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

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

Given ellipse may be written as $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\Rightarrow a^2=4,b^2=3$

$\Rightarrow e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
$\therefore$ Focus of the ellipse $=(\pm ae,0)=(\pm 1,0)$
Given required hyperbola is confocal to the ellipse
Let $a^{\prime },b^{\prime },e^{\prime }$ are transverse axis, conjugate axis an eccentricity of the hyperbola
$a^{\prime }{e}^{\prime }=1\Rightarrow \sin \theta .e^{\prime }=1\Rightarrow e^{\prime }=\frac{1}{\sin \theta }$
Using $b^{\prime 2}=a^{\prime 2}(e^2-1)\Rightarrow b^{\prime 2}=1-{\sin }^2\theta ={\cos }^2\theta$
Therefore required hyperbola is $\frac{x^2}{a^{\prime 2}}-\frac{y^2}{b^{\prime 2}}=1$
$\Rightarrow \frac{x^2}{{\sin }^2\theta }-\frac{y^2}{{\cos }^2\theta }=1$
$\Rightarrow x^2{\text{cosec}}^2\theta -y^2{\sec }^2\theta =1$

Hence, option 'A' is correct.