Four. (20 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with an inscribed $\triangle A B C$ where sides $A B$ and $A C$ pass through the left and right foci $F_{1}$ and $F_{2}$ respectively, and the left and right vertices of the ellipse are $D$ and $E$ respectively. The lines $D B$ and $C E$ intersect at point $P$. When point $A$ varies on the ellipse, find the locus of point $P$.
As shown in Figure 5, let \( A(2 \cos \theta, \sqrt{3} \sin \theta), B(2 \cos \alpha, \sqrt{3} \sin \alpha), C(2 \cos \beta, \sqrt{3} \sin \beta) \), where \(\alpha, \beta, \theta\) are distinct. The line passing through points \(A\) and \(B\) is given by
\[
l_{A E}: y=\frac{\sqrt{3}(\sin \theta-\sin \alpha)}{2(\cos \theta-\cos \alpha)}(x-2 \cos \theta)+\sqrt{3} \sin \theta.
\]

Since the line \(A B\) passes through the point \(F_{1}(-1,0)\), substituting gives
\[
\begin{array}{l}
\sqrt{3}(\sin \theta-\sin \alpha) \cdot \\
(1+2 \cos \theta) \\
=\sqrt{3} \sin \theta \cdot \\
2(\cos \theta-\cos \alpha),
\end{array}
\]

which simplifies to
\[
\begin{array}{l}
2 \sin (\alpha-\theta) \\
=\sin \theta-\sin \alpha \\
=2 \sin \frac{\theta-\alpha}{2} \cdot \cos \frac{\theta+\alpha}{2}.
\end{array}
\]

Thus,
\[
2 \cos \frac{\alpha-\theta}{2}+\cos \frac{\alpha+\theta}{2}
=3 \cos \frac{\alpha}{2} \cdot \cos \frac{\theta}{2}+\sin \frac{\alpha}{2} \cdot \sin \frac{\theta}{2}=0,
\]

which implies \(\tan \frac{\alpha}{2} \cdot \tan \frac{\theta}{2}=-3\).
Similarly, for the line \(l_{B D}\):
\[
y=\frac{\sqrt{3} \sin \alpha}{2(1+\cos \alpha)}(x+2)
=\frac{\sqrt{3}}{2} \tan \frac{\alpha}{2}(x+2)=-\frac{3 \sqrt{3}}{2 \tan \frac{\theta}{2}}(x+2).
\]

By the same reasoning, \(\tan \frac{\beta}{2} \cdot \tan \frac{\theta}{2}=-\frac{1}{3}\).
For the line \(l_{C E}\):
\[
\begin{aligned}
y & =\frac{\sqrt{3} \sin \beta}{2(\cos \beta-1)}(x-2)=\frac{-\frac{\sqrt{3}}{2}(x-2)}{\tan \frac{\beta}{2}} \\
& =\frac{3 \sqrt{3}}{2} \tan \frac{\theta}{2}(x-2).
\end{aligned}
\]

Solving the system of equations for the intersection point \(P\):
\[
P=\left(\frac{2 \tan ^{2} \frac{\theta}{2}-2}{\tan ^{2} \frac{\theta}{2}+1}, \frac{-6 \sqrt{3} \tan \frac{\theta}{2}}{\tan ^{2} \frac{\theta}{2}+1}\right).
\]

Eliminating \(\tan \frac{\theta}{2}\), we find that \(P(x, y)\) lies on the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{27}=1\). Upon inspection, the points \((-2,0)\) and \((2,0)\) should be excluded.