Q. An ellipse has $OB$as semi minor axis, $F$ and $F$ it’s foci and the angle $FBF'$ 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) $\frac{1}{\sqrt{2}}$
Explanation
(d)
Since, $\angle FBF^{''} = 90{^\circ}$, then
$\angle OBF^{''} = 45$° and $\angle BF^{''}O = 45{^\circ}$
[image:11_image82.emf]
$\Rightarrow \ ae = b$
$\lbrack\ \because\mathrm{\Delta}BOF^{''}is\ an\ isosceles\ traiangle\rbrack$
and $e^{2}\ = 1\ –\frac{b^{2}}{a^{2}}$
$\ \Rightarrow \ \ e^{2} = 1 - \frac{a^{2}e^{2}}{a^{2}}$
$\Rightarrow \ e = \frac{1}{\sqrt{2}}\ \ \ \lbrack\therefore e\ cannot\ be\ negative\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.