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 value of the area of $\triangle A B C$ is $\qquad$ .
Given $C(3 \cos \theta, 2 \sin \theta), A B: x=n y+1$,
then $|C D|=\frac{3 \cos \theta-2 n \sin \theta-1}{\sqrt{n^{2}+1}}$
$=\frac{\left|\sqrt{4 n^{2}+9} \sin (\theta+\alpha)-1\right|}{\sqrt{n^{2}+1}} \leqslant \frac{\sqrt{4 n^{2}+9}+1}{\sqrt{n^{2}+1}}$.
Solving the system $\left\{\begin{array}{l}x=n y+1, \\ 4 x^{2}+9 y^{2}=36\end{array} \Rightarrow\left(4 n^{2}+9\right) y^{2}+8 n y-32=0\right.$
$\Rightarrow|A B|=\sqrt{n^{2}+1} \cdot \frac{\sqrt{64 n^{2}+4 \cdot 32\left(4 n^{2}+9\right)}}{4 n^{2}+9}$. Thus, $S_{\triangle A B C}=\frac{1}{2}|C D| \cdot|A B|$
$=\frac{6\left(\sqrt{4 n^{2}+9}+1\right) \sqrt{4 n^{2}+8}}{4 n^{2}+9} \xlongequal{t=\sqrt{4 n^{2}+9}} \frac{6(t+1) \sqrt{t^{2}-1}}{t^{2}}=6 \sqrt{\left(1+\frac{1}{t}\right)^{2}\left(1-\frac{1}{t^{2}}\right)}$
$=6 \sqrt{\left(1+\frac{1}{t}\right)^{3}\left(1-\frac{1}{t}\right)}$. Let $u=\frac{1}{t} \in\left(0, \frac{1}{3}\right], y=(1+u)^{3}(1-u)$, then $y^{\prime}=3(1+u)^{2}(1-u)-(1+u)^{3}$ $=2(1+u)^{2}(1-2 u)>0 \Rightarrow y$ is a monotonically increasing function, thus $y_{\max }=\frac{128}{81}$.
Therefore, when $n=0$, the maximum value of $S_{\triangle A B C}$ is $6 \cdot \frac{8 \sqrt{2}}{9}=\frac{16 \sqrt{2}}{3}$.