7. Given the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, which intersects the $x$-axis at points $A$ and $B$. A tangent line $l$ is drawn through a point $P$ on the ellipse (where $P$ does not coincide with $A$ or $B$). Vertical lines are drawn from points $A$ and $B$ to the $x$-axis, intersecting the tangent line $l$ at points $C$ and $D$, respectively. Let $C B$ and $A D$ intersect at point $Q$, and the symmetric point of $Q$ with respect to $P$ is $S$. Then the equation of the trajectory of $S$ is $\qquad$
7. $\frac{x^{2}}{5}+\frac{y^{2}}{9}=1(x \neq \pm \sqrt{5})$.

As shown in Figure 3, let $P\left(x_{0}, y_{0}\right)$.
It is easy to know that $l_{C D}: \frac{x_{0} x}{5}+\frac{y_{0} y}{4}=1$.
Then, $C\left(-\sqrt{5}, \frac{4\left(\frac{x_{0}}{\sqrt{5}}+1\right)}{y_{0}}\right)$,
$D\left(\sqrt{5}, \frac{4\left(1-\frac{x_{0}}{\sqrt{5}}\right)}{y_{0}}\right)$.
Thus, $\frac{A Q}{Q D}=\frac{\sqrt{5}+x_{0}}{\sqrt{5}-x_{0}}$.
Therefore, $P Q / / A C$.
It is easy to know that $Q\left(x_{0}, \frac{y_{0}}{2}\right)$. Hence, $S\left(x_{0}, \frac{3 y_{0}}{2}\right)$.
Therefore, the equation of the trajectory of point $S$ is $\frac{x^{2}}{5}+\frac{y^{2}}{9}=1(x \neq \pm \sqrt{5})$.