8. Given that $A$, $B$, and $C$ are three points on the parabola $y^{2}=x$, and $\left(2, \frac{4}{3}\right)$ is the centroid of $\triangle A B C$. Then the maximum value of the area of $\triangle A B C$ is $\qquad$ .
8. $\frac{\sqrt{3}}{9}$.

Let $A\left(a^{2}, a\right), B\left(b^{2}, b\right), C\left(c^{2}, c\right)$.
Given that $F$ is the centroid of $\triangle A B C$
$$
\Rightarrow a+b+c=4, a^{2}+b^{2}+c^{2}=6 \text {. }
$$

Then $a b=\frac{1}{2}\left((a+b)^{2}-a^{2}-b^{2}\right)=c^{2}-4 c+5$.
By $(b-c)^{2}=2\left(b^{2}+c^{2}\right)-(b+c)^{2}$
$=2\left(6-c^{2}\right)-(4-c)^{2}$
$=-4+8 c-3 c^{2}>0$
$$
\Rightarrow \frac{2}{3}<c<2 \text {. }
$$

Notice that,
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2}\left|a^{2} b+b^{2} c+c^{2} a-a b^{2}-b c^{2}-c a^{2}\right| \\
=\frac{1}{2}|a-b||(a-c)(b-c)| \\
=\frac{1}{2} \sqrt{-4+8 c-3 c^{2}}\left|5-8 c+3 c^{2}\right| . \\
\text { Let } t=-4+8 c-3 c^{2}=-3\left(c-\frac{4}{3}\right)^{2}+\frac{4}{3} . \text { Then } \\
S_{\triangle A B C}=\frac{1}{2} \sqrt{t}|1-t| .
\end{array}
$$

Let $t=-4+8 c-3 c^{2}=-3\left(c-\frac{4}{3}\right)^{2}+\frac{4}{3}$. Then

When $t \leqslant 1$, we have
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \sqrt{t}|1-t| \\
\quad=\frac{1}{2 \sqrt{2}} \sqrt{2 t(1-t)(1-t)} \\
\quad \leqslant \frac{1}{2 \sqrt{2}} \sqrt{\left(\frac{2 t+1-t+1-t}{3}\right)^{3}}=\frac{\sqrt{3}}{9} .
\end{array}
$$

When $1<t \leqslant \frac{4}{3}$, we have
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \sqrt{t}|1-t| \\
\leqslant \frac{1}{2} \times \sqrt{\frac{4}{3}}\left(\frac{4}{3}-1\right)=\frac{\sqrt{3}}{9} .
\end{array}
$$

In particular, when $a=\frac{4-\sqrt{3}}{3}, b=\frac{4+\sqrt{3}}{3}, c=\frac{4}{3}$, $\left(S_{\triangle A B C}\right)_{\max }=\frac{\sqrt{3}}{9}$.