Question

A hyperbola, having the transverse axis of length $2\sin \theta$ is confocal with the ellipse $3x^2+4y^2=12.$ Its equation is

• A. $x^2{\sin }^2\theta -y^2{\cos }^2\theta =1$
• B. $x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$
• C. $(x^2+y^2){\sin }^2\theta =1+y^2$
• D. $x^2{\text{cosec}}^2\theta =x^2+y^2+{\sin }^2\theta$

Answer 1

The correct option is B $x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$

The given ellipse is,
$\frac{x^2}{4}+\frac{y^2}{3}=1$

$e^2=1-\frac{b^2}{a^2}$
$\Rightarrow e^2=1-\frac{3}{4}$
$\Rightarrow e=\frac{1}{2}$
$\therefore S(ae,0)\Rightarrow S(1,0)$


$2A=2\sin \theta$
$\Rightarrow A=\sin \theta$

For hyperbola foci are same
$A{e}_1=ae=1$
$\Rightarrow \left(\sin \theta \right)e_1=1$

$\Rightarrow B^2=A^2(e_1^2-1)=(A{e}_1{)}^2-A^2$
$\Rightarrow B^2=1-{\sin }^2\theta ={\cos }^2\theta$

Hence, equation of hyperbola is,
$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1\Rightarrow \frac{x^2}{{\sin }^2\theta }-\frac{y^2}{{\cos }^2\theta }=1\Rightarrow x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$