2. Let an ellipse be centered at the origin, with its major axis on the $x$-axis, and eccentricity $e$. If the maximum distance between a point on the circle $C: x^{2}+\left(y-\frac{3}{2}\right)^{2}=1$ and a point on this ellipse is $1+\sqrt{7}$, find the equation of this ellipse.
2. Let point $P$ be on the circle and $Q$ on the desired ellipse, and $|P Q|=1+\sqrt{7}$, then points $P$, $C$, and $Q$ are collinear. The fixed point $C\left(\frac{3}{2}, 0\right)$ has a maximum distance of $\sqrt{7}$ to the point on the ellipse.
Assume the ellipse equation is $\left\{\begin{array}{l}x=a \cos \theta \\ y=b \sin \theta\end{array} \quad(\theta\right.$ is the parameter), then $|C Q|^{2}=a^{2} \cos \theta^{2}+\left(b \sin \theta-\frac{3}{2}\right)^{2}$.
Since $e=\frac{\sqrt{3}}{2}$, then $a=2 b$. Therefore, $|C Q|^{2}=-3 b^{2}\left(\sin \theta+\frac{1}{2 b}\right)^{2}+4 b^{2}+3$. If $b<\frac{1}{2}$, it leads to a contradiction. Thus, $b \geqslant \frac{1}{2}$, and $|C Q|_{\max }^{2}=4 b^{2}+3=7$, solving for $b=1$. Hence, $a=2$. Therefore, the ellipse equation is $\left\{\begin{array}{l}x=2 \cos \theta \\ y=\sin \theta\end{array}\right.$ ( $\theta$ is the parameter).