Question

If the tangent at a point $(a\cos \theta ,b\sin \theta )$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the auxiliary circle in two points, the chord joining them subtends a right angle at the centre; then the eccentricity of the ellipse is given by

• A. ${(1+{\cos }^2\theta )}^{-1/2}$
• B. $1+{\sin }^2\theta$
• C. ${(1+{\sin }^2\theta )}^{-1/2}$
• D. $1+{\cos }^2\theta$

Answer 1

Answer: (C)

${(1+{\sin }^2\theta )}^{-1/2}$

Equation of the tangent at $(a\cos \theta ,b\sin \theta )$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

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

The joint equation of the lines joining the points of intersection of (i) and the auxiliary circle $x^2+y^2=a^2$ to the origin, which is the center of the circle, is

$x^2+y^2=a^2{[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta ]}^2$

Since these lines are at right angles Co-efficient of $x^2+$ Co-efficient of $y^2=0$

$\Rightarrow$ $1-a^2\left(\frac{{\cos }^2\theta }{a^2}\right)+1-a^2\left(\frac{{\sin }^2\theta }{b^2}\right)=0$

$\Rightarrow$ ${\sin }^2\theta (1-\frac{a^2}{b^2})+1=0$

$\Rightarrow$ ${\sin }^2\theta (b^2-a^2)+b^2=0$

$\Rightarrow$ ${\sin }^2\theta [a^2(1-e^2)-a^2]+a^2(1-e^2)=0$

$\Rightarrow$ $(1+{\sin }^2\theta )a^2{e}^2=a^2$

$\Rightarrow$ $e={(1+{\sin }^2\theta )}^{-\frac{1}{2}}$