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An ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ and the parabola $\mathrm{x}^{2}=4(\mathrm{y}+\mathrm{b})$ are such that the two foci of the ellipse and the end points of the latus rectum of parabola are the vertices of a square. The eccentricity of the ellipse is
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Verified Answer
The correct answer is:
$\frac{2}{\sqrt{13}}$
As it is square
$$
\begin{array}{l}
\mathrm{ae}=2 \\
4=\sqrt{(\mathrm{ae}-2)^{2}+(1-\mathrm{b})^{2}} \\
\mathrm{~b}=-3 \quad \quad \mathrm{~b}=5 \\
\mathrm{a}^{2}=13 \quad \mathrm{a}^{2}=29 \\
\quad\left(\text { from realation } \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)\right. \\
\mathrm{ae}=2 \\
\mathrm{e}=\frac{2}{\sqrt{13}}
\end{array}
$$
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