An ellipse has $O B$ as semi minor axis, $F$ and $F^{\prime}$ its focii and the angle $F B F^{\prime}$ is a…
An ellipse has $O B$ as semi minor axis, $F$ and $F^{\prime}$
its focii and the angle $F B F^{\prime}$ is a right angle. Then the eccentricity of the ellipse is
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{\sqrt{3}}$
Solution
$\begin{aligned} & \because \angle F B F^{\prime}=90^{\circ} \Rightarrow F B^{2}+F^{\prime} B^{2}=F F^{\prime 2} \\ \therefore &\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}=(2 a e)^{2} \\ & \Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2} \Rightarrow e^{2}=\frac{b^{2}}{a^{2}} \ldots(1) \end{aligned}$
Also, $e^{2}=1-b^{2} / a^{2}=1-e^{2}$
(By using equation (i)) $\Rightarrow 2 e^{2}=1 \Rightarrow e=\frac{1}{\sqrt{2}}$
