Question

The tangent at a point whose eccentric angle is ${60}^{\circ }$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b),$ meets the auxiliary circle at $L$ and $M$. If $LM$ subtends a right angle at the centre, then eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{7}}$
• B. $\frac{2}{\sqrt{7}}$
• C. $\frac{3}{\sqrt{7}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{2}{\sqrt{7}}$

Equation of the tangent is
$\frac{x}{a}\cdot \left(\frac{1}{2}\right)+\frac{y}{b}\cdot \left(\frac{\sqrt{3}}{2}\right)=1\dots \left(1\right)$
Auxiliary circle is $x^2+y^2=a^2\dots \left(2\right)$
$C$ is the centre.
Combined equation of $CL,CM$ is obtained by homogenising $\left(2\right)$ with $\left(1\right)$, i.e.,
$x^2+y^2-a^2{(\frac{x}{2a}+\frac{\sqrt{3}y}{2b})}^2=0\Rightarrow x^2+y^2-a^2[{\left(\frac{x}{2a}\right)}^2+{\left(\frac{\sqrt{3}y}{2b}\right)}^2+2\left(\frac{x}{2a}\right)\left(\frac{\sqrt{3}y}{2b}\right)]=0$
$\Rightarrow x^2(1-\frac{1}{4})+y^2(1-\frac{3a^2}{4b^2})-\frac{xa\sqrt{3}y}{2b}=0$
Since $\angle LCM={90}^{\circ },$
coefficient of $x^2+$coefficient of $y^2=0$
$\Rightarrow 1-\frac{1}{4}+1-\frac{3a^2}{4b^2}=0\Rightarrow \frac{3a^2}{4b^2}=\frac{7}{4}$


$\Rightarrow 7b^2=3a^2\Rightarrow 7a^2(1-e^2)=3a^2\Rightarrow e=\frac{2}{\sqrt{7}}$