Example 1 (2002 National High School Competition Question) The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A$ and $B$. There is a point $P$ on the ellipse such that the area of $\triangle PAB$ is 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4
Solution: Choose B. Reason: Assume $P_{1}(4 \cos \alpha, 3 \sin \alpha)\left(0<\alpha<\frac{\pi}{2}\right)$, which means point $P_{1}$ is the point on the ellipse in the first quadrant as required. As shown in Figure 6-1, consider the area $S$ of quadrilateral $P_{1} A O B$, we have
$$
\begin{aligned}
S & =S_{\triangle\left(A P_{1}\right.}+S_{\triangle O B P_{1}} \\
& =\frac{1}{2} \times 4(3 \sin \alpha)+\frac{1}{2} \times 3(4 \cos \alpha) \\
& =6(\sin \alpha+\cos \alpha)=6 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$

Thus, $S_{\text {max }}=6 \sqrt{2}$ (at this point $\alpha=\frac{\pi}{4}$ ).
Since $S_{\triangle \triangle A B}=\frac{1}{2} \times 4 \times 3=6$ is a constant, the maximum value of $S_{\triangle P_{1} A B}$ is $6 \sqrt{2}-6$.
Since $6 \sqrt{2}-6<3$, point $P$ cannot be on the 1 side of line $A B$, and there are clearly two points $P$ below line $A B$.