Question

The latusrectum LL’ subtends a right angle at the centre of the ellipse, then its eccentricity is

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{2}+1}{3}$
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{\sqrt{3}-\sqrt{2}}{2}$

Answer 1

The correct option is C $\frac{\sqrt{5}-1}{2}$

From the right angled triangle OSL,
$\tan {45}^{\circ }=\frac{SL}{OS}\Rightarrow \frac{b^2}{a}=ae$
$\Rightarrow a^2(1-e^2)=a^{2}e\Rightarrow e^2+e-1=0$
$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}\Rightarrow e=\frac{\sqrt{5}-1}{2}$