14. In the Cartesian 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 $A C$.
( I ) Find the equation of the locus of point $P$;
(II) 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 $l$.
( I ) $2 x^{2}+2 y^{2}+3 y-2=0$ and the hyperbola $T$:
$$
8 x^{2}-17 y^{2}+12 y-8=0 \text { ) }
$$
( II ) $\left\{0, \pm \frac{1}{2}, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\}$
14. 【Analysis and Solution】( I ) The equations of the lines $A B$, $A C$, and $B C$ are sequentially $y=\frac{4}{3}(x+1)$, $y=-\frac{4}{3}(x-1)$, and $y=0$. The distances from point $P(x, y)$ to $A B$, $A C$, and $B C$ are sequentially $d_{1}=\frac{1}{5}|4 x-3 y+4|$, $d_{2}=\frac{1}{5}|4 x+3 y-4|$, and $d_{3}=|y|$. According to the problem, $d_{1} d_{2}=d_{3}^{2}$, which gives $\left|16 x^{2}-(2 y-4)^{2}\right|=25 y^{2}$, or $16 x^{2}-(3 y-4)^{2}+25 y^{2}=0$, or $16 x^{2}-(3 y-4)^{2}-25 y^{2}=0$. Simplifying, we get the trajectory equation of point $P$ as the circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$ and the hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$.
( II) From the above, the trajectory of point $P$ consists of two parts:
Circle $\mathrm{S}: 2 x^{2}+2 y^{2}+3 y-2=0$
and hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$
Since: $B(-1,0)$ and $C(1,0)$ satisfy the conditions of the problem, points $B$ and $C$ are on the trajectory of point $P$, and the common points of the trajectory curves $S$ and $T$ are only $B$ and $C$.
The incenter $D$ of $\triangle A B C$ also satisfies the conditions of the problem. By $d_{1}=d_{2}=d_{3}$, we solve to get $D\left(0, \frac{1}{2}\right)$, and it is known that it lies on the circle $S$. The line $l$ passes through $D$, and intersects the trajectory of point $P$ at 3 points, so the slope of $l$ exists. Let the equation of $l$ be $y=k x+\frac{1}{2}$.
(i) When $k=0$, $l$ is tangent to the circle $S$ and has a unique common point $D$; at this time, the line $y=\frac{1}{2}$ is parallel to the $x$-axis, indicating that $l$ intersects the hyperbola at two points different from $D$, so $l$ intersects the trajectory of point $P$ at exactly 3 points. .10 points
(ii) When $k \neq 0$, $l$ intersects the circle $S$ at two different points. At this time, $l$ intersects the trajectory of point $P$ at exactly 3 points in only two cases:
Case 1: The line $l$ passes through point $B$ or point $C$, at this time the slope of $l$ is $k= \pm \frac{1}{2}$, and the equation of the line $l$ is $x= \pm(2 y-1)$.
Substituting into equation (2) gives $y(3 y-4)=0$, solving for $E\left(\frac{5}{3}\right), \frac{4}{3}$ or $F\left(-\frac{5}{3}, \frac{4}{3}\right)$. This indicates that the line $B D$ intersects the curve $T$ at 2 points $B$ and $E$; the line $C D$ intersects the curve $T$ at 2 points $C$ and $F$.
Therefore, when $k= \pm \frac{1}{2}$, $L$ intersects the trajectory of point $P$ at exactly 3 points.
Case 2: The line $l$ does not pass through points $B$ and $C$ (i.e., $\left.k \neq \pm \frac{1}{2}\right)$, since $l$ intersects $S$ at two different points, $l$ intersects the hyperbola $T$ at exactly one point, i.e., 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 gives $\left(8-17 k^{2}\right) x^{2}-5 k x-\frac{25}{4}=0$,
The necessary and sufficient condition for this equation to have a unique real solution is $8-17 k^{2}=0$
or $(-5 k)^{2}+4\left(8-17 k^{2}\right) \frac{25}{4}=0$
Solving equation (4) gives $k= \pm \frac{2 \sqrt{34}}{17}$, solving equation (5) gives $k= \pm \frac{\sqrt{2}}{2}$.
In summary, the range of the slope $k$ of the line $l$ is the finite set $\left\{0, \pm \frac{1}{2}, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\}$
.20 points