# Solve this

Question:

Let $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$. Let $\mathrm{E}_{2}$ be another ellipse such that it touches the end points of major axis of $E_{1}$ and the foci of $E_{2}$ are the end points of minor axis of $E_{1}$. If $E_{1}$ and $E_{2}$ have same eccentricities, then its value is :

1. $\frac{-1+\sqrt{5}}{2}$

2. $\frac{-1+\sqrt{8}}{2}$

3. $\frac{-1+\sqrt{3}}{2}$

4. $\frac{-1+\sqrt{6}}{2}$

Correct Option: 1,

Solution:

$e^{2}=1-\frac{b^{2}}{a^{2}}$

$e^{2}=1-\frac{a^{2}}{c^{2}}$

$\Rightarrow \frac{b^{2}}{a^{2}}=\frac{a^{2}}{c^{2}}$

$\Rightarrow \mathrm{c}^{2}=\frac{\mathrm{a}^{4}}{\mathrm{~b}^{2}} \Rightarrow \mathrm{c}=\frac{\mathrm{a}^{2}}{\mathrm{~b}}$

Also $b=c e$

$\Rightarrow \mathrm{c}=\frac{\mathrm{b}}{\mathrm{e}}$

$\frac{\mathrm{b}}{\mathrm{e}}=\frac{\mathrm{a}^{2}}{\mathrm{~b}}$

$\Rightarrow \mathrm{e}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=1-\mathrm{e}^{2}$

$\Rightarrow \mathrm{e}^{2}+\mathrm{e}-1=0$

$\Rightarrow \mathrm{e}=\frac{-1+\sqrt{5}}{2}$