$\mathrm{S}$ and ${\mathrm{S}}^1$ foci of an ellipse. $\mathrm{B}$ is one end of the minor axis. If $\angle SBS\text{'}$ is a right angled isosceles triangle, then $\mathrm{e}=$

Let  the ellipse be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 (\mathrm{a}>\mathrm{b})$

from ${∆}^{\mathrm{le}} \mathrm{BCS},$

$$\begin{gathered} \tan {45}^0=\frac{\mathrm{ae}}{\mathrm{b}} \\ \Rightarrow 1=\frac{\mathrm{ae}}{\mathrm{b}} \\ \Rightarrow \mathrm{b}=\mathrm{ae} \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2{\mathrm{e}}^2 \\ \Rightarrow {\mathrm{a}}^2(1-{\mathrm{e}}^2)={\mathrm{a}}^2{\mathrm{e}}^2 \\ \Rightarrow 1-{\mathrm{e}}^2={\mathrm{e}}^2 \\ \Rightarrow 2{\mathrm{e}}^2=1 \\ \Rightarrow {\mathrm{e}}^2=\frac{1}{2} \\ \Rightarrow \mathrm{e}=\frac{1}{\sqrt{2}} \end{gathered}$$