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An ellipse and a hyperbola have their principal axes along the coordinate axes and have common foci separated by a distance 2 \sqrt{13}. The distance of their focal semi-axes is equal to 4 . If the ratio of their eccentricities is \frac{3 }{7}, find the equation of hyperbola is
 

Option: 1

\mathrm{\frac{x^2}{4}-\frac{y^2}{9}=1}


Option: 2

\mathrm{ \frac{x^2}{16}-\frac{y^2}{12}=1}


Option: 3

\mathrm{\frac{x^2}{12}-\frac{y^2}{16}=1}


Option: 4

\mathrm{\frac{x^2}{9}-\frac{y^2}{4}=1}


Answers (1)

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(d) : Let the semi-major axis, semi-minor axis and eccentricity of the ellipse be a, b and e respectively and those of hyperbola be \mathrm{a^{\prime}, b^{\prime} } and \mathrm{e^{\prime}} respectively.
Given that \mathrm{a-a^{\prime}=4\quad \quad \dots(1)}

\begin{aligned} &\mathrm{ 2 a c=2 \sqrt{13} \text { or } a e=\sqrt{13} \quad \quad \dots(2) }\\ &\mathrm{ 2 a^{\prime} c^{\prime}=2 \sqrt{13} \text { or } a^{\prime} c^{\prime}=\sqrt{13} \quad \quad \dots(3)}\\ &\mathrm{and ~\frac{e}{e^{\prime}}=\frac{3}{7}\quad \quad \dots(4)}\\ &\mathrm{From ~(2), ~a=\frac{\sqrt{13}}{c} \quad \quad \dots(5)}\\ &\mathrm{From ~(3),~ a^{\prime}=\frac{\sqrt{13}}{e^{\prime}} \quad \quad \dots(6)}\\ \end{aligned}
Putting in (1), we get

\mathrm{\frac{\sqrt{13}}{e}-\frac{\sqrt{13}}{e^{\prime}}=4 \text { or } \sqrt{13}\left(e^{\prime}-e\right)=4 e e^{\prime}}

Putting \mathrm{e=\frac{3 e^{\prime}}{7}}, we get


\begin{aligned} &\mathrm{ \sqrt{13}\left(e^{\prime}-\frac{3}{7} e^{\prime}\right)=4 \times \frac{3}{7}\left(e^{\prime}\right)^2 }\\ &\mathrm{ \therefore e^{\prime}=\frac{\sqrt{13}}{3} \text { and } e=\frac{3}{7} e^{\prime}=\frac{\sqrt{13}}{7} }\\ &\mathrm{ a=\frac{\sqrt{13}}{e}=\frac{\sqrt{13}}{\sqrt{13}} \times 7=7 }\\ &\mathrm{ a^{\prime}=\frac{\sqrt{13}}{e^{\prime}}=\frac{\sqrt{13}}{\sqrt{13}} \times 3=3 }\\ &\mathrm{ b^2=a^2\left(1-c^2\right)=49\left(1-\frac{13}{49}\right)=36 }\\ &\mathrm{ \left(b^{\prime}\right)^2=\left(a^{\prime}\right)^2\left\{\left(e^{\prime 2}\right)-1\right\}=9\left(\frac{13}{9}-1\right)=4} \end{aligned}

The equation of ellipse is \mathrm{\frac{x^2}{49}+\frac{y^2}{36}=1} and the equation of hyperbola is \mathrm{\frac{x^2}{9}-\frac{y^2}{4}=1.}

Posted by

Ajit Kumar Dubey

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