Question

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

• A. $x^{2}cose{c}^2\theta -y^{2}se{c}^2\theta =1$
• B. $x^{2}se{c}^2\theta -y^{2}cose{c}^2\theta =1$
• C. $x^{2}si{n}^2\theta -y^{2}co{s}^2\theta =1$
• D. $x^{2}co{s}^2\theta -y^{2}si{n}^2\theta =1$

Answer 1

The correct option is A

$x^{2}cose{c}^2\theta -y^{2}se{c}^2\theta =1$

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