Question

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

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

Answer 1

Answer: (A)

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

Answer 2

$\because \angle F B F'=90^{\circ}$
$\Rightarrow F B^{2}+F^{\prime} B^{2}=F F'2$
$\therefore\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}+\left(\sqrt{a^{2} e^{2}+b^{2}}\right)^{2}$
$=(2 a e)^{2}$
$\Rightarrow 2\left(a^{2} e^{2}+b^{2}\right)=4 a^{2} e^{2}$
$\Rightarrow e^{2}=\frac{b^{2}}{a^{2}}$ ... (i)
Also, $e ^{2}=1-b^{2} / a^{2}=1-e^{2}$
(By using equation (i))
$\Rightarrow 2 e^{2}=1$
$\Rightarrow e=\frac{1}{\sqrt{2}}$