Let $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be he equation of ellipse.

Given that  ${F_1}B$ and ${F_2}B$ are perpandicluar to each other.

Slope of ${F_1}B \times $ alop of ${F_2}B =  – 1$

$\left( {\frac{{0 – b}}{{ – ae – 0}}} \right) \times \left( {\frac{{0 – b}}{{ae – 0}}} \right) =  – 1$

$\left( {\frac{b}{{ae}}} \right) \times \left( {\frac{{ – b}}{{ae}}} \right) =  – 1$

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

${e^2} = \frac{{{b^2}}}{{{a^2}}}\,\,\,$         {$\because $ $\,\,{e^2} = 1 – \frac{{{b^2}}}{{{a^2}}}\,\,$}

$1 – \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{{{b^2}}}{{{a^2}}}\,$

$1 = 2\frac{{{b^2}}}{{{a^2}}}\,\,\, \Rightarrow \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{1}{2}$

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

${e^2} = \frac{1}{2}$