Question

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. $x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta=1$

Answer 1

Answer: (A)

The length of transverse axis is
$2 \sin \theta=2 a$
or $a=\sin \theta$
Also, for the ellipse
$3 x^{2}+4 y^{2}=12$
or $ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
$ a^{2}=4, b^{2}=3 $
$\therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Hence, the focus of ellipse is $(2 \times 1 / 2,0) \equiv(1,0)$.
As the hyperbola is confocal with the ellipse, the focus of the hyperbola is $(1,0)$.
Now, $ae = 1$
or $\sin\theta \times e = 1$
or $e = \text{cosec}\,\theta$
$\therefore b^2 = a^2(e^2 - 1) = \sin^2 \theta (\text{cosec}^2 \theta -1) = \cos^2 \theta$
Therefore, the equation of hyperbola is
$\frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$
or $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$