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. $x^2{\mathrm{cosec}}^2\theta -y^2{\sec }^2\theta =1$
• B. $x^2{\sec }^2\theta -y^2{\mathrm{cosec}}^2\theta =1$
• 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
$2\sin \theta =2a$
or $a=\sin \theta$
Also, for the ellipse
$3x^2+4y^2=12$
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 $x^2{\mathrm{cosec}}^2\theta -y^2{\sec }^2\theta =1$