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

Answer 1

The correct option is B $\sqrt{2}-1$

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now, distance between foci $=2ae$.
Length of latus rectum $=\frac{2b^2}{a}$
According to question
$2ae=\frac{b^2}{a}$
$\Rightarrow 2a^{2}e=b^2$
$\Rightarrow 2a^{2}e=a^2(1-e^2)$
$\Rightarrow e^2+2e-1=0$
$\Rightarrow e=-1\pm \sqrt{2}$
$\Rightarrow e=\sqrt{2}-1$ as $0<e<1$