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., $x^2+y^2-a^2{\left(\frac{x}{2a}+\frac{\sqrt{3}y}{2b}\right)}^2=0$
Since $\angle LCM=90^{\circ }$
$\Rightarrow 1-\frac{1}{4}+1-\frac{3a^2}{4b^2}=0\Rightarrow \frac{3a^2}{4b^2}=\frac{7}{4}$
$\Rightarrow 7b^2=3a^2\Rightarrow 7a^2\left(1-e^2\right)=3a^2$
Hence $e=\frac{2}{\sqrt{7}}$