Question

The tangent at any point P on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the auxiliary circle in two points which subtend right angle at the centre of the ellipse. If $\theta$ be the eccentric angle of $P$ and $e$ be the eccentricity of the ellipse, then $\sin \theta =$

• A. $\sqrt{1-e^2}$
• B. $e/\sqrt{1-e^2}$
• C. $\sqrt{1-e^2}/e$
• D. None of these

Answer 1

The correct option is C $\sqrt{1-e^2}/e$

Equation of the tangent at $P\left(\theta \right)$ on the given ellipse is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\cdots \left(1\right)$

Equation of the auxiliary circle of the given ellipse is $x^2+y^2=a^2\cdots \left(2\right)$


If $A,B$ be the intersection points of the above tangent and the auxiliary circle, then equation of the pair of straight lines $OA,OB$ ($O$ is the origin) is $x^2+y^2-a^2{(\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b})}^2=0\cdots \left(3\right)$

According to the given condition, equation $\left(3\right)$ must represent a pair of perpendicular straight lines. Therefore, we have coeff. of $x^2+$ coeff. of $y^2=0$

i.e. $(1-{\cos }^2\theta )+(1-\frac{a^2}{b^2}{\sin }^2\theta )=0$

i.e. ${\sin }^2\theta (\frac{a^2}{b^2}-1)=1$

i.e. ${\sin }^2\theta =\left(\frac{b^2}{a^2-b^2}\right)=\frac{1-e^2}{e^2}$

i.e. $\sin \theta =\frac{\sqrt{1-e^2}}{e}$