4. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A, B$. 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
(B)
4. 【Analysis and Solution】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. Consider the area $S$ of quadrilateral $P_{1} A O B$.
$$
\begin{array}{l}
S=S_{\triangle O A P_{1}}+S_{\triangle O B P_{1}}=\frac{1}{2} \times 4(3 \sin \alpha)+\frac{1}{2} \times 3(4 \cos \alpha) \\
\quad=6(\sin \alpha+\cos \alpha)=6 \sqrt{2} \sin \alpha+\frac{\pi}{4}, \\
\therefore S_{\max }=6 \sqrt{2} \text { when } \alpha=\frac{\pi}{4} . \\
\because S_{\triangle O A B}=\frac{1}{2} \times 4 \times 3=6 \text { is a constant, }
\end{array}
$$
$\therefore S_{\text {max }}=6 \sqrt{2}$ when $\alpha=\frac{\pi}{4}$.
$\because S_{\triangle O A B}=\frac{1}{2} \times 4 \times 3=6$ is a constant,
$\therefore$ The maximum value of $S_{\triangle P_{1} A B}$ is $6 \sqrt{2}-6$.
$$
\because 6 \sqrt{2}-6<3 \text {, }
$$
$\therefore$ Point $P$ cannot be above line $A B$, and there are clearly two points $P$ below line $A B$, so the answer is $B$.
Alternative solution: Considering that the ellipse is derived from compressing a circle, and the corresponding area is obtained proportionally. Therefore, we can consider this problem in the context of a circle. If there exists a point $P$ that meets the problem's requirements, then the area of the triangle in the corresponding circle is $\frac{3}{3} \times 4=4$. If point $P$ is on $A B$, the maximum area of triangle $P A B$ is $(4-2 \sqrt{2}) \times \frac{4 \sqrt{2}}{2}<4$, so $P$ cannot be above $A B$. Below $A B$, there are clearly two points that satisfy the problem's requirements.