Question

A hyperbola, having the transverse axis of length $2\sin \theta$, is confocal with the ellipse $3x^2+4y^2=12$. 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 $e^{\prime }$ be the eccentricity of hyperbola. The transverse axis is
$2\sin \theta =2a^{\prime }$
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,
$x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$