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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
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Verified Answer
The correct answer is:
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$
The given ellipse is
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\
& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
$$
$$
\therefore \quad a e=1
$$
Hence the eccentricity $e_1$, of the hyperbola is given by
$$
\begin{array}{rlrl}
& & 1 & =e_1 \sin \theta \\
\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\
\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta
\end{array}
$$
Hence, the hyperbola is
$$
\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
$$
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\
& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
$$
$$
\therefore \quad a e=1
$$
Hence the eccentricity $e_1$, of the hyperbola is given by
$$
\begin{array}{rlrl}
& & 1 & =e_1 \sin \theta \\
\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\
\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta
\end{array}
$$
Hence, the hyperbola is
$$
\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
$$
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