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  • A hyperbola passes through the foci of the ellipse  and its transverse and con

A hyperbola passes through the foci of the ellipse \frac{x^2}{25}+\frac{y^2}{16}=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :
Option: 1 \frac{x^2}{9}-\frac{y^2}{25}=1
Option: 2 \frac{x^2}{9}-\frac{y^2}{16}=1
Option: 3 x^2-y^2=1
Option: 4 \frac{x^2}{9}-\frac{y^2}{4}=1

Answers (1)

best_answer

Given equation of an ellipse is

\frac{x^{2}}{25}+\frac{y^{2}}{16}=1

\text { For ellipse } \mathrm{e}_{1}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\frac{3}{5}

\text { for hyperbola } \mathrm{e}_{2}=\frac{5}{3}

Since,  transverse and conjugate axes coincide with major and minor axes of the ellipse, so e_1\times e_2=1

\\\text{foci of the ellipse is }(ae,0)=(3,0)

\\\text{Let hyperbola be} \;\; \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

These pass-through foci of the ellipse

\frac{9}{a^2}-0=1\Rightarrow a=3

\\\Rightarrow b^{2}=a^{2}\left(e^{2}-1\right)\\b^2=9\left(\frac{25}{9}-1\right)=16 

So, the equation of the hyperbola is

\frac{x^2}{9}-\frac{y^2}{16}=1

Posted by

himanshu.meshram

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