9. Given the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with two vertices at $A(-2,0), B(2,0)$, and the eccentricity is $\frac{1}{2}$. Point $P$ (different from $A, B$) lies on the ellipse $\Gamma$, and point $D$ has coordinates $(-4,0), \overrightarrow{D E}=\frac{3}{5} \overrightarrow{D P}$, the line $A P$ intersects with $B E$ at point $Q$. Find the equation of the trajectory of point $Q$.
Given $a=2, e=\frac{c}{a}=\frac{1}{2} \Rightarrow c=1, b=\sqrt{3}$, then
the equation of the ellipse $\Gamma$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
Let $P(2 \cos \theta, \sqrt{3} \sin \theta), \sin \theta \neq 0$, then
$$
\begin{array}{l}
\overrightarrow{D P}=(2 \cos \theta+4, \sqrt{3} \sin \theta) \Rightarrow \overrightarrow{D E}= \\
\frac{3}{5} \overrightarrow{D P}=\left(\frac{6}{5} \cos \theta+\frac{12}{5}, \frac{3 \sqrt{3}}{5} \sin \theta\right) \\
\Rightarrow E\left(\frac{6}{5} \cos \theta-\frac{8}{5}, \frac{3 \sqrt{3}}{5} \sin \theta\right). \text { Thus } A P: y=\frac{\sqrt{3} \sin \theta}{2 \cos \theta+2}(x+2),
\end{array}
$$
$\Rightarrow E\left(\frac{6}{5} \cos \theta-\frac{8}{5}, \frac{3 \sqrt{3}}{5} \sin \theta\right)$. Thus $A P: y=\frac{\sqrt{3} \sin \theta}{2 \cos \theta+2}(x+2)$,
$B E: y=\frac{\sqrt{3} \sin \theta}{2 \cos \theta-6}(x-2)$. Solving the equations of lines $A P$ and $B E$ simultaneously gives $\frac{x+2}{\cos \theta+1}=\frac{x-2}{\cos \theta-3}$
$\Rightarrow x=\cos \theta-1, y=\frac{\sqrt{3}}{2} \sin \theta$. Since point $P$ is different from points $A$ and $B$, the equation of the trajectory of point $Q$ is
$(x+1)^{2}+\frac{4 y^{2}}{3}=1(y \neq 0)$.