Example 8. As shown in Figure 5, the endpoints of the base of $\triangle \mathrm{ABC}$ are $B(0,6)$ and $C(0,-6)$, and the product of the slopes of the other two sides is $-\frac{4}{9}$. Find the equation of the locus of vertex $\mathrm{A}$.
Let the coordinates of vertex $\mathrm{A}$ be $(x, y)$,
then $k_{A B}=\frac{y-6}{x}, k_{A C}=\frac{y+6}{x}$.
According to the problem, we have
$$
\frac{y-6}{x} \cdot \frac{y+6}{x}=-\frac{4}{9},
$$

Simplifying, we get
$$
\frac{x^{2}}{81}+\frac{y^{2}}{36}=1
$$

Is this the equation of the trajectory of vertex $\mathrm{A}$? Examining the simplification process, it can be seen that transforming (1) into (2) is not an equivalent transformation; the denominator $x$ in the left side of (1) should not be zero. Therefore, the trajectory equation of vertex $\mathrm{A}$ is
$$
\frac{x^{2}}{81}+\frac{y^{2}}{36}=1 \quad(x \neq 0),
$$

which means the trajectory of vertex $\mathrm{A}$ is an ellipse excluding points $\mathrm{B}$ and $\mathrm{C}$.