Question

If the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ at the point $(a\cos \theta ,b\sin \theta )$ meets the auxiliary circle in two points $A,B$ such that the chord $AB$ subtends a right angle at the centre, then the eccentricity of the ellipse is given by $\frac{1}{\sqrt{\alpha +\beta {\sin }^2\theta }},$ then the value of $(\alpha +\beta {)}^2$ is equal to

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Tangent to the ellipse is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Auxiliary circle is $x^2+y^2=a^2$
Homogenising, we get
$x^2+y^2-a^2{(\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b})}^2=0$
$\Rightarrow x^2(1-\frac{a^2{\cos }^2\theta }{a^2})+y^2(1-\frac{a^2{\sin }^2\theta }{b^2})-2a^2\cdot \frac{\cos \theta }{a}\cdot \frac{\sin \theta }{b}xy=0$
which represents a pair of straight lines.
These lines are perpendicular.
$\Rightarrow a^2(\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2})=2$
$\Rightarrow {\cos }^2\theta +\frac{a^2{\sin }^2\theta }{b^2}=2$
$\Rightarrow \frac{a^2{\sin }^2\theta }{b^2}-{\sin }^2\theta =1$
$\Rightarrow \frac{a^2}{b^2}=\frac{1+{\sin }^2\theta }{{\sin }^2\theta }$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\left(\frac{{\sin }^2\theta }{1+{\sin }^2\theta }\right)}=\sqrt{\frac{1}{1+{\sin }^2\theta }}$
On comparing with the given expression,
$\alpha =1,\beta =1$
$\therefore (\alpha +\beta {)}^2=4$