1. (2002 National High School Competition Question) The line $\frac{x}{3}+\frac{y}{4}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A, B$. A point $P$ on the ellipse makes the area of $\triangle P A B$ equal to 3. How many such $P$ points are there?
A. 1
B. 2
C. 3
D. 4
1. B. Let $P(4 \cos \alpha, 3 \sin \alpha)\left(0<\alpha<\frac{\pi}{2}\right)$, which means $P$ is on the ellipse in the first quadrant, then $S_{P A C B}=S_{\triangle P A O}+S_{\triangle B P P}=\frac{1}{2} \times 4 \cdot$ $(3 \sin \alpha)+\frac{1}{2} \times 3(4 \cos \alpha)=6(\sin \alpha+\cos \alpha)=6 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant 6 \sqrt{2}$, thus $S_{\triangle P A B}=S_{P A O B}-S_{\triangle A O B} \leqslant$ $6 \sqrt{2}-\frac{1}{2} \times 3 \times 4=6(\sqrt{2}-1)<3$, so point $P$ cannot be above the line $A B$, and there are clearly 2 points $P$ below $A B$, hence the answer is B.