1. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A, B$. There is a point $P$ on the ellipse such that the area of $\triangle P A B$ equals 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4
1. B Let $P_{1}(4 \cos \alpha, 3 \sin \alpha)\left(0<\alpha<\frac{\pi}{2}\right), P_{1}$ is on the ellipse in the first quadrant. Consider the area $S$ of quadrilateral $P_{1} A O B$. $S=S_{\triangle O A P_{1}}+S_{\triangle O B P_{1}}=\frac{1}{2} \times 4 \cdot 3 \sin \alpha+\frac{1}{2} \times 3 \cdot 4 \cos \alpha=6 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right), S_{\max }=6 \sqrt{2}$, and $S_{\triangle O A B}=$ $6, S_{\triangle P_{1} A B}$ has a maximum value of $6 \sqrt{2}-6,6 \sqrt{2}-6<3$, which means point $P$ cannot be above line $A B$, and there are clearly two points $P$ below line $A B$.