Example 9 (2004 National High School Competition Question) In the plane rectangular coordinate system $x O y$, given three points $A\left(0, \frac{4}{3}\right), B(-1,0), C$ $(1,0)$, the distance from point $P$ to line $B C$ is the geometric mean of the distances from the point to lines $A B$ and $A C$.
(1) Find the equation of the locus of point $P$;
(2) If line $l$ passes through the incenter of $\triangle A B C$ (denoted as $D$), and intersects the locus of point $P$ at exactly 3 points, find the range of the slope $k$ of line $l$.
(1) The equations of lines $AB$, $AC$, and $BC$ are
$$
y=\frac{4}{3}(x+1), y=-\frac{4}{3}(x-1), y=0 \text {. }
$$

The distances from point $P(x, y)$ to $AB$, $AC$, and $BC$ are
$$
\begin{array}{l}
d_{1}=\frac{1}{5}|4 x-3 y+4|, d_{2}=\frac{1}{5}|4 x+3 y-4|, \\
d_{3}=|y|, \text { and } d_{1} d_{2} d_{3} \neq 0 .
\end{array}
$$

According to the problem, we have $\left|16 x^{2}-(3 y-4)^{2}\right|=25 y^{2}$,
which simplifies to $16 x^{2}-(3 y-4)^{2}+25 y^{2}=0$,
or $16 x^{2}-(3 y-4)^{2}-25 y^{2}=0$.
Simplifying further, the equation of the trajectory of point $P$ is
the circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$ and the hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$, excluding points $B$ and $C$.
(2) From the above, the trajectory of point $P$ consists of two parts:
$$
2 x^{2}+2 y^{2}+3 y-2=0 \text {, }
$$

and $8 x^{2}-17 y^{2}+12 y-8=0$.
(Excluding points $B$ and $C$.)
Given $d_{1}=d_{2}=d_{3}$, the incenter $D$ of $\triangle ABC$ satisfies the conditions of the problem, and is found to be $D\left(0, \frac{1}{2}\right)$, which lies on circle $S$. The line $l$ passes through $D$ and intersects the trajectory of point $P$ at 3 points, so the slope of line $l$ exists. Let the equation of line $l$ be
$$
y=k x+\frac{1}{2} \text {. }
$$
(i) When $k \doteq 0$, line $l$ is tangent to circle $S$ and has only one common point $D$; in this case, the line $y=\frac{1}{2}$ is parallel to the $x$-axis, indicating that line $l$ intersects the hyperbola $T$ at two points different from $D$, so line $l$ intersects the trajectory of point $P$ at exactly 3 points.
(ii) When $k \neq 0$, since points $B$ and $C$ are not on the trajectory, $k \neq \pm \frac{1}{2}$. In this case, for line $l$ to intersect the trajectory of point $P$ at exactly 3 points, line $l$ must intersect circle $S$ at two distinct points and intersect the hyperbola $T$ at exactly one point, meaning the system of equations
$$
\left\{\begin{array}{l}
8 x^{2}-17 y^{2}+12 y-8=0, \\
y=k x+\frac{1}{2}
\end{array}\right.
$$

has exactly one real solution. Eliminating $y$ and simplifying, we get
$$
\left(8-17 k^{2}\right) x^{2}-5 k x-\frac{25}{4}=0 \text {. }
$$

or $(-5 k)^{2}+4\left(8-17 k^{2}\right) \frac{25}{4}=0$.
Solving (4) gives $k= \pm \frac{2 \sqrt{34}}{17}$, and solving (5) gives $k= \pm \frac{\sqrt{2}}{2}$.
In summary, the set of possible values for the slope $k$ of line $l$ is
$$
\left\{0, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\} .
$$