Question

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If $(1, a)$ lies on $C$, then $10 \alpha^2$ is equal to

Answer 1

Equation of normal of ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ at any point $P (6 \cos \theta, 4 \sin \theta)$ is
$3 \sec \theta x-2 \operatorname{cosec} \theta y=10$ this normal is also the normal of the circle passing through the point $(2,0)$ So,
$6 \sec \theta=10$ or $\sin \theta=0$ (Not possible) $\cos \theta=\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$ so point $P =\left(\frac{18}{5}, \frac{16}{5}\right)$
So the largest radius of circle
$r=\frac{\sqrt{320}}{5}$
So the equation of circle $(x-2)^2+y^2=\frac{64}{5}$
Passing it through $(1, \alpha)$
Then $\alpha^2=\frac{59}{5}$
$10 \alpha^2=118$