2fy + c = 0 ………… (2)
Its centre is (-g, -f) = (2, 3) given
g = -2, f = -3
The two circles (1) and (2) are cutting each other orthogonally.
So 2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(-2)(-2) + 2(-3)(1) = – 7 + c
8 – 6 = -7 + c
+ 2 = -7 + c
c = 7 + 2 = 9 ⇒ c = 9
Hence the required circle is,
x² + y² – 4x – 6y + 9 = 0

III.

Question 1.
Find the equation of the circle which intersects the circle
x² + y² – 6x + 4y – 3 = 0 orthogonally and passes through the point (3, 0) and touches Y-axis.
Solution:
Let circle be (x – h)² + (y – k)² = r²
If circle touches Y-axis then co-ordinate of centre (h, k); radius = |h|
(x – h)² + (y – k)² = h²
x² – 2hx + h² + y²
-2ky + k² = h²
Orthogonal to
x² + y² – 6x + 4y – 3 = 0
2(-h) (-3) + 2(-k) (2) = -3 + k²
6h – 4k = -3 + k²
x² – 2hx + y²
-2ky + k² = h
Passes through (3, 0) 6h – 4k + 3 – k² = 0
9 – 6h + k² = 0 ……… (ii)
Adding .(i) and (ii) we get
c = 9
12 – 4k = 0 or k = 3, h = 3,
Equation of circle be y² + x² – 6x – 6y + 9 =0.

Question 2.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 =0 and x² + y² – 10x – 4y + 2t = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7.
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 (v)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 3.
If P, Q are conjugate points with respect to a circle S ≡ x² + y² + 2gx + 2fy + c = 0 then prove that the circle PQ as diameter cuts the circles S = 0 orthogonally.
Solution:
Let P = (x₁, y₁) and Q(x₂, y₂) be the conjugate points w.r.t. the circle
S ≡ x² + y² – a² = 0 …….. (i)
Polar of P w.r.t.(1) is xx₁ + yy₁ – a² = 0 ………… (ii)
Given P and Q are conjugate points ⇒ Q lies on (ii)
x₁x₂ + y₁y₂ – a² = 0 …….. (iii)
Equation of the circle on PQ as diameter is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ x² + y² – (x₁ + x₂) x – (y₁ + y₂)y + (x₁x₂ + y₁y₂) = 0 …………… (iv)
(i) and (iv) are orthogonal.
2g₁g₂ + 2f₁f₂ = 2[0(\(\frac{-x_{1}+x_{2}}{2}\)) + 0(\(\frac{-y_{1}+y_{2}}{2}\))]
c₁ + c₂ = -a² + a²
⇒ 2g₁g₂ + 2f₁f₂ = c₁ + c₂
∴ The circle on PQ as diameter cuts S orthogonally.

Question 4.
If the equations of two circles whose radii are a, a’ are S = 0 and S’ = 0, then show that the circles \(\frac{S}{a}+\frac{S’}{a’}\) = 0 and \(\frac{S}{a}-\frac{S’}{a’}\) = 0 intersect orthogonally.
Solution:
Let 2d be the distance between the centres of the circles S = 0 and S’ =. 0. Take the line joining the centres as X-axis and the point midway between the centres as origin.
Then the equations of the circles are
S ≡ (x – d)² + y² – a² = 0
S’ ≡ (x + d)² + y² – a² = 0
∴ \(\frac{S}{a}+\frac{S’}{a’}\) =0 becomes Sa’ + S’a = 0
(or) [(x – d)² + y² -a²]a’ + [(x + d)² + y² – a’²]a = 0
(or) x² + y² + 2{\(\frac{(a-a’)d}{(a+a’)}\)}x + (d² – aa’) = 0 ………… (i)
Putting -a’ for a’ in we get \(\frac{S}{a}-\frac{S’}{a’}\) = 0
reduces to x² + y² + 2{\(\frac{(a+a’)d}{(a-a’)}\)}x + (d² + aa’) = 0 ………… (ii)
If (i) and (ii) cut orthogonally then
L.H.S. = 2gg’ + 2ff = c + c’
= 2{\(\frac{(a+a’)d}{(a-a’)}\)}{\(\frac{(a+a’)d}{(a-a’)}\)} + 2(0). (0) = 2d²
= (d² – aa’) + (d² + aa’) = 2d²
Which being true the circles (i) and (ii) the circles \(\frac{S}{a}±\frac{S’}{a’}\) = 0 cut each other at right angles.

Question 5.
Find the equation of the circle which intersects each of the following circles orthognally
i) x² + y² + 2x + 4y + 1 = 0, x² + y² – 2x + 6y – 3 = 0, 2(x² + y²) +6x + 8y – 3 = 0.
Solution:
Let equation of circle be
x² + y² + 2gx + 2fy + c = 0
Given circle is orthogonal to all 3 circles then
2g(1) + 2f(2) = c + 1 ……….. (i)
2g(\(\frac{3}{2}\)) = 2f(2) = c – \(\frac{3}{2}\) ………… (ii)
2g (-1) +2f(3) = c – 3 ………… (iii)
(iii) – (ii)
-5g + 2f = \(\frac{-3}{2}\) (or) – 10g + 4f = -3 ……….. (iv)
(iii) – (i)
-4g + 2f = – 4
f – 2g = -2 ……….. (v)
Solving (iv) and (v) we get
f = -7, g = -5/2, c = -34
∴ Equation of circle be
x² + y² – 5x – 14y – 34 = 0

ii) x² + y² + 4x + 2y + 1 = 0, 2(x² + y²) + 8x + 6y – 3 = 0, x² + y² + 6x – 2y – 3 = 0.
Solution:
Let required circle equation be
x² + y² + 2gx + 2fy + c = 0 this circle is orthogonal to above three circles.
∴ 2g(2) + 2f(1) = c + 1 ……….. (i)
2g(2) + 2f(\(\frac{3}{2}\)) = c – \(\frac{3}{2}\) ……….. (ii)
2g(3) + 2f(-1) = c – 3 ……….. (iii)
(i) – (ii) we get (ii) – (iii) we get
– f = \(\frac{5}{2}\) then – 2g + 5f = \(\frac{3}{2}\)
We get
g = -7 and f = \(\frac{-5}{2}\)
Substituting ‘g’ and ‘f’ in (i) we get
4(-7) + 2(\(\frac{-5}{2}\)) = c + 1
c = -34
Required equation of circle be
x² +y² – 5y – 14x – 34 = 0

Question 6.
If the straight line 2x + 3y = 1 intersects the circle x² + y² = 4 at the points A and B, then find the equation of the circle having AB as diameter.
Solution:
Equation of circle passing through x² + y² – 4 and 2x + 3y – 1 =0 can be written as
(x² + y² – 4) + λ(2x + 3y – 1) = 0
x² + y² + 2λx + 3λy – 4 – λ = 0
Centre : (-λ, \(\frac{-3 \lambda}{2}\))
Centre lies on 2x + 3y – 1 =0
∴ 2(-λ) + 3(\(\frac{-3 \lambda}{2}\)) – 1 = 0
λ = \(\frac{-2}{13}\)
∴ Equation of circle be
13 (x² + y²) – 4 x 13 – 2(2x + 3y- 1) = 0
13 (x² + y²) – 4x – 6y – 50 = 0.

Question 7.
If x + y = 3 is 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.
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

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 8.
Find the equation of the circle passing through the intersection of the circles x² + y² = 2ax and x² + y² = 2by and having its centre on the line \(\frac{x}{a}-\frac{y}{b}\) = 2.
Solution:
Equation of circle passes through
x² + y² – 2ax = 0 and x² + y² – 2by = 0 can be written as
x² + y² – 2ax + X (x² + y² – 2by) = 0
x²(1 + λ) + y²(1 + λ) + x(-2a) – (2bλ)y = 0

1- λ = (1 + λ)
λ = -1/3
Equation of circle be
3x² + 3y² – 6ax – x² – y² + 2by = 0
⇒ 2x² + 2y² – 6ax + 2by = 0
⇒ x² + y² – 3ax + by = 0

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(e)

I.

Question 1.
Discuss the relative position of the following pair of circles.
i) x² + y² – 4x – 6y – 12 = 0
x² + y² + 6x + 18y + 26 = 0.
Solution:
Centres of the circles are A (2, 3), B(-3, -9)
radii are r₁ = \(\sqrt{4+9+12}\) = 5
r₂ = \(\sqrt{9+81-26}\) = 8
AB = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)
= \(\sqrt{25+144}\) = 13 = r₁ + r₂
∴ The circles touch externally.

ii) x² + y² + 6x + 6y + 14 = 0,
x² + y² – 2x – 4y – 4 = 0.
Solution:
Centres are A (-3, -3), B (1, 2)
r₁ = \(\sqrt{9+9-14}\) = 2,
r₂ = \(\sqrt{1+4+4}\) = 3
AB = \(\sqrt{(-3-1)^{2}+(-3-2)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\) > r₁ + r₂
∴ Each circle lies exterior to the other circle.

iii) (x – 2)² + (y + 1)² = 9, (x + 1)² + (y – 3)²= 4
Solution:
Centres are A( 2, -1), B(-1, 3)
r₁ = \(\sqrt{4+1+4}\) = 3, r₂ = \(\sqrt{1+9-6}\) = 2
AB = \(\sqrt{(2+1)^{2}+(-1-3)^{2}}\)
= \(\sqrt{9+16}\)
= 5 = r₁ + r₂
∴ The circles touch each other externally.

iv) x² + y² – 2x + 4y – 4 = 0, x² + y² + 4x – 6y – 3 = 0
Solution:
Centres are A (1, -2), B (-2, 3)
r₁ = \(\sqrt{1+4+4}\) = 3, r₂ = \(\sqrt{4+9+3}\) = 4
AB = \(\sqrt{(1+2)^{2}+(-2-3)^{2}}=\sqrt{9+16}\)
= \(\sqrt{9+25}=\sqrt{34}\) < r₁ + r₂
r₁ – r₂ < AB < r₂ + r₁
∴ The circles intersect each other.

Question 2.
Find the number of possible common tangents that exist for the following pairs of circles.
i) x² + y² + 6x + 6y + 14 = 0,
x² + y² – 2x – 4y – 4 = 0
Solution:
c₁ ( -3, -3) c₂ = (1, 2)

The circles are externally No. of common tangents = 4

ii) x² + y² – 4x – 2y + 1 = 0;
x² + y² – 6x – 4y + 4 = 0.
Solution:
C₁ (2, 1), C₂ = (3, 2)
r₁ = \(\sqrt{4+1-1}\) = 2
r₂ = \(\sqrt{9+4-4}\) =3

C₁ C₂ = \(\sqrt{(2-3)^{2}+(1-2)^{2}}=\sqrt{2}\)
C₁ C₂ < r₁ + r₂ intersect each other 2 tangents (direct)

iii) x² + y² – 4x + 2y – 4 = 0;
x² + y² + 2x – 6y + 6 = 0.
Solution:
C₁ (2, -1) C₂ = (-1, 3)

C₁ C₂ = r₁ + r₂ touch each other externally; No. of common tangents = 3.

iv) x² + y² = 4; x² + y² – 6x – 8y + 16 = 0
Solution:

Two circles touch each other.
2 – direct tangents
1 – transverse tangent (at common point)

v) x² + y² + 4x – 6y – 3 = 0
x² + y² + 4x – 2y + 4 = 0.
Solution:
C₁ (-2, 3) C₂ = (-2, 1)
r₁ = \(\sqrt{4+9+3}\) = 4 r₂ = \(\sqrt{4+1-4}\) = 1
C₁ C₂ = \(\sqrt{(-2+2)^{2}+(3-1)^{2}}\)
C₁ C₂ = 2 < 3 = r₁ – r₂
One circle is inside the other.
∴ No common tangent.

Question 3.
Find the internal centre of similitude for the circles x² + y² + 6x – 2y + 1 =0 and x² + y² – 2x – 6y + 9 = 0.
Solution:

The internal centre of similitude S divides
C₁ C₂ internally in the ratio r₁ : r₂ = 3 : 1
Co-ordinates of S¹ are

Question 4.
Find the external centre of similitude for the circles x² + y² – 2x – 6y + 9 = 0 and x² + y² = 4.
Solution:
Centre of the circle C₁ (1, 3) and C2 (0,0)
r₁ = \(\sqrt{1+9-9}\), r₂ = 2
External centre of similitude S¹ divides C₁ C₂ externally in the ratio
r₁ : r₂ = 1 : 2

II.

Question 1.
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.
Solution:
Equations of the circles are
S₁ ≡ x² + y² – 6x – 2y + 1 =0
S₂ ≡ x² + y² + 2x – 8y + 13 = 0

∴ 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
(\(\frac{3(-1)+2.3}{5}\), \(\frac{3.4+2.1}{5}\) i.e., P(\(\frac{3}{5}\), \(\frac{14}{5}\))
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)

∴ The circles touch each other internally. The point of contact ‘P’ divides AB externally in the ratio r₁ : r₂ = \(\frac{\sqrt{65}}{2}\) : √65
= 1 : 2 Co-ordinates of P are

P = (5, 1)
∴ Equation of the common tangent is
S₁ – S₂ = 0
– 4x + 7y +13 = 0
4x – 7y – 13 = 0

Question 2.
Find the equation of the circle which touches the circle x² + y² – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5.
Solution:
x² + y² – 2x – 4y – 20 = 0 ,
C = (1, 2)

Equation of circle be
(x – 9)² + (y – 8)²
x² + y² – 18x – 16y + 120 = 0
If (h, k) is the centre of the required circle (5, 5) is the mid-point of (1, 2) and (h, k).

Question 3.
Find the direct common tangents of the circles.
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


c = -22 m – 4
∴ y = mx – 22m – 4
This line is a tangent to the second circle
∴ \(\frac{|11m+2-22m-4|}{\sqrt{m^{2}+1}}\) = 5
Squaring and cross multiplying
25(1 + m²) = (11m + 2 – 22m – 4)²
96m² + 44m – 21 = 0
⇒ 96m² + 72m – 28m – 21 = 0

⇒ 24y = 7x – 250
⇒ 7x – 24y – 250 = 0

Question 4.
Find the transverse common tangents of the circles x² + y² – 4x – 10y + 28 = 0 and x² + y² + 4x – 6y + 4 = 0.
Solution:
C₁ = (2, 5), C₂ = (-2, 3)

= [\(\frac{4}{4}\), \(\frac{9}{2}\)] = [1, \(\frac{9}{2}\)]
Equation of the pair transverse of the common tangents is

Question 5.
Find the pair of tangents from (4, 10) to the circle x² + y² = 25.
Solution:
SS₁₁ = S²₁
(x² + y² – 25) (16 + 100 – 25)
= (4x + 10 y -25)²
91x² + 91y² – 2275
= [16x² + 100y² + 625 + 80xy – 200x – 500y]
75x² – 9y² – 80xy + 500y + 200x – 2900 = 0

Question 6.
Find the pair of tangents drawn from (0,0) to x2 + y2 + 10x + 10y + 40 = 0.
Solution:
S.S₁₁ = S²₁
(x² + y² + 10x + 10y + 40) (40) = [5x + 5y + 40]²
8(x² + y² + 10x + 10y + 40) = (x + y + 8)²5
8x² + 8y² + 80x + 80y + 320
= 5x² + 5y² + 10xy + 80x + 80y + 320
3x² + 3y² – 10xy = 0

III.

Question 1.
Find the equation of the circle which touches x² + y² – 4x + 6y – 12 = 0 at (-1, 1) internally with a radius of 2.
Solution:
C₁ = (2, -3)
r₁ = \(\sqrt{4+9+12}\) = 5

Let C₂ = (h, k)
point of contact – (x, y) = (-1, 1)
since the two circles touch internally

Equation of a circle with centre (\(\frac{1}{5}\), \(\frac{-3}{5}\)) and radius 2 is given by
(x – \(\frac{1}{5}\))² + (y + \(\frac{1}{5}\))² = 4
5x² + 5y² – 2x + 6y – 18 = 0

Question 2.
Find all common tangents of the following pairs of circles.
(i) x² + y² = 9 and x² + y² – 16x + 2y + 49 = 0
Solution:
The circles are x² + y² = 9
x² + y² – 16x + 2y + 49 = 0
Centre are A (0, 0), B(8, -1)

The circles lie outside each other.
A (0, 0), B (8, -1)
External center of similitude S divides AB externally in this ratio 3 : 4
Co-ordinates of are (-24, + 3)
Suppose m is the slope of the direct common tangents
y – 3 = m(x + 24)
= mx + 24m
mx – y 4- (24m + 3) = 0 ………… (1)
This is a tangent to the circle x² + y² = 9
3 = \(\frac{|24m+3|}{\sqrt{m^{2}+1}}\)
9(m² +1 = 9(m² + 1) = 9(8m + 1)²
= 64m² + 16m + 1
63m² + 16m = 0
m(63m + 16) = 0
m = 0 or \(\frac{-16}{63}\).

Case (i): m = 0
Substituting in (1), equation of the tangent is
-y + 3 = 0
y – 3 = 0

Case (ii) : m = \(\frac{-16}{63}\).
Equation of the tangent is

Internal center of similitude S’ divides AB internally in the ratio 3 : 4
Co-ordinates of S’ are (\(\frac{24}{7}\), \(\frac{-3}{7}\))
Equation of the transverse common tangent is

7y + 3 = 7mx – 24m
7mx – 7y – (24m + 3) = 0 ………… (2)
This is a tangent to the circle x² + y² = 9
3 = \(\frac{|24m+3|}{\sqrt{49m^{2}+49}}=\frac{3}{\sqrt{7}} \frac{|28m+1|}{\sqrt{m^{2}+1}}\)
49 (m² + 1) = (8m + 1)²
49m² + 49 = 64m² +16m + 1
15m² + 16m – 48 = 0
(3m – 4) (5m + 12) = 0
m = \(\frac{4}{3}\) or \(\frac{-12}{5}\)

Case (i): Substituting in (2), equation of the tangent is

Case (ii): m = \(\frac{-12}{5}\)
Equation of the transverse common tangent is ie

∴ Equation of direct common tangents are
y – 3 = 0; 16x + 63y + 195 = 0
Equation of transverse common tangents are
4x – 3y – 15 = 0 and 12x + 5y – 39 = 0

ii) x² + y² + 4x + 2y – 4 = 0 and x² + y² – 4x – 2y + 4 = 0
Solution:
r₁ = \(\sqrt{4+1+4}\) = 3, r₂ = \(\sqrt{4+1+4}\) = 1
External center of similitude S divides AB externally in the ratio 3 : 1
3 : 1 A (-2,-1), S’ B (2, 1) S
Co-ordinates of S are
(\(\frac{3.2-(-2)1}{3-1}\), \(\frac{3.1-1(-1)}{3-1}\)) = (4, 2)
Suppose m is the slope of the tangent Equation of the tangent can be taken as
y – 2 = m(x – 4)
= mx – 4m
mx – y + (2 – 4m) = 0 ………… (1)
This is a tangent to the circle
x² + y² + 4x + 2y – 4 = 0

Squaring and cross-multiplication is
(1 – 2m)² = (m² + 1)
4m² – 4m + 1 = m² + 1
3m² – 4m = 0
m(3m-4) = 0
m,= 0 or = \(\frac{4}{3}\)
m = 0
Substituting in (1), equation of the tangent is y + 2 = 0 or y -2 = 0
m = \(\frac{4}{3}\)
Substituting in (1), equation of the tangent is
\(\frac{4}{3}\)x – y + (2 – \(\frac{16}{3}\)) = 0
⇒ \(\frac{4}{3}\)x – y –\(\frac{10}{3}\) = 0
4x – 3y – 10 = 0
Internal centre of similitude S’ divides AB internally in the ratio 3 : 1
Co-ordinates of S’ are
(\(\frac{6-2}{3+1}\), \(\frac{3-1}{3+1}\)x) = (1, \(\frac{1}{2}\))
Equation of the tangent can be taken as
y – \(\frac{1}{2}\) = m(x – 1)
= mx – m
mx – y + (\(\frac{1}{2}\) – m) = 0 ………. (2)
This is a tangent to the circle
x² + y² + 4x + 2y – 4 = 0

Squaring and cross-multiply is
(1 – 2m)² = 4(m² + 1)
1 + 4m² – 4m = 4m² + 4
Here then value of m is a, so that the tangent is a vertical line, equation of the tangent is
x = \(\frac{1}{2}\)
x – 1 = 0
4m + 3 = 0
m = \(\frac{-3}{4}\)
Substituting in (2), equation of the tangent is
\(\frac{-3}{4}\)x – y + (\(\frac{1}{2}+\frac{3}{4}\)) = 0
-3x – 4y + 5 = 0
3x + 4y – 5 = 0
Equation of direct common tangents are
y – 2 = 0, 4x – 3y – 10 = 0
Equation of transvere common tangents are
x – 1 = 0, 3x + 4y – 5 = 0

Question 3.
Find the pair of tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.
Solution:
S.S₁₁ = S²₁
(x² + y² – 6x + 4y – 2)
(9+ 4 – 6 × 3 + 4 × 2 – 2)
= (3x + 2y – 3(x + 3) + 2(y + 2) – 2)²
(x² + y² – 6x + 4y – 2) = (4y – 7)²
x² + y² – 6x + 4y – 2 = 16y² – 56y + 49
x² – 15y² – 6x + 60y – 51 = 0

Question 4.
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.
Solution:
S.S₁₁ = S²₁
(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 – 60 y
9x² – 16y² – 18x + 96y – 135 = 0

Question 5.
Find the pair of tangents from the origin to the circle x² + y² + 2gx + 2fy + c = 0 and hence deduce a condition for these tangents to be perpendicular.
Solution:
S.S₁₁ = S²₁
(x² + y² + 2gx + 2fy + c) (c) = [gx + fy + c]²
= g²x² +f²y²+ 2gfxy + 2gcx + 2fyc + c²
(gx + fy)² = c(x² + y²)
g²x² + f²y² + 2fg xy = cx² + cy²
(g² – c)x + 2fgxy + (f² – c)y² = 0
co-efficient of x² + co-efficient of y² = 0
g² – c + f² – c = 0
g² + f² = 2c

Question 6.
From a point on the circle x² + y² + 2gx + 2fy + c = 0, two tangents are drawn to the circle x² + y² + 2gx + 2fy + c sin² α + (g² + f²) cos² α = 0 (0 < a < π/2). Prove that the angle between them is 2α.
Solution:
[x²₁ + y²₁ + 2gx₁ + 2fy₁ + c sin² α + (g² + f²) cos² α] (S)
= (xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c sin² α + (g² + f²) cos² α)²
[(-c + c sin² α)+ (g² + f²) cos² α]S
= (x (x₁ + g) + y (y₁ + f) + gx₁ + fy₁ + c sin² α + (g² + f²) cos² α)²
[cos² α (g² + f² – c)] S
= [x (x₁ + g) + y (y₁ + f) + gx₁ + fy₁ + c sin² α + (g² + f²)cos² α
Let g² + f² – c = r²
= [x (x₁ + g) + y (y₁ + f) + gx₁ + fy₁ + gx₁ + fy₁ + c + (cos² α).r²)²
Coefficient of x² is r² cos² α – (x₁ + g)²
Coefficient of y² is r² cos2 α – (y₁ + f)²
Coefficient of xy is
h. cos² α r² – 2 (x₁ + g) (y₁ + f)

cos θ = cos 2 α
θ = 2α
Hence proved.

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(d)

I.

Question 1.
Find the condition that the tangents drawn from (0, 0) to S ≡ x² + y² + 2gx + 2fy + c = 0 be perpendicular to each other.
Solution:
If θ is the angle between the pair of tangents from P₁ then tan \(\frac{\theta}{2}=\frac{r}{\sqrt{s_{11}}}\)

g² + f² – c = c
g² + f² = 2c
This is the required condition.

Question 2.
Find the chord of contact of (0, 5) with respect to the circle x² + y² – 5x + 4y – 2 = 0
Solution:
Equation of the chord of contact is S₁ = 0
i.e., xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
Equation of the circle is
S = x² + y² – 5x + 4y – 2 = 0
Equation of the chord of contact is
x. 0 + y. 5 – \(\frac{5}{2}\) (x + 0) + 2(y + 5) – 2 = 0
Multiplying with 2
10y – 5x + 4y + 20 – 4 = 0 – 5x + 14y +16 = 0
or 5x – 14y – 16 = 0

Question 3.
Find the chord of contact of (1, 1) to the circle x² + y² = 9.
Solution:
Equation of the circle is x² + y² = 9
Equation of the chord of contact is
x.1 + y. 1 =9
i.e. x + y = 9

Question 4.
Find the polar of (1, 2) with respect to x² + y² = 7.
Solution:
Polar of P(x₁, y₁) w.r. to S = 0
xx₁ + yy₁ = a²
x + 2y – 7 = 0 is the polar equation.

Question 5.
Find the polar of (3, -1) with respect to 2x² + 2y² = 11.
Solution:
Equation of circle is 2x² + 2y² = 11
x² + y² = \(\frac{11}{2}\)
Equation of polar is xx₁ + yy₁ = a²
x(3) + y(-1) = \(\frac{11}{2}\)
Equation-of polar is 6x – 2y -11 = 0

Question 6.
Find the polar of (1, -2) with respect to x² + y² – 10x – 10y + 25 = 0
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 pole of ax + by + c = 0 (c + 0) with respect to x² + y² = r².
Solution:
Let (x₁, y₁) be pole, the polar equation be
xx₁ + yy₁ – r² = 0 ………… (i)
ax + by + c = 0 ………….. (ii)
(i) and (ii) both are same (polar) equations.

Question 8.
Find the pole of 3x + 4y – 45 = 0 with respect to x² + y² – 6x – 8y + 5 = 0.
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

Pole is (6, 8).

Question 9.
Find the pole of x – 2y + 22 = 0 with respect to x² + y² – 5x + 8y + 6 = 0.
Solution:
Polar equation is
xx₁ + yy₁ – \(\frac{5}{2}\)(x + x₁) + 4(y + y₁) + 6 = 0
(or) x(x₁ – \(\frac{5}{2}\)) + y(y₁ + 4) – \(\frac{5}{2}\)
x₁ + 4y₁ + 6 = 0 …………. (i)
x – 2y + 22 = 0 ………….. (ii)
(i) and (ii) both are equations of the same line comparing it we get


y₁ = -3
∴ Pole is (2, -3)

Question 10.
Show that the points (-6, 1) and (2, 3) are conjugate points with respect to the circle x² + y² – 2x + 2y + 1 =0.
Solution:
Polar equation w.r.t. (2, 3) is
2x + 3y – 1(x + 2) + 1(y + 3) + 1 = 0
x + 4y + 2 = 0 ………. (i)
Now (-6, 1) should satisfy equation (i) if it is conjugate point
(-6) + 4(1) + 2 = 0
∴ (-6, 1) and (2, 3) are conjugate w.r.t. circle.

Question 11.
Show that the points (4, 2) (3, -5) are conjugate points with respect to the circle x² + y² – 3x – 5y + 1 = 0.
Solution:
Polar equation w.r.t. (4, 2) is
4x + 2y – \(\frac{3}{2}\)(x + 4) – \(\frac{5}{2}\)(y + 2) + 1 = 0 ………. (i)
Now (3, -5) should satisfy equation (i) if it is conjugate points.
4(3) – 5 × 2 – \(\frac{3}{2}\)(3 + 4) – \(\frac{5}{2}\)(-5 + 2) + 1
12 – 10 – \(\frac{21}{2}\) + \(\frac{15}{2}\) + 1 = 0
Hence given points are conjugate.

Question 12.
Find the value of k if kx + 3y – 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to the circle x² + y² – 2x – 4y – 4 = 0.
Solution:
Let pole be (x₁, y₁) polar equation w.r.t.
x² + y² – 2x – 4y – 4 = 0 be
xx₁ + yy₁ -1 (x + x₁) – 2(y + y₁) – 4 = 0
x(x₁ – 1) + y(y₁ – 2) – x₁ – 2y₁ – 4 = 0 …….. (i)
2x + y + 5 = 0 is polar so comparing with equation (i) we get

x₁ = -1, y₁ = 1 Pole (-1, 1)
kx + 3y – 1 = 0 is polar so it should satisfy
(1, 1)
k(-1) + 3(1) – 1 = 0
-k + 2 = 0
k = 2

Question 13.
Find the value of k if x + y – 5 = 0, 2x + ky – 8 = 0 are conjugate with respect to the circle x² + y² – 2x – 2y – 1 = 0.
Solution:
If l₁x + m₁y + n₁ = 0; l₂x + m₂y + n₂ = 0 are conjugate, S = 0 then,
r²(l₁l₂ + m₁m₂)
= (l₁g + m₁f – n₁) (l₂g + m₂f – n₂)
l₁ = 1, m₁ = 1, n₁ = -5
l₂ = 2, m₂ = k, n₂ = – 8
g = -1, f = -1, r² = 3
∴ 3(1.2 + k) = (-1 -1 + 5) (-2 – k + 8)
6k = 18 – 6 = 12 ⇒ k = 2

Question 14.
Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35.
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 15.
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.
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 = \(\frac{28}{2}\)

II.

Question 1.
Find the angle between the tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.
Solution:
Equation of the circle is
S ≡ x² + y² – 6x + 4y – 2 = 0

Angle between the tangent at P = cos^-1 (\(\frac{7}{8}\))

Question 2.
Find the angle between the pair of tangents dravyn from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0.
Solution:
Equation of the circle is
S₁ ≡ x² + y² – 2x + 4y – 11 = 0
r = \(\sqrt{1+4+11}\) = 4
S₁₁ =1 +9-2 + 12-11 = 9

Angle between the tangents = cos^-1 (\(\frac{7}{25}\))

Question 3.
Find the angle between the pair of tangents drawn from (0,0) to the circle x² + y² – 14x + 2y + 25 = 0.
Solution:
Equation of the circle is
x² + y² – 14x + 2y + 25 = 0
r = \(\sqrt{49+1-25}\) = 5
S₁₁ = 0 + 0 – 0 + 0 + 25 = 25 ⇒ √S₁₁ = 5
tan θ = \(\frac{r}{\sqrt{s_{11}}}\) = \(\frac{5}{5}\) = 1 ⇒ θ = \(\frac{\pi}{4}\)
Angle between the tangents = 2θ = \(\frac{\pi}{2}\)

Question 4.
Find the locus of P if the tangents drawn from P to x² + y²= a² include an angle α.
Solution:
Equation of the circle is x² + y² = a²
r = radius = a
Suppose θ (x₁, y₁) is any point as the locus

Question 5.
Find the locus of P if the tangents drawn from P to x2 + y2= a2 are perpendicular to each other.
Solution:
Equation of the circle is x² + y² = a²
r = a
Let P(x₁, y₁) be any point on the locus
S₁₁ = x²₁ + y²₁ – a²

Squaring and cross-multiplying
a² = x² + y² – a²
x² + y² = 2a²
Locus of P (x₁, y₁) is x² + y²= 2a²

Question 6.
Find the slope of the polar of (1, 3) with respect to the circle x² + y² – 4x – 4y – 4= 0. Also find the distance from the centre to it.
Solution:
Equation of the circle is x² + y² – 4x – 4y – 4 =0
Polar of P (1, 3) is
x.1 + y.3 – 2(x + 1) – 2(y + 3) – 4 = 0
x + 3y – 2x – 2 – 2y – 6 – 4 = 0
-x + y – 12 = 0
Slope of the polar is – \(\frac{a}{b}=\frac{1}{1}\) = 1
Distance from the centre

Question 7.
If ax + by + c = 0 is the polar of (1, 1) with respect to the circle x² + y² – 2x + 2y +1=0 and H.C.F. of a, b, c is equal to one then find a² + b² + c².
Solution:
Equation of the circle is
x2+ y2-2x + 2y +1 = 0
Polar of (1, 1) w.r.to the circle is
x.1 + y.1 -(x + 1) + (y + 1) + 1 = 0
x + y – x – 1 + y + 1 + 1 = 0
2y + 1 = 0 …………. (1)
Given equation of the line ax + by + c = 0 ……….. (2)
Comparing (1) and (2)
\(\frac{a}{0}=\frac{b}{2}=\frac{c}{1}\) = k
a = 0, b = 2k, c = k
a² + b² + c² = 0 + 4k² + k² = 5k²
H.C.F of (a, b, c) = 1 ⇒ k = 1
a² + b² + c2 = 5(1)² = 5

III.

Question 1.
Find the coordinates of the point of in-tersection of tangents at the points where x + 4y – 14 = 0 meets the circle x² + y² – 2x + 3y – 5 = 0.
Solution:
Equation of the given circle is
x² + y² – 2x + 3y – 5 = 0
polar of P(x₁, y₁) is

2xx₁ + 2yy₁ – 2x – 2x₁ + 3y + 3y₁ – 10 = 0
2(x₁ – 1)x + (2y₁ + 3)y – (2x₁ – 3y₁ + 10) = 0 ……….. (1)
Equation of QR is x + 4y – 14 = 0 ……… (2)
Comparing (1) and (2)

∴ Co-ordinates of p are (\(\frac{109}{76}\), \(\frac{9}{38}\))

Question 2.
If the polar of the points on the circle x² + y² = a² with respect to the circle x² + y² = b² touches the circle x² + y² = c² then prove that a,b,c are in Geometrical progression.
Solution:

P(x₁, y₁) is a point on the circle x² + y² = a²
∴ x²₁ + y²₁ = a² ……. (1)
Polar of P w.r.to the circle x²+ y² = b² is xx₁ + yy₁ = b²
There is a tangent to the circle x² + y² = c²

Cross – multiplying we get b2 = ac
∴ a, b, c are in Geometric progression.

Question 3.
Tangents are drawn to the circle x² + y² = 16 from the point P(3, 5). Find the area of the triangle formed by these tangents and the chord of contact of P.
Solution:
Equation of the circle is x² + y² = 16

Polar of P (3, 5) is 3x + 5y = 16
PL = Length of the perpendicular from P
\(\frac{|9+25-16|}{\sqrt{9+25}}=\frac{18}{\sqrt{34}}\)
Centre of the circle = C (0, 0)
P = Length of the perpendicular from c

Question 4.
Find the locus of the point whose polars with respect to the circles x² + y² – 4x – 4y – 8 = 0 and x² + y² – 2x + 6y – 2 = 0 are mutually perpendicular.
Solution:
Suppose P(x, y) is any position the locus Equation of the circles are
x² + y² – 4x – 4y – 8 = 0 ………….. (1)
x² + y² – 2x + 6y – 2 = 0 …………. (2)
Equation of the polar of P w.r.to circle (1) is
xx₁ + yy₁ – 2(x + x₁) – 2(y + y₁) – 8 = 0
x(x₁ – 2) + y(y₁ – 2) – (2x₁ +2y₁ + 8) = 0 ………… (3)
Polar of P w.r. to circle (2) is
xx₁ + yy₁ – 1(x + x₁) – 3(y + y₁) – 2 = 0
xx₁ + yy₁ – x – x₁ + 3y + 3y₁ – 2 = 0
x(x₁ – 1) + y(y₁ + 3) – (x₁ + 3y₁ + 2) = 0 ………. (4)
(3) and (4) are perpendicular 0
⇒ a₁a₂ + b₁b₂
(x₁ – 2) (x₁ – 1) + (y₁ – 2) (y₁ + 3) = 0
x²₁ – 3x₁ + 2 + y²₁ + y₁ – 6 = Q
x²₁ + y²₁ – 3x₁ + y₁ – 4 = 0
Locus of p(x₁, y₁) is x² + y² – 3x+ y – 4 = 0

Question 5.
Find the locus of the foot of the perpen¬dicular drawn from the origin to any chord of the circle S ≡ x² + y² + 2gx + 2fy + c = 0 which subtends a right angle at the origin.
Solution:

Suppose L (x₁, y₁) is the foot of the per-pendicular from the origin on the chord QL
Slope a QL = \(\frac{y_{1}}{x_{1}}\)
Slope of QR = \(\frac{x_{1}}{y_{1}}\)
Equation of QR is y – y₁ = \(\frac{x_{1}}{y_{1}}\)(x – x₁)
yy₁ – y²₁ = -xx₁ + x²₁
xx₁ + yy₁ = x²₁ + y²₁
or \(\frac{xx_{1}+yy_{1}}{x_{1}^{2}+y_{1}^{2}}\) = 1 …………. (1)
Equation of the circle is
x² + y² + 2gx + 2fy + c = 0 …………. (2)
Hamogenising (2) with the help of (1), combined equation of
QQ, OR is x² + y² + (2gx + 2fy)

OQ, OR are perpendicular
Co – eff. of x² + co- eff of y² = 0

2(x²₁ + y²₁) + 2gx₁ + 2fy₁ + c =
Locus of L (x₁, y₁)is
2(x² + y²) + 2gx + 2fy + c = 0

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(c)

I.

Question 1.
Find the equation of the tangent at P of the circle S = 0 where P and S are given by
i) P = (7, -5), S ≡ x² + y² – 6x + 4y – 12
Solution:
Equation of the circle
S ≡ x² + y² – 6x + 4y – 12 = 0
Equation of the tangent at P(7, -5) is
S₁ = xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x. 7 + y(-5) – 3(x + 7) + 2(y – 5) – 12 = 0
⇒ 7x – 5y – 3x – 21 + 2y – 10 – 12 = 0
4x – 3y – 43 = 0

ii) P = (-1, 1), S ≡ x² + y² – 6x + 4y – 12
Solution:
Equation of the tangent at P is
x(-1) + y . 1 – 3(x – 1) + 2(y + 1) – 12 = 0
⇒ x + y – 3x + 3 + 2y + 2 – 12 = 0
⇒ -4x + 3y – 7 = 0
⇒ 4x – 3y + 7 = 0

iii) P = (-6, -9), S ≡ x² + y² + 4x + 6y – 39
Solution:
Equation of the tangent at P is S₁ =0
(i.e.,) x(-6) + y(-9) + 2(x – 6) + 3(y – 9) – 39 = 0
⇒ -6x – 9y + 2x – 12 + 3y – 27 – 39 = 0
⇒ – 4x- 6y- 78 = 0
⇒ 4x + 6y + 78 = 0
⇒ 2x + 3y + 39 = 0

iv) P = (3, 4), S ≡ x² + y² – 4x – 6y + 11
Solution:
Equation of the tangent at P is S₁ = 0
⇒ x(3) + y(4) – 2 (x + 3) – 3(y + 4) + 11 =0
3x + 4y – 2x – 6 – 3y – 12 + 11 =0
x + y – 7 = 0

Question 2.
Find the equation of the normal at P of the circle S = 0 where P and S are given by
i) P = (3, -4), S ≡ x² + y² + x + y – 24
Solution:
Equation of the normal is
(x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0
(x – 3) (-4 + \(\frac{1}{2}\)) – (y + 4) (3 + \(\frac{1}{2}\)) = 0
–\(\frac{7}{2}\)(x – 3) – \(\frac{7}{2}\) (y + 4) = 0
⇒ (x – 3) + (y + 4) = 0
x – 3 + y + 4 = 0
x + y + 1 = 0

ii) P = (3, 5), S ≡ x² + y² – 10x – 2y + 6
Solution:
Equation of the normal is
(x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0
(x – 3) (5 – 1) – (y – 5) (3 – 5) = 0
4x – 12 + 2y – 10 = 0
4x + 2y – 22 = 0
or
2x + y – 11 = 0

iii) P = (1, 3), S ≡ 3(x² + y²) – 19x – 29 y + 76
Solution:
Equation of the circle is
x² + y² – \(\frac{19}{3}\)x – \(\frac{29}{3}\)y + \(\frac{76}{3}\) = 0
Equation of the normal is
(x – 1)(3 – \(\frac{29}{6}\)) – (y – 3) (1 – \(\frac{19}{6}\) ) = 0
– \(\frac{11}{6}\)(x – 1) + \(\frac{13}{6}\)(y – 3) = 0
11(x – 1) – 13(y – 3) = 0
11x – 11 – 13y + 39 = 0
11x – 13y + 28 = 0

iv) P = (1, 2), S ≡ x² + y² – 22x – 4y + 25
Solution:
Equation of the normal at P is
(x – 1) (2 – 2) – (y – 2) (1 – 11) = 0
10(y – 2) = 0 ⇒ y – 2 = 0
or y = 2

II.

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

Equation of the line is y = x – 3 ⇒ x – y – 3 = 0
P distance from the centre
\(=\frac{\left|\frac{1}{2}+\frac{3}{2}-3\right|}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\)
Length of the chord = \(\sqrt{r^{2}-P^{2}}\)

Question 2.
Find the length of the chord intercepted by the circle x² + y² – 8x – 2y – 8 = 0 on the line x + y + 1 = 0
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 Sine 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 3.
Find the length of the chord formed by x² + y² = a² on the line x cos α + y sin α = p.
Solution:
Equation of the circle is x² + y² = a²
Centre C(0, 0), r = a
Equation of the line is
x cos α + y sin α – p = 0
P = distance from the centre
= \(\frac{|0+0-p|}{\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}}\) = p
Length of the chord = 2\(\sqrt{r^{2}-p^{2}}\)

Question 4.
Find the equation of circle with centre (2, 3) and touching the line 3x – 4y + 1 =0.
Solution:

Equation of circle (x – h)² + (y – k)² = r²
(x – 2)² + (y – 3)² = 1
x² + y² – 4x – 6y + 12 = 0

Question 5.
Find the equation of the circle with centre (-3, 4) and touching y – axis.

Solution:
Centre of the circle is C(-3, 4)
The circle touches y – axis
r = x co – ordinates of c = |-3| = 3
Equation of the circle is (x + 3)² + (y – 4)² = 9
x² + 6x + 9 + y² – 8y + 16 – 9 = 0
x² + y² + 6x – 8y + 16 = 0

Question 6.
Find the equation of tangents of the circle x² + y² – 8x – 2y + 12 = 0 at the points whose ordinates are 1.
Solution:
Equation of the circle is
x² + y² – 8x – 2y + 12
Suppose co – ordinates of P are (x₁, 1)
P is a point on the circle
x²₁ + 1 – 8x₁ – 2 + 12 = 0
x²₁ – 8x₁ + 11 = 0

Co – ordinates of P are (4 + √5, 1) and Q(4 – √5, 1)
Equation of the tangent at P(4 + √5, 1) is
x(4 + √5) + y • 1 – 4(x + 4 + √5) – (y + 1) + 12 = 0
⇒ 4x + √5x + y – 4x – 16 – 4√5 – y – 1 + 12 = 0
⇒ √5x – 5 – 4√5 = 0 ⇒ √5 (x – √5 – 4) = 0
⇒ x – √5 – 4 = 0
x = 4 + √5
Equation of the tangent at Q(4 – √5, 1) is
⇒ x(4 – √5) + y. 1 – 4(x + 4 – √5) – (y + 1) + 12 = 0
⇒ 4x – √5x + y – 4x – 16 + 4√5 – y- 1 + 12 = 0
⇒ -√5x + 4√5 – 5 = 0
⇒ -√5 (x – 4 + √5) = 0
⇒ x – 4 + √5 = 0
x = 4 – √5

Question 7.
Find the equation of tangents of the circle x² + y² – 10 = 0 at the points whose abscissae are 1.
Solution:
Equation of the circle is x² + y² = 10
Suppose co-ordinates of P are (1, y₁)
1 + y²₁ = ,10 ⇒ y²₁ =9
y₁ = ±3
Co-ordinates of P are (1, 3) and (1, -3)
Equation of the tangent at P(1, 3) is
x. 1 + y. 3 = 10
x + 3y – 10 = 0
Equation of the tangent of P(1, -3) is
x. 1 + y(-3) = 10 ⇒ x – 3y – 10 = 0

III.

Question 1.
If x² + y² = c² and \(\frac{x}{a}+\frac{y}{b}\) = 1 intersect at A and B, then find \(\overline{\mathrm{AB}}\). Hence deduce the condition, that the line touches the circle.
Solution:
x² + y² = c²
C = (0, 0) r = c

Length of chord = 2\(\sqrt{r^{2}-d^{2}}\)

Line touches circle when r² = d² or r = d
Length of the chord = 0
c = \(\sqrt{\frac{a^2 b^2}{a^2+b^2}}\)

Question 2.
The line y = mx + c and the circle x² + y² = a² intersect at A and B. If AB = 2λ, then show that c² =(1 + m²) (a² – λ²)
Solution:
C = (0, 0) r = a
Length of chord 2\(\sqrt{r^{2}-d^{2}}\)
2\(\sqrt{r^{2}-d^{2}}\) = 2λ
r² – d² = λ²
y = mx + c ⇒ mx – y + c = 0
d = \(\frac{|0-0+c|}{\sqrt{m^{2}+1}}\)
d² = \(\frac{c^{2}}{m^{2}+1}\)
r² – d² = λ² and r = a
a² – d² = λ²
a² – λ² = d² = \(\frac{c^{2}}{m^{2}+1}\)
∴ c² = (a² – λ²) (1 + m²) is the required condition.

Question 3.
Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0
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 4.
Find the equation of tangent and normal at (3, 2) of the circle x² + y² – x – 3y – 4 = 0.
Solution:
Equation of the circle is x² + y² – x – 3y – 4 = 0
Equation of the tangent at P (3, 2) is
x. 3 + y. 2 – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0
6x + 4y – x – 3 – 3y – 6 – 8 = 0
5x + y – 17 = 0
The normal is perpendicular to the tangent Equation of the normal can be taken as x – 5y + c = 0
The normal passes through P(3, 2)
3 – 10 + c = 0 ⇒ c = 7
Equation of the normal is x – 5y + 7 = 0

Question 5.
Find the equation of the tangent and normal at (1, 1) to the circle 2x² + 2y² – 2x – 5y + 3 = 0.
Solution:
Equation of the circle is
2x² + 2y² – 2x – 5y + 3 = 0
x² + y² – x – \(\frac{5}{2}\)y – \(\frac{3}{2}\) = 0
Equation of the tangent at P(1, 1) is
x. 1 + y.1 – \(\frac{1}{2}\)(x + 1) – \(\frac{5}{4}\)(y + 1) + \(\frac{3}{2}\) = 0
4x + 4y – 2 (x + 1) – 5(y +1) + 6 = 0
4x + 4y – 2x – 2-5y- 5 + 6 = 0
2x – y – 1 = 0
Equation of the normal can be taken as
x + 2y + k = 0
The normal passes through P(1, 1)
⇒ 1 + 2 + k = 0 ⇒ k = -3
Equation of the normal is x + 2y – 3 = 0

Question 6.
Prove that the tangent at (3, -2) of the cjrcle x² + y² = 13 touches the circle x² + y² + 2x – 10y – 26 = 0 and find its point of contact.
Solution:
Equation of the circle is x² + y² = 13
Equation of the tangent at P(3, -2) is
x. 3 + y (-2) = 13
3x – 2y – 13 = 0
Equation of the second circle is
x² + y² + 2x – 10y – 26 = 0
Centre is C(-1, 5) r = \(\sqrt{1+25+26}\)
= √52 = 2√3
P = length of the perpendicular from

= radius
∴ The tangent to the first circle also touches the second circle.
Equation of the circle is
x² + y² + 2x – 10y – 26 = 0
⇒ 3x – 2y – 13 = 0

⇒ 13x² – 130 x + 325 = 0
x² – 10x + 25 = 0 ⇒ (x – 5)² = 0
x – 5 = 0 ⇒ x = 5
x = 5, y = 1 (5, 1) is point of contact

Question 7.
Show that the tangent at (-1, 2) of the circle x² + y² – 4x – 8y + 7 = 0 touches the circle x² + y² + 4x + 6y = 0 and also find its point of contact.
Solution:
Tangent at (-1, 2) to
x² + y² – 4x – 8y + 7 = 0 is
⇒ x(-1) + y(2) – 2(x – 1) – 4(y + 2) + 7 = 0
⇒ -3x – 2y + 1 = 0 (or) 3x + 2y – 1 = 0
If 3x + 2y – 1 = 0 is tangent to
x² + y² + 4x + 6y = 0, then radius of circle should be equal to perpendicular from centre to line 3x + 2y – 1 = 0
C: (-2, -3)

d = r
Hence 3x + 2y – 1 = 0 is also tangent to x² + y² + 4x + 6y = 0
point of contact (foot of perpendicular)
Let (h, k) be foot of perpendicular

h = 1, k = -1
(1, -1) is point of contact.

Question 8.
Find the equations of the tangents to the circle x² + y² – 4x + 6y – 12 = 0 which are parallel to x + y – 8 = 0. .
Solution:
Equation of the circle is
x² + y² – 4x + 6y – 12 = 0
Centre is C(2, -3); r = radius
= \(\sqrt{4+9+12}\) = 5
Equation of the given line is x + y – 8 = 0
The tangent is parallel to this line
Equation of the tangent can be taken as x + y + k = 0
Length of the perpendicular from
c = \(\frac{|2-3+k|}{\sqrt{1+1}}\)
⇒ |k – 1| = 5√2
⇒ k – 1 = ± 5√2
⇒ k = 1 ± 5√2
Equation of the required tangent is
x + y + 1 ± 5√2 = 0

Question 9.
Find the equations of the tangents to the circle x² + y² + 2x – 2y – 3 = 0 which are perpendicular to 3x – y + 4 = 0.
Solution:
x² + y² + 2x – 2y – 3 = 0, C : (-1, 1)
r = \(\sqrt{1+1+3}\) = √5
y = –\(\frac{1}{3}\)x + k is line ⊥ to 3x – y + 4 = 0
x + 3y – 3k = 0
√5 = \(\frac{|-1+3+3k|}{\sqrt{1+9}}\)
Squaring on both sides
(3k + 2)² = 50
9k² + 12k + 4 – 50 = 0
9k² + 12k – 46 = 0
k = \(\frac{-12 \pm \sqrt{144+1656}}{18}=\frac{-12 \pm 30 \sqrt{2}}{18}\)
= \(\frac{6(-2 \pm 5 \sqrt{2})}{18}\) ⇒ 3k = -2 ± 5√2
Equation of the required tangent is
x + 3y – 2 ± 5√2 = 0

Question 10.
Find the equation of the tangents to the circle x² + y² – 4x – 6y + 3 = 0 which makes an angle 45° with X – axis.
Solution:
Equation of the circle is x² + y² – 4x – 6y + 3 = 0
Centre C(2, 3), r = \(\sqrt{4+9-3}\) = √10
Slope of the tangent m = tan 45° = 1
Equation of the tangent can be taken as
y = x + c
x – y + c = 0
Length of the perpendicular from centre =

Equation of the tangents
x – y + 1 ± 2 √5 = 0

Question 11.
Find the equation of the circle passing through (-1, 0) and touching x + y – 7 = 0 at (3, 4).
Solution:
Suppose equation of the circle is
x² + y² + 2gx + 2fy + c = 0
It passes through A (-1, 0)
1 + 0 – 2g + 0 + c = 0
2g – c = 1 ……….. (1)
Equation of the tangent at P(3, 4) is
3x + 4y + g(x + 3) + f(y + 4) + c = 0
(3 + g)x + (4 + f)y + (3g + 4f + c) = 0 ………….. (2)
Given equation of the tangent is
x + y – 7 = 0 ……….. (3)
Comparing (2) and (3)

From (1)- 2 – c = 1 ⇒ c = -3
Equation of the circle is
x² + y² – 2x – 4y – 3 = 0

Question 12.
Find the equations of the circles passing through (-1, 1), touching the lines 4x + 3y + 5 = 0 and 3x – 4y – 10 = 0.
Solution:
Suppose equation of the circle is
x² + y² + 2gx + 2fy + c = 0
Let centre be (-g, -f) 1 from centre to lines be equal (radius)

Case (i):
If g = -1; f = -2
Circle is passing through (1, -1)
∴ x² + y² + 2gx + 2fy + c = 0
1 + 1 + 2g – 2f + c = 0
2 – 2 + 4 + c = 0 (or) c = – 4
Required equation of circle be
x² + y² – 2x + 4y – 4 = 0
Case (ii):

∴ Required equation of circle be
25(x² + y²) – 26x +68y + 44 = 0

Question 13.
Show that x + y + 1 =0 touches the circle x² + y² – 3x + 7y + 14 = 0 and find its point of contact.
Solution:
x² + y² – 3x + 7y + 14 = 0

Perpendicular distance from centre on x + y + 1 = 0 if equals to radius, then x + y + 1 = 0 is tangent

∴ x + y + 1 is tangent y = -x – 1
Substituting value of ‘y’ in equation
x² + y² – 3x + 7y + 14 = 0, We get
x² + (-x – 1)² – 3x + 7 (-x – 1) + 14 = 0
⇒ 2x² – 8x + 8 = 0
⇒ x² – 4x + 4 = 0 (or) x = 2
y – x – 1, y = -3
Point of contact is (2, -3)