Question 60
Straight line $l$ with slope $\frac{1}{3}$ intercepts ellipse $C: \frac{x^2}{36} + \frac{y^2}{4} = 1$ at points $A, B$, and point $P(3\sqrt{2}, \sqrt{2})$ is in the top-left of $l$ (as shown in Fig. 11.1).
Fig. 11.1
a. Prove that the center of the inscribed circle of $\triangle PAB$ is on the line $x = 3\sqrt{2}$.
b. When $\angle APB = 60^{\circ}$, find the area of $\triangle PAB$.
Fig. 11.1
a. Prove that the center of the inscribed circle of $\triangle PAB$ is on the line $x = 3\sqrt{2}$.
b. When $\angle APB = 60^{\circ}$, find the area of $\triangle PAB$.