4. 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 $\triangle P A B$ is 3. How many such points $P$ are there?
(A) 1
(B) 2
(C) 3
(D) 4
4.(B).

Let $P_{1}(4 \cos \alpha, 3 \sin \alpha)$ $\left(0<\alpha<\frac{\pi}{2}\right)$, i.e., point $P_{1}$ is on the ellipse in the first quadrant. As shown in Figure 7, consider the area $S$ of quadrilateral $P_{1} A O B$:
$$
\begin{aligned}
S & =S_{\text {Cax }_{1}}+S_{\text {COses }_{1}} \\
& =-\frac{1}{2} \times 4(3 \sin \alpha)+\frac{1}{2} \times 3(4 \cos \alpha) \\
& =6(\sin \alpha+\cos \alpha)=6 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) . \\
\therefore & S_{\text {max }}=6 \sqrt{2}\left(\text { at this point } \alpha=\frac{\pi}{4}\right) .
\end{aligned}
$$

Since $S_{\triangle O B B}=\frac{1}{2} \times 4 \times 3=6$ 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 above the line $A B$, and there are clearly two points $P$ below the line $A B$.