A hyperbola, having the transverse axis of le | vedclass

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Question

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) \Ri

Vedclass · Accepted Answer

$x^2 \operatorname{cosec}^2 	heta-y^2 \sec ^2 	heta=1$

Vedclass · Answer

The given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\Rightarrow \mathrm{a}=2, \mathrm{~b}=\sqrt{3} \Rightarrow 3=4\left(1-\mathrm{e}^2ight) \Rightarrow \mathrm{e}=\frac{1}{2}$

so that $\mathrm{ae}=1$

Hence the eccentricity $e_1$, of the hyperbola is given by

$ 1=e_1 \sin 	heta \Rightarrow e_1=\operatorname{cosec} 	heta $

$ \Rightarrow b^2=\sin ^2 	heta\left(\operatorname{cosec}^2 	heta-1ight)=\cos ^2 	heta$

Hence the hyperbola is $\frac{x^2}{\sin ^2 	heta}-\frac{y^2}{\cos ^2 	heta}=1$ or $x^2 \operatorname{cosec}^2 	heta-y^2 \sec ^2 	heta=1$

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

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