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 $A C$.
(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) $A B: y=\frac{4}{3}(x+1), A C: y=-\frac{4}{3}(x-1)$, let $P(x, y)$,

Thus, $y^{2}=\frac{\left|y-\frac{4}{3}(x+1)\right|}{\frac{5}{3}} \cdot \frac{\left|y+\frac{4}{3}(x-1)\right|}{\frac{5}{3}} \Rightarrow \frac{25 y^{2}}{9}=\left|\left(y-\frac{4}{3}\right)^{2}-\frac{16 x^{2}}{9}\right|$. When $\left(y-\frac{4}{3}\right)^{2} \geqslant \frac{16 x^{2}}{9}$, $\frac{16 x^{2}}{9}+\frac{16 y^{2}}{9}+\frac{8}{3} y-\frac{16}{9}=0 \Rightarrow x^{2}+y^{2}+\frac{3}{2} y-1=0$; when $\left(y-\frac{4}{3}\right)^{2}<\frac{16 x^{2}}{9}$, $\frac{16 x^{2}}{9}-\frac{34 y^{2}}{9}+\frac{8}{3} y-\frac{16}{9}=0 \Rightarrow 8 x^{2}-17 y^{2}+12 y-8=0$. Therefore, the trajectory equation of point $P$ is the circle $x^{2}+y^{2}+\frac{3}{2} y-1=0$ and the hyperbola $8 x^{2}-17 y^{2}+12 y-8=0$. (2) As shown in the figure, point $D\left(0, \frac{1}{2}\right)$ is on the trajectory of point $P$. Then the lines passing through $D$ and intersecting the trajectory of point $P$ at exactly 3 points have the following four cases:
(1) Parallel to the $x$-axis: $y=\frac{1}{2}$, corresponding slope $k=$
(2) Lines $D B$ and $D C: y= \pm \frac{1}{2} x+1$,

corresponding slope $k= \pm \frac{1}{2}$;
(3) Parallel to the asymptotes of the hyperbola:
$$
y= \pm \frac{2 \sqrt{34}}{17} x+\frac{1}{2} \text {, }
$$

corresponding slope $k= \pm \frac{2 \sqrt{34}}{17}$;
(4) Tangent to the hyperbola:

Solving the system $\left\{\begin{array}{l}y=k x+\frac{1}{2}, \\ 8 x^{2}-17 y^{2}+12 y-8=0\end{array} \Rightarrow\left(8-17 k^{2}\right) x^{2}-5 k x-\frac{25}{4}=0\right.$.
Thus, $\Delta=25 k^{2}+25\left(8-17 k^{2}\right)=25\left(8-16 k^{2}\right)=0 \Rightarrow k= \pm \frac{\sqrt{2}}{2}$.
In summary, the range of the slope $k$ of line $l$ is $\left\{0, \pm \frac{1}{2}, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\}$.