10. Given the ellipse $\frac{x^{2}}{20}+\frac{y^{2}}{16}=1$ intersects the $x$-axis at points $A, B$, and a tangent line $l$ is drawn through a moving point $M$ on the ellipse (where $M$ does not coincide with $A, B$). Perpendicular lines are drawn from points $A, B$ to the $x$-axis, intersecting the tangent line $l$ at points $C, D$ respectively. The lines $C B, A D$ intersect at point $Q$, and the symmetric point of $Q$ with respect to $M$ is $P$. Find the equation of the trajectory of point $P$.
Given $M(m, n)$, then the tangent line $l: \frac{m x}{20}+\frac{n y}{16}=1$.
Substitute $x_{A}=-2 \sqrt{5}$, we get $C\left(-2 \sqrt{5}, \frac{16}{n}+\frac{8 \sqrt{5} m}{5 n}\right)$,
Similarly, $D\left(2 \sqrt{5}, \frac{16}{n}-\frac{8 \sqrt{5} m}{5 n}\right)$.
Thus, $k_{A D}=\frac{4 \sqrt{5}-2 m}{5 n}, k_{B C}=-\frac{4 \sqrt{5}+2 m}{5 n}$.
Solving the system $\left\{\begin{array}{l}x=\frac{5 n}{4 \sqrt{5}-2 m} y-2 \sqrt{5}, \\ x=-\frac{5 n}{4 \sqrt{5}+2 m} y+2 \sqrt{5}\end{array} \Rightarrow\left\{\begin{array}{l}x=m, \\ y=\frac{40-2 m^{2}}{5 n}\end{array} \Rightarrow Q\left(m, \frac{40-2 m^{2}}{5 n}\right)\right.\right.$.
Therefore, $P\left(m, \frac{2 m^{2}+10 n^{2}-40}{5 n}\right)$. Substituting $m=2 \sqrt{5} \cos \theta, n=4 \sin \theta$ into the coordinates of point $P$, we get $P(2 \sqrt{5} \cos \theta, 6 \sin \theta)$. Thus, the trajectory equation of point $P$ is $\frac{y^{2}}{36}+\frac{x^{2}}{20}=1(y \neq 0)$.