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

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Aicharya · Accepted Answer

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

Also $e^{2}=1-\frac{b^{2}}{a^{2}}=1-e^{2}$ $\begin{array}{l} \Rightarrow 2 e^{2}=1 \\ \Rightarrow e=\frac{1}{\sqrt{2}} \end{array}$

An ellipse has $\mathrm{OB}$ as semi minor axis, $\mathrm{F}$ and $\mathrm{F}^{\prime}$ its focii and the…

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