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 ${\mathrm{E}}_1$ and the foci of ${\mathrm{E}}_2$ are the end points of minor axis of ${\mathrm{E}}_1$. If ${\mathrm{E}}_1$ and ${\mathrm{E}}_2$ have same, eccentricities, then its value is : 

${\mathrm{E}}_1 : \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

${\mathrm{E}}_2 : \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{c^2}=1$

$$\begin{gathered} Given both the ellipse having same eccentricity and they are \\ {e}^2=1-\frac{b^2}{a^2} \\ {e}^2=1-\frac{a^2}{c^2} \\ \Rightarrow \frac{b^2}{a^2}=\frac{a^2}{c^2} \\ \Rightarrow c^2=\frac{a^4}{b^2} \\ \Rightarrow c=\frac{a^2}{b}\to \left(1\right) \\ also b=ce \\ \Rightarrow c=\frac{b}{e} \\ from \left(1\right) \\ \frac{b}{e}=\frac{a^2}{b} \\ \Rightarrow b^2=a^{2}e \\ \Rightarrow a^2(1-e^2)=a^{2}e \\ \Rightarrow e^2+e-1=0 \\ \Rightarrow e=\frac{-1\pm \sqrt{5}}{2} \\ \Rightarrow e=\frac{\sqrt{5}-1}{2} \end{gathered}$$