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A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then its equation is
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$
$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$
$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$
$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$
Solution
The given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\Rightarrow \mathrm{a}=2, \mathrm{~b}=\sqrt{3} \Rightarrow 3=4\left(1-\mathrm{e}^2\right) \Rightarrow \mathrm{e}=\frac{1}{2}$
so that $\mathrm{ae}=1$
Hence the eccentricity $e_1$, of the hyperbola is given by
$ 1=e_1 \sin \theta \Rightarrow e_1=\operatorname{cosec} \theta $
$ \Rightarrow b^2=\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta$
Hence the hyperbola is $\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1$ or $x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$
