Question

An ellipse has $OB$ as semi minor axis, $F$ and $F^1$ are its focii and the angle $FB{F}^1$ is a right angle. Then the eccentricity of the ellipse is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{6}$

Answer 1

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

Given that $\angle {FBF}^1={90}^0$

${BF}^1\perp BF$

Thus slope of $BF\times B{F}^1=-1$

$\Rightarrow \frac{b}{ae}\times \frac{b}{-ae}=-1\Rightarrow b^2=a^2{e}^2$.......1

We know that $e=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow \sqrt{1-\frac{{e^{2}a}^2}{a^2}}=\sqrt{1-e^2}$

$e^2=1-e^2\Rightarrow e^2=\frac{1}{2}$

Thus $e=\frac{1}{\sqrt{2}}$