Question

If the distance between the foci of an ellipse is equal to the length of the latusrectum 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)

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

Given In ellipse the distance between the foci $=$ Length of latusrectum
$\Rightarrow$ $2ae=\frac{2b^2}{a}$
$\Rightarrow$ $a^{2}e=b^2$ $\Rightarrow$ $e=\frac{b^2}{a^2}.....\left(i\right)$
$\because$ $e^2=1-\frac{b^2}{a^2}$
$\Rightarrow$ $e^2=1-e$ ......(from eq. $\left(i\right)$)
$\Rightarrow$ $e^2+e-1=0$
$\Rightarrow$ $e=\frac{-1\pm \sqrt{1+4}}{2}$
By quadratic formula:
$\Rightarrow$ $e=\frac{-1\pm \sqrt{5}}{2}$
$\Rightarrow$ $e=\frac{\sqrt{5}-1}{2}$ ($\because 0<e<1$)