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

Answer: 118

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
$3\sec \theta x-2\mathrm{cosec}\theta y=10$ 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 ${\alpha }^2=\frac{59}{5}$
$10{\alpha }^2=118$