Question

A hyperbola having the transverse axis of length $2\sin \theta$ unit, 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

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

Ellipse is $3x^2+4y^2=12$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$
$b^2=a^2(1-e^2)$
$\Rightarrow e=\frac{1}{2}$
Foucs of ellipse $(\pm ae,0)\equiv (\pm 1,0)$
Let equation of confocal hyperbola is
$\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1$
and eccentricity $=e_1$
Given, length of transvese axis $2\sin \theta =2a_1$
$\Rightarrow a_1=\text{sin}\theta$
Now, focus of hyperbola is same as the ellipse
$\Rightarrow a_1{e}_1=1$
$\Rightarrow e_1^2{a}_1^2=a_1^2+b_1^2$
$\Rightarrow 1=a_1^2+b_1^2$
$\Rightarrow b_1^2={\cos }^2\theta$
Thus, the equation of hyperbola is
$\frac{x^2}{{\text{sin}}^2\theta }-\frac{y^2}{{\text{cos}}^2\theta }=1$
$\Rightarrow x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$