Three. (20 points) Given the ellipse $\frac{x^{2}}{20}+\frac{y^{2}}{16}=1$ intersects the $x$-axis at points $A$ and $B$. A tangent line $l$ is drawn through a moving point $M$ on the ellipse (where $M$ 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. The lines $CB$ and $AD$ intersect at point $Q$, and the symmetric point of $Q$ with respect to $M$ is $P$. Find the equation of the trajectory of point $P$.

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The ellipse $\frac{x^{2}}{20}+\frac{y^{2}}{16}=1$ intersects the $x$-axis at points $A$ and $B$. A tangent line $l$ is drawn through a moving point $M$ on the ellipse (where $M$ 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. The lines $CB$ and $AD$ intersect at point $Q$, and the symmetric point of $Q$ with respect to $M$ is $P$. Find the equation of the trajectory of point $P$.
Three, as shown in Figure 4, $A(-2 \sqrt{5}, 0), B(2 \sqrt{5}, 0)$.
Let the moving point $M\left(x_{0}, y_{0}\right)$. Then
the tangent line $l: \frac{x_{0} x}{20}+\frac{y_{0} y}{16}=1$.
Let $C\left(-2 \sqrt{5}, y_{c}\right), D\left(2 \sqrt{5}, y_{D}\right)$, substituting into the equation of $l$ gives
$$
y_{C}=\frac{8\left(10+\sqrt{5} x_{0}\right)}{5 y_{0}}, y_{D}=\frac{8\left(10-\sqrt{5} x_{0}\right)}{5 y_{0}} .
$$

Since $A C / / B D$, the point $Q$ divides $C B$ in the ratio
$$
\lambda=\frac{A C}{B D}=\frac{y_{c}}{y_{D}}=\frac{10+\sqrt{5} x_{0}}{10-\sqrt{5} x_{0}} \text {. }
$$

Thus, $y_{Q}=\frac{y_{C}+\lambda y_{B}}{1+\lambda}=\frac{2\left(100-5 x_{0}^{2}\right)}{25 y_{0}}$,
$$
x_{Q}=\frac{-2 \sqrt{5}+2 \sqrt{5} \lambda}{1+\lambda}=x_{0} \text {. }
$$

Since point $Q$ is symmetric to $P$ about $M\left(x_{0}, y_{0}\right)$, and $\frac{x_{0}^{2}}{20}+\frac{y_{0}^{2}}{16}=1$, then
$$
\begin{array}{l}
x_{P}=x_{0}, \\
y_{P}=2 y_{0}-y_{Q}=2 y_{0}-\frac{2\left(100-5 x_{0}^{2}\right)}{25 y_{0}} \\
=\frac{2\left(x_{0}^{2}+5 y_{0}^{2}-20\right)}{5 y_{0}}=\frac{3}{2} y_{0},
\end{array}
$$

i.e., $x_{0}=x_{P}, y_{0}=\frac{2}{3} y_{P}$.
Therefore, $1=\frac{x_{0}^{2}}{20}+\frac{y_{0}^{2}}{16}=\frac{x_{P}^{2}}{20}+\frac{2^{2} y_{P}^{2}}{3^{2} \times 16}$
$$
=\frac{x_{P}^{2}}{20}+\frac{y_{P}^{2}}{36} \text {. }
$$

By the problem, we know $y_{0} \neq 0 \Rightarrow y_{P} \neq 0$.
Thus, the equation of the trajectory of point $P$ is
$$
\frac{x^{2}}{20}+\frac{y^{2}}{36}=1(y \neq 0) .
$$