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. $x^{2} \text{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$
• B. $x^{2} \sec ^{2} \theta-y^{2} \text{cosec}^{2} \theta=1$
• 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 given equation of ellipse can be written as
$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
Now, $a=2, b=\sqrt{3} \Rightarrow e=\frac{1}{2}$
The foci of ellipse is $(\pm 1,0)$.
Since the hyperbola is confocal with ellipse, the foci of Hyperbola is $(\pm 1,0)$.
Now, let $e'$ be the eccentricity of hyperbola. The transverse axis is
$2 \sin \theta=2 a ^{\prime}$
Therefore, the focus is
$\left(e^{\prime} \sin \theta, 0\right)=(1,0)$
$\Rightarrow e^{\prime}=\text{cosec} \theta$
Now, $b^{\prime 2}=a^{\prime 2}\left(e^{\prime 2}-1\right)$
$=\sin ^{2} \theta\left(\text{cosec}^{2} \theta-1\right)$
$=1-\sin ^{2} \theta$
$=\cos ^{2} \theta$
Therefore, the equation of hyperbola is
$\frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$
That is,
$x^{2} \text{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$