Question 170, The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity of $\frac{\sqrt{3}}{2}$, a line with a slope of 1 passing through point $M(b, 0)$ intersects the ellipse at points $\mathrm{A}$ and $\mathrm{~B}$. Let $\mathrm{O}$ be the origin, if $\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=\frac{32}{5} \cot \angle \mathrm{AOB}$, find the equation of the ellipse.
Question 170, Answer: Since the eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, we have $a=2b$, thus the equation of the ellipse is $\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1$.
As shown in the figure, it is easy to know that point $\mathrm{A}$ is the lower vertex of the ellipse, so the coordinates of point $\mathrm{A}$ are $(0,-b)$. Construct $B C \perp y$-axis at $C$, then $\triangle$ $A B C$ is an isosceles right triangle. Since $\frac{32}{5} \cot \angle A O B=\overrightarrow{O A} \cdot \overline{O B}=|O A| \cdot|O B| \cos \angle A O B=$ $\mathrm{b} \cdot|O B| \cos \angle A O B$, we have $|O B| \sin \angle A O B=\frac{32}{5 b}$, which means $|B C|=\frac{32}{5 b}$, thus $|O C|=\frac{32}{5 b}-b$. Therefore, the coordinates of point $B$ are $\left(\frac{32}{5 b}, \frac{32}{5 b}-b\right)$. Substituting into the equation of the ellipse, we get $\frac{\left(\frac{32}{5 b}\right)^{2}}{4 \mathrm{~b}^{2}}+\frac{\left(\frac{32}{5 b} b\right)^{2}}{\mathrm{~b}^{2}}=1$, solving for $\mathrm{b}^{2}=4$. Therefore, the equation of the ellipse is $\frac{x^{2}}{16}+$ $\frac{y^{2}}{4}=1$.