Question

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

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

Answer 1

Answer: (C)

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

Answer 2

Given, In ellipse the distance between the foci = Length of the latusrectum $\Rightarrow 2 a e=\frac{2 b^{2}}{a}$ $\Rightarrow a^{2} \theta=b^{2}$ $\Rightarrow e=\frac{b^{2}}{a^{2}}$....(i) $\because e^{2}=1-\frac{b^{2}}{a^{2}}$ $\Rightarrow e^{2}=1-e \quad$ [from E (i)] $\Rightarrow e^{2}+e-1=0$ $\Rightarrow \theta=\frac{-1 \pm \sqrt{1+4}}{2}$ (by quadratic formula) $\Rightarrow e=\frac{-1 \pm \sqrt{5}}{2}$ $\Rightarrow e=\frac{\sqrt{5}-1}{2} (\because 1 > e > 0)$