Let C be the largest circle centred at (2, 0) and inscribed in the...

Given:

$\dfrac{{x}^{2}}{36} + \dfrac{{y}^{2}}{16} = 1$

Let any point on the ellipse be $P\left(6cos\theta, 4sin\theta\right).</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5c57cd7a-c11f-49bb-b866-14d7d979bbda-9893882.PNG" style="width: 400px; height: 186px;" /></p><p>Equation of normal of ellipse at any point$P$is</p><p>$3sec\thetax – 2cosec\thetay = 10

This normal is also the $normal of the circle passing through the point \left(2, 0\right).</p><p>So,$6sec\theta = 10 or sin\theta = 0$(Not possible)</p><p>$cos\theta = $\dfrac{3}{5} and$sin\theta =$\dfrac{4}{5} </p><p>So, point$P ≡ \left(\dfrac{18}{5}, \dfrac{16}{5}\right)$.</p><p>So, the largest radius of circle$r = \dfrac{\sqrt{320}}{5} = \sqrt{\dfrac{64}{5}}$</p><p>So the equation of circle${\left(x - 2\right)}^{2} + {y}^{2} = \dfrac{64}{5}$</p><p>This circle is passing through \left(1, \alpha\right)$, hence

${\left(1 - 2\right)}^{2} + {\alpha}^{2} = \dfrac{64}{5}$

$⇒ {\alpha}^{2} = \dfrac{59}{5}$

$⇒ 10{\alpha}^{2} = 118$.