Question

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

• A. $\frac{1}{4}(\sqrt{5}-1)$
• B. $\frac{1}{2}(\sqrt{5}+1)$
• C. $\frac{1}{2}(\sqrt{5}-1)$
• D. $\frac{1}{4}(\sqrt{5}+1)$

Answer 1

Answer: (C)

$\frac{1}{2}(\sqrt{5}-1)$

Distance between foci = length of latus rectum

$2ae=\frac{{2b}^2}{a}$

$e=\frac{b^2}{a^2}⟶\left(1\right)$

and, $e=\sqrt{1-\frac{b^2}{a^2}}⟶\left(2\right)$

putting (1) in (2)

$e=\sqrt{1-e}$

Squaring both sides

$e^2+e-1=0$

$e=\frac{-1\pm \sqrt{1+4}}{2}$

$0<e<1$

Therefore, $e=\frac{\sqrt{5}-1}{2}$