Question

An ellipse has $OB$ as semi-minor axis, $F$ and $F^{{}^{\prime }}$ 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

Answer: (D)

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

Given, $F$ and $F^{{}^{\prime }}$ are foci of an ellipse, whose coordinates are $(ae,0)$ and $(-ae,0)$ respectively and coordinates of $B$ are $(0,b)$.
Slope of $BF=\frac{b}{-ae}$
and slope of $B{F}^{{}^{\prime }}=\frac{b}{ae}$
Since, $\angle FB{F}^{{}^{\prime }}={90}^o$
Thus $-\frac{b}{ae}\left(\frac{b}{ae}\right)=-1$
$\Rightarrow b^2=a^2{e}^2$
Here $e^2=1-\frac{b^2}{a^2}=1-\frac{a^2{e}^2}{a^2}$
$\Rightarrow 2e^2=1$
$\Rightarrow e=\frac{1}{\sqrt{2}}$