Question

If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is :

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

Answer 1

Answer: (C)

$S_1{S}_2=$ distance between foci
$=2ae=\frac{1}{2}($ length of latus rectum $=\frac{2b^2}{a})$
$\therefore 2ae=\frac{1}{2}\left(\frac{2b^2}{a}\right)$
$\Rightarrow 2a^{2}e=b^2$
$\Rightarrow e=\frac{b^2}{2a^2}=\frac{a^2\left(1-e^2\right)}{2a^2}=\frac{1-e^2}{2}$
$\Rightarrow e^2+2e-1=0$
$\Rightarrow e=\frac{-2+\sqrt{4+4}}{2}=\frac{\sqrt{8-2}}{2}=\sqrt{2}-1$