Four. (20 points) As shown in Figure 3, given that $A$ and $B$ are the endpoints of the major axis of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

$P$ is a point on the ellipse different from $A$ and $B$. Lines $l_{1} \perp P A$ and $l_{2} \perp P B$ are drawn through points $A$ and $B$ respectively, and the lines $l_{1}$ and $l_{2}$ intersect at point $M$. Find the equation of the locus of point $M$ as point $P$ moves along the ellipse.
From the problem, we know that points $A, P, B, M$ lie on the circle $\odot N$, where the center $N$ is on the perpendicular bisector of chord $AB$, the $y$-axis, with coordinates $N(0, t)$.
Let $P\left(x_{0}, y_{0}\right), M(x, y)$.
Since $N$ is the midpoint of diameter $PM$, we have
$$
x_{0}=-x, y_{0}=2 t-y.
$$

Since point $P$ lies on the ellipse, we have $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$.
Thus, $\frac{x^{2}}{a^{2}}+\frac{(2 t-y)^{2}}{b^{2}}=1$.
Also, since $N A=N M$, we get
$$
\begin{array}{l}
t^{2}+a^{2}=(t-y)^{2}+x^{2} \\
\Rightarrow t=\frac{x^{2}+y^{2}-a^{2}}{2 y}.
\end{array}
$$

From equations (1) and (2), we have
$$
\begin{array}{l}
\frac{x^{2}}{a^{2}}+\left(\frac{x^{2}-a^{2}}{b y}\right)^{2}=1 \\
\Rightarrow \frac{x^{2}-a^{2}}{a^{2}}+\frac{\left(x^{2}-a^{2}\right)^{2}}{(b y)^{2}}=0.
\end{array}
$$

Since $|x|=\left|x_{0}\right|<a$, we have $x^{2}-a^{2} \neq 0$.
Therefore, from equation (3), we get
$$
\begin{array}{l}
\frac{1}{a^{2}}+\frac{x^{2}-a^{2}}{b^{2} y^{2}}=0 \Rightarrow \frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=\frac{a^{2}}{b^{2}} \\
\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{\left(\frac{a^{2}}{b}\right)^{2}}=1(|x|<a),
\end{array}
$$

which represents an ellipse (excluding its two intersection points with the $x$-axis $(\pm a, 0)$).