Question

The tangent at the point ' $\alpha$ ' on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{h^2}=1$ meets the auxiliary circle in two points which subtend a right angle at the centre. The eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{1+{\sin }^2\alpha }}$
• B. $\frac{1}{\sqrt{1+{\cos }^2\alpha }}$
• C. $\sqrt{1+{\sin }^2\alpha }$
• D. none of these

Answer 1

Answer: (A)

Given ellipse is $\frac{x^3}{a^2}+\frac{y^2}{b^2}=\mathrm{1...}$(1)

Its auxiliary circle is $x^2+y^2=a^2...$(2)
Let $P\equiv (a\cos \alpha ,b\sin \alpha )$
Equation of tangent to the ellipse at $P(a\cos \alpha ,b\sin \alpha )$ is
$\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b}=\mathrm{1.....}$(3)
Making equation (2) homogeneous with the help of (3), we get
$x^2+y^2-a^2{\left(\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b}\right)}^2=0$
$\Rightarrow \left(1-{\cos }^2\alpha \right)x^2+\left(1-\frac{a^2}{b^2}{\sin }^2\alpha \right)y^2$
$-2\frac{a}{b}\cos \alpha \sin \alpha xy=\mathrm{0....}$(4)
(4) is the joint equation of $OL$ and $OM$.
Since $\angle LOM=90^{\circ },$
$\therefore$ coefficient of $x^2+$ coefficient of $y^2=0$
$\Rightarrow 1-{\cos }^2\alpha +1-\frac{a^2}{b^2}{\sin }^2\alpha =0$
$\Rightarrow {\sin }^2\alpha \left(\frac{a^2}{b^2}-1\right)=1$
or, ${\sin }^2\alpha \left(\frac{1}{1-e^2}-1\right)=1\left[\because b^2=a^2\left(1-e^2\right)\right]$
$\Rightarrow e^2{\sin }^2\alpha =1-e^2\text{ or }{e}^2\left(1+{\sin }^2\alpha \right)=1$
$\therefore e=\frac{1}{\sqrt{11{\sin }^2\alpha }}$