8. Given that $\triangle A B C$ is an inscribed triangle in the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and $A B$ passes through the point $P(1,0)$. Then the maximum area of $\triangle A B C$ is $\qquad$ .
8. $\frac{16 \sqrt{2}}{3}$.

After the scaling transformation $\left\{\begin{array}{l}x=3 X, \\ y=2 Y,\end{array}\right.$, we get $\triangle A^{\prime} B^{\prime} C^{\prime}$ inscribed in the circle $X^{2}+Y^{2}=1, A^{\prime} B^{\prime}$ passing through the point $P^{\prime}\left(\frac{1}{3}, 0\right)$,
$$
S_{\triangle A B C}=6 S_{\triangle A^{\prime} B^{\prime} C^{\prime}} .
$$

Let the distance from point $O^{\prime}$ to $A^{\prime} B^{\prime}$ be $t$. Then
$$
\begin{array}{l}
0 \leqslant t \leqslant \frac{1}{3}, \\
\left|A^{\prime} B^{\prime}\right|=2 \sqrt{1-t^{2}}, \\
S_{\triangle A^{\prime} B^{\prime} C^{\prime}} \leqslant \sqrt{1-t^{2}}(1+t) .
\end{array}
$$

It is easy to see that when $t=\frac{1}{3}$, $S_{\triangle A^{\prime} B^{\prime} C}$ has a maximum value of $\frac{8 \sqrt{2}}{9}$. Therefore, the maximum value of $S_{\triangle A B C}$ is $\frac{16 \sqrt{2}}{3}$.