Let C be the largest circle centred at (2,0) | vedclass

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Question

Equation of normal of ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ at any point $P (6 \cos 	heta, 4 \sin 	heta)$ is

$3 \sec 	heta x-2 \operatorname

Vedclass · Accepted Answer

$118$

Vedclass · Answer

Equation of normal of ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ at any point $P (6 \cos 	heta, 4 \sin 	heta)$ is

$3 \sec 	heta x-2 \operatorname{cosec} 	heta y=10$ this normal is also the normal of the circle passing through the point $(2,0)$ So,

$6 \sec 	heta=10$ or $\sin 	heta=0$ (Not possible) $\cos 	heta=\frac{3}{5}$ and $\sin 	heta=\frac{4}{5}$ so point $P=\left(\frac{18}{5}, \frac{16}{5}ight)$

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 $\alpha^2=\frac{59}{5}$

$10 \alpha^2=118$

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 $(1, \alpha)$ lies on $C$, then $10 \alpha^2$ is equal to $.........$

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