9. Given the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

with an eccentricity of $\frac{\sqrt{3}}{2}$, a line with a slope of 1 passing through the point $M(b, 0)$ intersects the ellipse at points $A$ and $B$. Let $O$ be the origin. If
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=\frac{32}{5} \cot \angle A O B,
$$

then the equation of the ellipse is $\qquad$
9. $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$.

From $e=\frac{\sqrt{3}}{2}=\frac{c}{a}$, we know $a=2 b, c=\sqrt{3} b$.
From $\left\{\begin{array}{l}y=x-b, \\ x^{2}+4 y^{2}=4 b^{2},\end{array}\right.$ we get $B(0,-b), A\left(\frac{8 b}{5}, \frac{3 b}{5}\right)$.
Also, $\angle A O B=\frac{\pi}{2}+\angle A O X$, so $\cot \angle A O B=-\tan \angle A O X=-k_{O A}=-\frac{3}{8}$. Noting that, $\overrightarrow{O A} \cdot \overrightarrow{O B}=\frac{32}{5} \cot \angle A O B$. Thus, $-\frac{3 b^{2}}{5}=-\frac{3}{8} \times \frac{32}{5}$.

Therefore, $b=2, a=4$.
Hence, the equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$.