Question

A hyperbola, having the transverse axis of length $2 \, \sin \, \theta$ is confocal with the ellipse $3x^2 + 4y^2 = 12$. Its equation is

• A. $x^2 \, \sin^2 \theta - y^2 \, \cos^2 \theta = 1 $
• B. $x^2 \, cosec^2 \theta - y^2 \, \sec^2 \theta = 1 $
• C. $(x^2 + y^2) \sin^2 \theta = 1 + y^2$
• D. $x^2 \, cosec^2 \theta = x^2 + y^2 \, \sin^2 \theta $

Answer 1

The Correct Option is B

Given, $2 a_{1}=2 \sin \theta$
$\Rightarrow a_{1}=\sin \theta $
and $ 3 x^{2}+4 y^{2}=12$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$
Here, $ a^{2}=4 $ and $ b^{2}=3$
$\therefore b^{2}=a^{2}\left(1-e^{2}\right)$
$ \Rightarrow 3 =4\left(1-e^{2}\right) $
$\Rightarrow e^{2} =1-\frac{3}{4}=\frac{1}{4} $
$\Rightarrow e =\frac{1}{2} $
Focus, $ F(a e, 0) =F\left(2 \times \frac{1}{2}, 0\right) $
$=F(1,0) $
For hyperbola foci are same
$\therefore a_{1} e_{1}=a e=1$
$\therefore (\sin \theta) e_{1}=1$
$\Rightarrow e_{1}=cosec\, \theta$
and $ b_{1}^{2}=a_{1}^{2}\left(e_{1}^{2}-1\right)=a_{1}^{2} e_{1}^{2}-a_{1}^{2} $
$ \Rightarrow b_{1}^{2}=1-\sin ^{2} \theta=\cos ^{2} \theta $
$ \frac{x^{2}}{a_{1}^{2}}-\frac{y^{2}}{b_{1}^{2}} =1 $
$ \Rightarrow \frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1 $
$ \Rightarrow x^{2} cosec^{2} \theta-y^{2} \sec ^{2} \theta=1$