Question

The tangent at a point P($\theta$) to the elllipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ cuts the auxiliary circle at point $Q$ and $R$. If $QR$ subtends a right angle at the centre $C$ of the ellipse, then the eccentricity of the ellipse is:

• A. $\frac{1}{\sqrt{1+{\cos }^2\theta }}$
• B. $\frac{1}{\sqrt{1+{\sin }^2\theta }}$
• C. $\sqrt{1+{\cos }^2\theta }$
• D. $\sqrt{1+{\sin }^2\theta }$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{1+{\sin }^2\theta }}$

The equation of the tangent at P$\left(\theta \right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
The combined equation of the lines $CQ$ and $CR$ is
$x^2+y^2=a^2{(\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta )}^2$
The lines represented by this equation are at right angle.
$\therefore 1-{\cos }^2\theta +1-\frac{a^2}{b^2}{\sin }^2\theta =0$
$\Rightarrow {\sin }^2\theta +1-\frac{1}{1-e^2}{\sin }^2\theta =0$
$\Rightarrow 1-e^2=\frac{{\sin }^2\theta }{1+{\sin }^2\theta }\Rightarrow e^2=\frac{1}{1+{\sin }^2\theta }\Rightarrow e=\frac{1}{\sqrt{1+{\sin }^2\theta }}$