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 $\frac{x^{3}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 ...$(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
$\frac{x \cos \alpha}{a}+\frac{y \sin \alpha}{b}=1.....$(3)
Making equation (2) homogeneous with the help of (3), we get
$x^{2}+y^{2}-a^{2}\left(\frac{x \cos \alpha}{a}+\frac{y \sin \alpha}{b}\right)^{2}=0$
$\Rightarrow \left(1-\cos ^{2} \alpha\right) x^{2}+\left(1-\frac{a^{2}}{b^{2}} \sin ^{2} \alpha\right) y^{2}$
$-2 \frac{a}{b} \cos \alpha \sin \alpha x y=0 ....$(4)
(4) is the joint equation of $O L$ and $O M$.
Since $\angle L O M=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 $
$\therefore e=\frac{1}{\sqrt{11 \sin ^{2} \alpha}}$