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 $\triangle A P B$ is 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4
Solve:
Given the system of equations
\[
\left\{\begin{array}{l}
b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}, \\
A x+B y+C=0
\end{array} \Rightarrow B^{2} b^{2} x^{2}+a^{2}(A x+C)^{2}=B^{2} a^{2} b^{2}\right.
\]
\[
\Rightarrow\left(A^{2} a^{2}+B^{2} b^{2}\right) x^{2}+2 A C a^{2} x+a^{2}\left(C^{2}-B^{2} b^{2}\right)=0 .
\]

Then, the discriminant \(\Delta\) is given by
\[
\Delta=4 a^{2}\left[A^{2} C^{2} a^{2}-\left(A^{2} a^{2}+B^{2} b^{2}\right)\left(C^{2}-B^{2} b^{2}\right)\right]
\]
\[
\begin{array}{l}
=4 a^{2}\left[A^{2} C^{2} a^{2}-\left(A^{2} C^{2} a^{2}+B^{2} C^{2} b^{2}-A^{2} B^{2} a^{2} b^{2}-B^{4} b^{4}\right)\right] \\
=4 B^{2} a^{2} b^{2}\left(A^{2} a^{2}+B^{2} b^{2}-C^{2}\right) .
\end{array}
\]

Thus, when \(A^{2} a^{2}+B^{2} b^{2}>C^{2}\), the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) intersects the line \(A x+B y+C=0\); when \(A^{2} a^{2}+B^{2} b^{2}=C^{2}\), the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is tangent to the line \(A x+B y+C=0\); and when \(A^{2} a^{2}+B^{2} b^{2}<C^{2}\), the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is separated from the line \(A x+B y+C=0\). Noting that \(A(0,4), B(3,0)\), then \(|A B|=5 \Rightarrow\) the distance \(d\) from point \(P\) to line \(A B\) is \(\frac{6}{5}\), thus point \(P\) lies on the line \(l_{1}: 3 x+4 y-6=0\) or the line \(l_{2}: 3 x+4 y-18=0\). By the above method, \(l_{1}\) intersects the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), and \(l_{2}\) is separated from the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), meaning there are exactly 2 points \(P\) that satisfy the condition. Therefore, the correct choice is \(B\).