Question

An ellipse has $OB$ as semi-minor axis, $F$ and $F^{{}^{\prime }}$ are its foci and the $\angle FB{F}^{{}^{\prime }}$, is a right angle. Then, the eccentricity of the ellipse is

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

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

Since, $\angle FB{F}^{{}^{\prime }}={90}^o$, then $\angle OB{F}^{{}^{\prime }}={45}^o$ and $\angle B{F}^{{}^{\prime }}O={45}^o$
$\Rightarrow ae=b$ ($\because △BO{F}^{{}^{\prime }}$ is an isosceles triangle)
and $e^2=1-\frac{b^2}{a^2}$
$\Rightarrow e^2=1-\frac{a^2{e}^2}{a^2}\Rightarrow e^2=1-e^2$
$\Rightarrow 2e^2=1\Rightarrow e=\frac{1}{\sqrt{2}}$ ($\because e$ cannot be negative)

Hence, option D is correct.