Question

The tangent drawn to the ellipse at the parametric point $q$, where $q={\tan }^{-1}2$ meets the auxiliary circle at $P$ and $Q$. $PQ$ subtends a right angle at the centre of the ellipse, then eccentricity is :

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

Answer 1

The correct option is D $\frac{\sqrt{5}}{3}$

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

And let the parametric point be $R(a\cos q,b\sin q)$

Equation of tangent at $R$ is $T=0$.

$\frac{x\cos q}{a}+\frac{y\sin q}{b}=1$ ......(i)

Equation of director circle is $x^2+y^2=a^2$.

Making the equation of circle homogeneous using equation of tangent,

$x^2+y^2=a^2{\left(1\right)}^2{x}^2+y^2=a^2{(\frac{x\cos q}{a}+\frac{y\sin q}{b})}^2{x}^2(1-{\cos }^{2}q)+y^2(1-\frac{a^2}{b^2}{\sin }^{2}q)-\frac{2a}{b}\cos q\times \frac{y\sin q}{b}xy=0$

It subtends right angle at the origin.

Therefore, $a+b=0$

$\Rightarrow 1-{\cos }^{2}q+1-\frac{a^2}{b^2}{\sin }^{2}q=0$

$\Rightarrow \tan q=2\Rightarrow \sin q=\frac{2}{\sqrt{5}},\cos q=\frac{1}{\sqrt{5}}$

$\Rightarrow 1-\frac{1}{5}+1-\frac{4}{5}\frac{a^2}{b^2}=0\Rightarrow \frac{9}{5}=\frac{4}{5}\frac{a^2}{b^2}\Rightarrow \frac{b^2}{a^2}=\frac{4}{9}$

We know $e^2=1-\frac{b^2}{a^2}$

$\Rightarrow e^2=1-\frac{4}{9}\Rightarrow e=\frac{\sqrt{5}}{3}$