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  • Let the eccentricity of an ellipse <img alt="\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2

Let the eccentricity of an ellipse \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1} is reciprocal to that of the hyperbola \mathrm{2x^{2}-2y^{2}=1}. If the ellipse
intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____:
 

Option: 1

2


Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

\begin{aligned} & \mathrm{E}: \frac{\mathrm{x}^4}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{b^2}=1 \rightarrow \mathrm{e} \\ & \mathrm{H}: \mathrm{x}^2-\mathrm{y}^2=\frac{1}{2} \Rightarrow \mathrm{e}^{\prime}=\sqrt{2} \\ & \mathrm{e}=\frac{1}{\sqrt{2}} \end{aligned}
\begin{aligned} & \because \mathrm{e}^2=\frac{1}{2} \\ & 1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2} \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2} \\ & \mathrm{a}^2=2 \mathrm{~b}^2 \end{aligned}

\mathrm{E} \: \& \: \mathrm{H} are at right angle they are confocal Focus of Hyperbola = focus of ellipse
\begin{aligned} & \left( \pm \frac{1}{\sqrt{2}} \cdot \sqrt{2}, 0\right)=\left( \pm \frac{\mathrm{a}}{\sqrt{2}}, 0\right) \\ & \mathrm{a}=\sqrt{2} \\ & \because \mathrm{a}^2=2 \mathrm{~b}^2 \Rightarrow \mathrm{b}^2=1 \end{aligned}

\mathrm{Length\: of \: L R=\frac{2 b^2}{a}=\frac{2(1)}{\sqrt{2}} }

=\sqrt{2}
\mathrm{Square\: of\: LR =2}

Posted by

Divya Prakash Singh

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