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. $x^2{\text{cosec}}^2\theta -y^{2}se{c}^2\theta =1$
• B. $x^{2}se{c}^2\theta -y^2{\text{cosec}}^2\theta =1$
• C. $x^{2}se{c}^2\theta -y^2{\cos }^2\theta =1$
• D. $x^2{\cos }^2\theta -y^2{\sin }^2\theta =1$

Answer 1

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

The given ellipse is
$\frac{x^2}{4}+\frac{y^2}{3}=1$
$a=2,b=\sqrt{3}$
$3=4(1-e^2)$
$e=\frac{1}{2}$
So that , $ae=1$
Hence , the eccentricity $e_1$ of the hyperbola is given by
$1=e_{1}sin\theta$
$e_1=cosec\theta$
$b^2=si{n}^2\theta (cose{c}^2\theta -1)$
$b^2=co{s}^2\theta$
Hence , the Hyperbola is
$\frac{x^2}{si{n}^2\theta }-\frac{y^2}{co{s}^2\theta }=1$
or
$x^{2}cose{c}^2\theta -y^{2}se{c}^2\theta =1$