Question

If tangents drawn to the ellipse at the parametric point $\theta$, where $\tan \theta =2$ meets the auxillary circle at $P$ and $Q$ and $PQ$ subtends rightangle at the centre of the ellipse, then eccentricity of the ellipse is

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

Answer 1

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

Let the equation of the ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$
So any point on the ellipse will be
$(a\cos \theta ,b\sin \theta )$
but $\tan \theta =2\Rightarrow \sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$
Hence equation of tangent will be
$\frac{x}{\sqrt{5}a}+\frac{2y}{\sqrt{5}b}=1\cdots \left(1\right)$
Now equation of auxillary circle will be
$x^2+y^2=a^2$
homogenizing with $\left(1\right)$ we get
$x^2+y^2=a^2{(\frac{x}{\sqrt{5}a}+\frac{2y}{\sqrt{5}b})}^2\Rightarrow \frac{4}{5}{x}^2+y^2(1-\frac{4a^2}{5b^2})-\frac{4axy}{5b}=0$
This will represent $\perp$ lines if
$\frac{4}{5}+1-\frac{4a^2}{5b^2}=0\Rightarrow \frac{b^2}{a^2}=\frac{4}{9}$
Hence $e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$