8. Given three points $A$, $B$, and $C$ on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, the line $BC$ with a negative slope intersects the $y$-axis at point $M$. If the origin $O$ is the centroid of $\triangle ABC$, and the ratio of the areas of $\triangle BMA$ to $\triangle CMO$ is $3:2$, then the slope of the line $BC$ is $\qquad$
8. $-\frac{3 \sqrt{3}}{2}$ or $-\frac{\sqrt{3}}{6}$.

Let $l_{B C}: y=k x+t$.
By eliminating $y$ after combining with the ellipse equation, we get
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k t x+4 t^{2}-4=0 \\
\Rightarrow x_{B} x_{C}=\frac{-4 k t \pm 2 \sqrt{4 k^{2}+1-t^{2}}}{1+4 k^{2}} .
\end{array}
$$

Notice that,
$$
\frac{3}{2}=\frac{S_{\triangle B M A}}{S_{\triangle C M O}}=\frac{3 B M}{C M} \Rightarrow 2 B M=C M .
$$

We consider two cases.
(1) Point $M$ is outside the ellipse. In this case, $x_{C}=2 x_{B}$.

Then $4 k^{2} t^{2}=36 k^{2}+9-9 t^{2}$
$$
\Rightarrow t^{2}=\frac{36 k^{2}+9}{9+4 k^{2}} \text {. }
$$
(2) Point $M$ is inside the ellipse. In this case, $2 x_{B}+x_{C}=0$.

Then $36 k^{2} t^{2}=4 k^{2}+1-t^{2}$
$$
\Rightarrow t^{2}=\frac{4 k^{2}+1}{36 k^{2}+1} \text {. }
$$

Since point $A$ is on the ellipse,
$$
\begin{array}{l}
x_{A}=-x_{B}-x_{C}=\frac{8 k t}{1+4 k^{2}}, \\
y_{A}=-y_{B}-y_{C}=\frac{-2 t}{1+4 k^{2}}, \\
\text { hence } 1=\frac{x_{A}^{2}}{4}+y_{A}^{2}=\frac{4 t^{2}}{1+4 k^{2}} \\
\Rightarrow t^{2}=k^{2}+\frac{1}{4} .
\end{array}
$$

Combining equations (1) and (3), we get $k=-\frac{3 \sqrt{3}}{2}$ or $-\frac{\sqrt{3}}{6}$.