Question 97
The straight line $\frac{x}{4} + \frac{y}{3} = 1$ intersects the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at two points $A$ and $B$. There is a point $P$ on this ellipse such that the area of $\triangle PAB$ is equal to $3$. There is/are ( ) such point/points $P$.
(A) 1 (B) 2 (C) 3 (D) 4
(A) 1 (B) 2 (C) 3 (D) 4