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  • A hyperbola, having the transverse axis of length  , is conf

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

Option: 1

\mathrm{x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1}


Option: 2

\mathrm{x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1}


Option: 3

\mathrm{x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1}


Option: 4

\mathrm{x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1}


Answers (1)

best_answer

The given ellipse is \mathrm{\frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow a=2, b=\sqrt{3}}

\mathrm{\begin{aligned} & \therefore \quad 3=4\left(1-\mathrm{c}^2\right) \Rightarrow \mathrm{c}=\frac{1}{2} \\ & \therefore \quad \mathrm{ae}=2 \times \frac{1}{2}=1 \end{aligned}}

Hence, the eccentricity \mathrm{e_1} of the hyperbola  is given by

\mathrm{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, equation of hyperbola is  \mathrm{\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1} or \mathrm{x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1}

Posted by

Rishabh

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