Mathematics - Ellipse Question with Solution | TestHub

$\begin{array}{rl}& \because \angle FB{F}^{\prime }=90^{\circ }\Rightarrow F{B}^2+F^{\prime }{B}^2=F{F}^{\prime 2}\\ \therefore & {\left(\sqrt{a^2{e}^2+b^2}\right)}^2+{\left(\sqrt{a^2{e}^2+b^2}\right)}^2=(2ae{)}^2\\ & \Rightarrow 2\left(a^2{e}^2+b^2\right)=4a^2{e}^2\Rightarrow e^2=\frac{b^2}{a^2}\dots (1)\end{array}$
Also, $e^2=1-b^2/a^2=1-e^2$ (By using equation (i)) $\Rightarrow 2e^2=1\Rightarrow e=\frac{1}{\sqrt{2}}$