In rectangle $ABCD$, $AB=2a, BC=2b, P$ is a moving point, the extensions of $CP$ and $DP$ intersect $AB$ (or its extension) at points $F$ and $E$ respectively. If $\overrightarrow{AE} \cdot \overrightarrow{BF}=-\overrightarrow{EF}^2$, find the equation of the trajectory of point $P$.
Solve: Establish a rectangular coordinate system as shown in Figure 3, with the line $AB$ as the $x$-axis and the midpoint of segment $AB$ as the origin. Then
$$
\begin{array}{l}
A(-a, 0), B(a, 0), \\
C(a, 2b), \\
D(-a, 2b).
\end{array}
$$

Let $P(x, y)$, $E\left(x_{1}, 0\right)$, and $F\left(x_{2}, 0\right)$.
First, from the collinearity of points $D$, $E$, $P$ and $C$, $F$, $P$, we have $\frac{x+a}{x_{1}+a}=\frac{y-2b}{-2b}$, $\frac{x-a}{x_{2}-a}=\frac{y-2b}{-2b}$.
Thus, $x_{1}=\frac{2bx+ay}{2b-y}$, $x_{2}=\frac{2bx-ay}{2b-y}$.
Also, $\overrightarrow{AE}=\left(x_{1}+a, 0\right)$, $\overrightarrow{BF}=\left(x_{2}-a, 0\right)$, so $\overrightarrow{AE} \cdot \overrightarrow{BF}=\left(x_{1}+a\right)\left(x_{2}-a\right)$.
And $\overrightarrow{EF}^{2}=\left(x_{2}-x_{1}\right)^{2}$, hence from $\overrightarrow{AE} \cdot \overrightarrow{BF}=-\overrightarrow{EF}^{2}$,
we get $\left(x_{1}+a\right)\left(x_{2}-a\right)=-\left(x_{1}-x_{2}\right)^{2}$.
Substituting the result (1) into equation (2) yields the trajectory equation of point $P$ as $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
(Xu Dao, Jiangsu Rugao Teacher Training School, 226500)