Let C be largest circle centred at (2,0) and inscribed in ellipse =frac{x^2}{36}+frac{y^2}{16}=1 .If

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10-2. Parabola, Ellipse, Hyperbola

hard

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 $.........$

$117$

$116$

$118$

$125$

(JEE MAIN-2023)

Solution

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 $\alpha^2=\frac{59}{5}$

$10 \alpha^2=118$

Standard 11

Mathematics

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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 $.........$

Solution

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