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Let $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \mathrm{a}\,>\,\mathrm{b} .$ Let $\mathrm{E}_{2}$ be another ellipse such that it touches the end points of major axis of $E_{1}$ and the foci $E_{2}$ are the end points of minor axis of $E_{1}$. If $E_{1}$ and $E_{2}$ have same eccentricities, then its value is :
$\frac{-1+\sqrt{3}}{2}$
$\frac{-1+\sqrt{6}}{2}$
$\frac{-1+\sqrt{5}}{2}$
$\frac{-1+\sqrt{8}}{2}$
Solution
$\mathrm{e}^{2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}} ….(1)$
$\mathrm{e}^{2}=1-\frac{\mathrm{a}^{2}}{\mathrm{c}^{2}} ….(2)$
$\Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{\mathrm{a}^{2}}{\mathrm{c}^{2}}$
$\Rightarrow \mathrm{c}^{2}=\frac{\mathrm{a}^{4}}{\mathrm{~b}^{2}} \Rightarrow \mathrm{c}=\frac{\mathrm{a}^{2}}{\mathrm{~b}}$
Also $b=c e$
$\therefore \frac{b}{e}=\frac{a^{2}}{b} \Rightarrow \frac{b^{2}}{a^{2}}=e$
$\quad e^{2}=1-b^{2} / a^{2} \Rightarrow e^{2}+e-1=0$
