Hyperbola 1 Question 10
11.
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^{2}+4 y^{2}=12$. Then, its equation is
(a) $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$
(b) $x^{2} \sec ^{2} \theta-y^{2} \operatorname{cosec}^{2} \theta=1$
(c) $x^{2} \sin ^{2} \theta-y^{2} \cos ^{2} \theta=1$
(d) $x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1$
$(2007,3 M)$
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Answer:
Correct Answer: 11. (a)
Solution:
- The given ellipse is
$ \frac{x^{2}}{4}+\frac{y^{2}}{3} =1 \Rightarrow a=2, b=\sqrt{3} $
$\therefore 3 =4\left(1-e^{2}\right) \Rightarrow \quad e=\frac{1}{2} $
$\text { So, } a =2 \times \frac{1}{2}=1$
Hence, the eccentricity $e _1$ of the hyperbola is given by
$e _1 =\operatorname{cosec} \theta \quad[\because a e=e \sin \theta] $
$\Rightarrow \quad b^{2} =\sin ^{2} \theta\left(\operatorname{cosec}^{2} \theta-1\right) =\cos ^{2} \theta$
Hence, equation of hyperbola is
$\frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$ or $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$.
