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) $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$
$(2007,3 M)$
(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) $x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1$
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Answer:
Correct Answer: 11. (b)
Solution:
- The given ellipse is
$$ \begin{aligned} & & \frac{x^{2}}{4}+\frac{y^{2}}{3} & =1 \Rightarrow a=2, b=\sqrt{3} \\ & \therefore & 3 & =4\left(1-e^{2}\right) \Rightarrow \quad e=\frac{1}{2} \\ & \text { So, } & a & =2 \times \frac{1}{2}=1 \end{aligned} $$
Hence, the eccentricity $e _1$ of the hyperbola is given by
$$ \begin{array}{rlrl} & e _1 & =\operatorname{cosec} \theta \quad[\because a e=e \sin \theta] \\ \Rightarrow \quad & b^{2} & =\sin ^{2} \theta\left(\operatorname{cosec}^{2} \theta-1\right) & =\cos ^{2} \theta \end{array} $$
Hence, equation of hyperbola is
$\frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$ or $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$.
