Let C be the largest circle centred at (2,0) and inscribed in the ellipse x^236+y^216=1.

Question

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$. If $(1,\alpha)$ lies on $C$, then $10\alpha^2$ is equal to \_\_\_\_.

Answer: 118

Accepted Answer

Normal to the ellipse at $P(6\cos\theta,4\sin\theta)$: $3x\sec\theta-2y\csc\theta=10$. For the largest circle centred at $(2,0)$ tangent internally to the ellipse, this normal must pass through $(2,0)$:
$$6\sec\theta=10\Rightarrow \cos\theta=\frac35,\ \sin\theta=\frac45\ \Rightarrow P=\left(\frac{18}5,\frac{16}5\right).$$

Radius$^2=\left(\dfrac{18}5-2\right)^2+\left(\dfrac{16}5\right)^2=\dfrac{64}{25}+\dfrac{256}{25}=\dfrac{64}{5}$.

Circle: $(x-2)^2+y^2=\dfrac{64}{5}$. At $(1,\alpha)$: $1+\alpha^2=\dfrac{64}5\Rightarrow\alpha^2=\dfrac{59}5\Rightarrow10\alpha^2=118.$

Solutions · 1