Let C be the largest circle centred at ( 2 , 0 ) and inscribed in the ellipse x 2 36 + y 2 16 = 1 . If ( 1 ,) lies on C , then 10 2 is equal to _______ . [2023]

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Solution

Q.
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 $(1, α)$ lies on $C$ , then $10 α^{2}$ is equal to _______ . [2023]

Ans.
(118)

Let P be the point where ellipse and circle touch each other.

Let $P$ be $(6 \cos θ, 4 \sin θ)$

Equation of tangent to ellipse at point P is

$\frac{x \cos θ}{6} + \frac{y \sin θ}{4} = 1$ $\Rightarrow y = \frac{4}{\sin θ} (\frac{- x \cos θ}{6} + 1)$

$\Rightarrow y = - \frac{2 \cos θ}{3 \sin θ} x + \frac{4}{\sin θ}$

Let $m_{1} = \frac{- 2 \cos θ}{3 \sin θ}$

Slope of normal to circle at $P = m_{2}$ .

$m_{2} = \frac{(4 \sin θ - 0)}{(6 \cos θ - 0)}$ $∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \sin θ = \frac{4}{5}$

So, $P \equiv (\frac{18}{5}, \frac{16}{5})$

$r^{2} = {(\frac{18}{5} - 2)}^{2} + {(\frac{16}{5})}^{2} = {(1 - 2)}^{2} + α^{2} = \frac{64}{25} + \frac{256}{25} = 1 + α^{2}$

$α^{2} = \frac{320}{25} - 1 = \frac{64}{5} - 1 = \frac{59}{5}$

$\Rightarrow 10 α^{2} = 59 \times 2 = 118$

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