Question

If the tangent at $\theta$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the auxiliary circle at two points which subtend a right angle at the centre, then $e^2(2-{\cos }^2\theta )=$

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $1$

Homogenising the equation of the auxiliary circle $x^2+y^2=a^2$ with the help of the tangent at $\theta$
i.e., $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1,$ we get $x^2+y^2-a^2{(\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta )}^2=0$
Since this equation represents a pair of perpendicular, lines, we have sum of the coefficients of $x^2$ and $y^2$ is zero
$\Rightarrow (1-{\cos }^2\theta )+1-\frac{a^2}{b^2}{\sin }^2\theta =0\Rightarrow b^2{\sin }^2\theta +b^2-a^2{\sin }^2\theta =0\Rightarrow a^2(1-e^2)(1+{\sin }^2\theta )-a^2{\sin }^2\theta =0$
$\Rightarrow (1-e^2)(1+{\sin }^2\theta )={\sin }^2\theta$
On simplification, we get, $e^2(2-{\cos }^2\theta )=1$