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 ____.
Step-by-Step Solution
Key Concept: The normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$ at $P(6\cos\theta, 4\sin\theta)$ passes through the centre $(2,0)$ of the circle. Use this condition to find $\theta$, then the radius $r$ of the largest such circle.
Normal at $P(6\cos\theta, 4\sin\theta)$ on ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$: $3x\sec\theta - 2y\csc\theta = 10$. Passes through $(2,0)$: $6\sec\theta = 10 \Rightarrow \cos\theta = 3/5$, $\sin\theta = 4/5$. $P = (18/5, 16/5)$. $r^2 = (18/5-2)^2 + (16/5)^2 = 64/5 + 256/25$... more precisely $r = \sqrt{320}/5$, so circle is $(x-2)^2 + y^2 = 64/5$. Substituting $(1,\alpha)$: $1 + \alpha^2 = 64/5 \Rightarrow \alpha^2 = 59/5 \Rightarrow 10\alpha^2 = 118$.
Correct Answer: 118