10. (20 points) Let the line $y=2 \sqrt{2} x+2 \sqrt{3}$ intersect the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ at points $A$ and $B$, with the origin of coordinates being $O$. If $\angle A O B=90^{\circ}$ and $A B=\frac{12 \sqrt{11}}{17}$, find the equation of the ellipse.
10. As shown in Figure 5.
Let $\angle x O A=\alpha$. Then $\angle x O B=\alpha+\frac{\pi}{2}$.
Thus, $A(O A \cos \alpha, O A \sin \alpha)$, $B(-O B \sin \alpha, O B \cos \alpha)$.
Substituting into the ellipse equation, we get
$\frac{1}{O A^{2}}=\frac{\cos ^{2} \alpha}{a^{2}}+\frac{\sin ^{2} \alpha}{b^{2}}$,
$\frac{1}{O B^{2}}=\frac{\sin ^{2} \alpha}{a^{2}}+\frac{\cos ^{2} \alpha}{b^{2}}$.
Therefore, $\frac{1}{O A^{2}}+\frac{1}{O B^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$.
Thus, the distance from point $O$ to $A B$ is
$\frac{2 \sqrt{3}}{\sqrt{(2 \sqrt{2})^{2}+1^{2}}}=\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{O A \cdot O B}{\sqrt{O A^{2}+O B^{2}}}=\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{1}{O A^{2}}+\frac{1}{O B^{2}}=\frac{3}{4}$
$\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{3}{4}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $x_{1} 、 x_{2}$ are the roots of the equation
$$
\frac{x^{2}}{a^{2}}+\frac{(2 \sqrt{2} x+2 \sqrt{3})^{2}}{b^{2}}=1 \text {, }
$$

which is $\left(\frac{1}{a^{2}}+\frac{8}{b^{2}}\right) x^{2}+\frac{8 \sqrt{6}}{b^{2}} x+\frac{12}{b^{2}}-1=0$,
or $\left(\frac{3}{4}+\frac{7}{b^{2}}\right) x^{2}+\frac{8 \sqrt{6}}{b^{2}} x+\frac{12}{b^{2}}-1=0$.
Thus, $x_{1}+x_{2}=\frac{-\frac{8 \sqrt{6}}{b^{2}}}{\frac{3}{4}+\frac{7}{b^{2}}}, x_{1} x_{2}=\frac{\frac{12}{b^{2}}-1}{\frac{3}{4}+\frac{7}{b^{2}}}$.
$$
\begin{array}{l}
\text { Hence } \frac{1}{9} A B^{2}=\frac{1}{9}\left(\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right) \\
=\left(x_{2}-x_{1}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2} \\
\Rightarrow \frac{1}{16} \times \frac{1}{9} A B^{2}=\frac{64 \times 6}{\left(3 b^{2}+28\right)^{2}}-\frac{12-b^{2}}{3 b^{2}+28} \\
\quad=\frac{3 b^{4}-8 b^{2}+48}{\left(3 b^{2}+28\right)^{2}} .
\end{array}
$$

According to the given $\frac{1}{144} A B^{2}=\frac{11}{289}$, then
$\frac{3 b^{4}-8 b^{2}+48}{\left(3 b^{2}+28\right)^{2}}=\frac{11}{289}$
$\Rightarrow 12 b^{4}-65 b^{2}+82=0$
$\Rightarrow\left(b^{2}-2\right)\left(12 b^{2}-41\right)=0$
$\Rightarrow b^{2}=2$ or $\frac{41}{12}$.
Since $\frac{2}{b^{2}}>\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{3}{4} \Rightarrow b^{2}<\frac{8}{3}$.
Thus, $b^{2}=2$. Therefore, $a^{2}=4$.
Hence, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$.