Question

The tangent at a point whose eccentric angle $60^{\circ}$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ meet 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

Answer: (B)

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