8. In $\triangle A B C$, it is known that $B C=2 \sqrt{3}$, and the sum of the lengths of the medians on sides $A B$ and $A C$ is 6. Establish a Cartesian coordinate system with line $B C$ as the $x$-axis and the perpendicular bisector of side $B C$ as the $y$-axis. Then the equation of the locus of vertex $A$ is
8. $\frac{x^{2}}{36}+\frac{y^{2}}{9}=1(y \neq 0)$.

Solution 1: Let the medians $C F$ and $B E$ of sides $A B$ and $A C$ intersect at point $G$. Then $G$ is the centroid of $\triangle A B C$.
$$
\begin{array}{l}
\text { By }|G B|+|G C|=\frac{2}{3}(|B E|+|C F|) \\
=\frac{2}{3} \times 6=4>2 \sqrt{3},
\end{array}
$$

we know that the locus of point $G$ is an ellipse with foci at $B$ and $C$ and a major axis length of 4 (excluding the two endpoints of the major axis).
Thus, the equation of this ellipse is $\frac{x^{2}}{4}+y^{2}=1(y \neq 0)$.
Let point $A(x, y)$. Then point $G\left(\frac{1}{3} x, \frac{1}{3} y\right)$.
Substituting into the above equation and simplifying, we get the locus equation of vertex $A$ as
$$
\frac{x^{2}}{36}+\frac{y^{2}}{9}=1(y \neq 0) \text {. }
$$

Solution 2: As shown in Figure 5, draw lines through point $A$ parallel to the medians $B E$ and $C F$, intersecting the $x$-axis at points $B^{\prime}$ and $C^{\prime}$.
$$
\text { Then }\left|A B^{\prime}\right|+\left|A C^{\prime}\right|=2(|B E|+|C F|)=12>6 \sqrt{3} \text {. }
$$

Thus, the locus of point $A$ is an ellipse with foci at $B^{\prime}(-3 \sqrt{3}, 0)$ and $C^{\prime}(3 \sqrt{3}, 0)$ and a major axis length of 12 (excluding the two endpoints of the major axis), whose equation is
$$
\frac{x^{2}}{36}+\frac{y^{2}}{9}=1(y \neq 0) .
$$