An ellipse has OB as semi minor axis, F and F^ its focii and the...

An ellipse has

\mathrm{OB}

as semi minor axis,

\mathrm{F}

and

\mathrm{F}^{\prime}

its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse is An ellipse has OB as semi minor axis, F and F^ its focii and the angle FBF' is a right angle. Then the eccentricity of

An ellipse has OB as semi minor axis, F and F^ its focii and the angle FBF' is a right angle. Then the eccentricity of

Options

A $\frac{1}{\sqrt{2}}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{\sqrt{3}}$

Correct answer

A. $\frac{1}{\sqrt{2}}$

Step-by-step solution

\because \angle \mathrm{FBF}^{\prime}=90^{{}^{\circ}}\Rightarrow \mathrm{FB}^{2}+\mathrm{F}^{\prime} \mathrm{B}^{2}=\mathrm{FF}^{\prime 2}\therefore\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}=(2 a e)^{2}\Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2}\Rightarrow e^{2}=\frac{b^{2}}{a^{2}}

Also

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

\begin{array}{l} \Rightarrow 2 e^{2}=1 \\ \Rightarrow e=\frac{1}{\sqrt{2}} \end{array}

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