Three. (Full marks 20 points) The two endpoints of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $A(a, 0)$ and $A_{1}(-a, 0)$. A perpendicular line is drawn from a point within the segment $A A_{1}$ intersecting the ellipse at points $C$ and $D$. Connect $A C$ and $A_{1} D$ intersecting at $P$. Find the equation of the curve on which point $P$ lies.
Three, let a perpendicular line be drawn through the point $\left(x_{1}, 0\right)\left(0<\left|x_{1}\right|<a\right)$ on $A A_{1}$, intersecting the ellipse at $C\left(x_{1}, y_{1}\right), D\left(x_{1},-y_{1}\right)$. From the intersection of $A C$ and $A_{1} D$, we know that $0<\left|y_{1}\right|<b$.
The equations of the lines $A C$ and $A_{1} D$ are
$$
\begin{array}{l}
\left(x_{1}-a\right) y=y_{1}(x-a), \\
\left(x_{1}+a\right) y=-y_{1}(x+a) .
\end{array}
$$

Multiplying them, we get
$$
\left(x_{1}^{2}-a^{2}\right) y^{2}=-y_{1}^{2}\left(x^{2}-a^{2}\right) .
$$

Since $C$ and $D$ are on the ellipse, we have
$$
x_{1}^{2}-a^{2}=-\frac{a^{2} y_{1}^{2}}{b^{2}} \text {. }
$$

Substituting into (3), we get
$$
-\frac{a^{2} y^{2}}{b^{2}} y^{2}=-y_{1}^{2}\left(x^{2}-a^{2}\right) .
$$

Cancelling $y_{1} \neq 0$, we obtain
$$
\frac{a^{2} y^{2}}{a^{2}}-x^{2}-a^{2},
$$

which means the equation of the curve on which point $P$ lies is
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text {. }
$$

Analysis: If we are to find the trajectory of point $P$, then the vertices $(a, 0),(-a, 0)$ should be removed from (4). This can be seen from solving (1) and (2):
$$
\left\{\begin{array}{l}
x=\frac{a^{2}}{x_{1}}, \\
y=-\frac{a y_{1}}{x_{1}},
\end{array}\right.
$$

We can see that when $0<|y|<b$, $y \neq 0$. Therefore, $|x| \neq a$.