10. The center of the ellipse $\Gamma$ is at the origin $O$, with foci on the $x$-axis, and the eccentricity $e=\sqrt{\frac{2}{3}}$. The line $l$ intersects the ellipse $\Gamma$ at points $A$ and $B$, satisfying $\overrightarrow{C A}=2 \overrightarrow{B C}$, where the fixed point $C(-1,0)$. When $\triangle O A B$ attains its maximum value, find the equation of the ellipse $\Gamma$.
10. Solution: Let the equation of $\Gamma$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. From $b^{2}=\left(1-e^{2}\right) a^{2}=\frac{1}{3} a^{2}$, we know that the equation of $\Gamma$ can be transformed into $x^{2}+3 y^{2}=a^{2}$. Given $\overrightarrow{C A}=2 \overrightarrow{B C}$, we know that $l$ is not parallel to the coordinate axes, so we can set $l: y=k(x+1)$, which can be transformed into $x=\frac{y}{k}-1$. Substituting this into the equation of $\Gamma$ and simplifying, we get $\left(\frac{1}{k^{2}}+3\right) y^{2}-\frac{2}{k} y+1-a^{2}=0$.

To ensure that $A, B$ exist, we should have $\Delta=\frac{4}{k^{2}}-4\left(\frac{1}{k^{2}}+3\right)\left(1-a^{2}\right)>0$; Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $y_{1}+y_{2}=\frac{2 k}{1+3 k^{2}}$. From $\overrightarrow{C A}=2 \overrightarrow{B C}$, we know that $y_{1}=-2 y_{2}$, substituting this into the previous equation gives $y_{2}=\frac{-2 k}{1+3 k^{2}}$, thus
$S_{\triangle O A B}=\frac{1}{2}|O C| \cdot\left|y_{1}-y_{2}\right|=\frac{3}{2}\left|y_{2}\right|=\frac{3|k|}{1+3 k^{2}} \leqslant \frac{3|k|}{2 \sqrt{3}|k|}=\frac{\sqrt{3}}{2}$,
where the equality holds when $k= \pm \frac{\sqrt{3}}{3}$. From $y_{2}=\frac{-2 k}{1+3 k^{2}}$, we get $y_{2}= \pm \frac{\sqrt{3}}{3}$. Substituting these two sets of $k, y$ values into (1), we can solve for $a^{2}=5$, hence the maximum area of $\triangle O A B$ is $\frac{\sqrt{3}}{2}$, and the corresponding equation of $\Gamma$ is $x^{2}+3 y^{2}=5$.