If S and S ' are the focii of an ellipse, B is one end of the minor axis and ∠ S B S ' = 90 °…

Let the coordinates of the focii of an ellipses are $S\left(ae,0\right)$ and $S^{\text{'}}\left(-ae,0\right)$.

Since, $\angle SB{S}^{\text{'}}=90$

$\therefore {\left(S{S}^{\text{'}}\right)}^2={\left(SB\right)}^2+{\left(S^{\text{'}}B\right)}^2$

${\left(2ae\right)}^2={\left[\sqrt{{\left(ae-0\right)}^2+{\left(0-b\right)}^2}\right]}^2+{\left[\sqrt{{\left(-ae-0\right)}^2+{\left(0-b\right)}^2}\right]}^2$

$\Rightarrow {\left(2ae\right)}^2=\left[{\left(ae\right)}^2+{\left(b\right)}^2\right]+{\left[{\left(ae\right)}^2+{\left(b\right)}^2\right]}^2$

$\Rightarrow 4a^2{e}^2=2a^2{e}^2+2b^2$

$\Rightarrow 2a^2{e}^2=2b^2$

$\Rightarrow \frac{b^2}{a^2}=e^2 ...\left(i\right)$

Also,

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

$\Rightarrow e^2=1-e^2; \left[\text{Using} \left(i\right)\right]$

$\Rightarrow e=\frac{1}{\sqrt{2}}$.