15. Given that point $F$ is the common focus of the parabola $C: x^{2}=4 y$ and the ellipse $\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$, and the maximum distance from a point $M$ on the ellipse to point $F$ is 3.
(1) Find the equation of the ellipse;
(2) Draw two tangent lines from point $M$ to $C$, and let the points of tangency be $A$ and $B$. Find the maximum value of the area of $\triangle M A B$.
(1) $F(0,1) \Rightarrow c=1$, and $a+c=3 \Rightarrow a=2, b=\sqrt{3}$.

So the equation of the ellipse is $\frac{y^{2}}{4}+\frac{x^{2}}{3}=1$.
(2) As shown in the figure, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
$M(\sqrt{3} \cos \theta, 2 \sin \theta)$, then the equations of the tangent lines to the parabola at
$A, B$ are
$$
x_{1} x=2\left(y+y_{1}\right), x_{2} x=2\left(y+y_{2}\right) \text {. }
$$

Substituting the point $M(\sqrt{3} \cos \theta, 2 \sin \theta)$, we get
$$
A B: \sqrt{3} x \cos \theta-2 y-4 \sin \theta=0 \text {. }
$$

Solving the system $\left\{\begin{array}{l}2 y=\sqrt{3} x \cos \theta-4 \sin \theta, \\ x^{2}=4 y\end{array}\right.$
$$
\Rightarrow x^{2}-2 \sqrt{3} x \cos \theta+8 \sin \theta=0 \text {. }
$$

Thus, $|A B|=\sqrt{1+\left(\frac{\sqrt{3} \cos \theta}{2}\right)^{2}} \cdot \sqrt{12 \cos ^{2} \theta-32 \sin \theta}$, and the distance from point $M$ to line $A B$ is $d=\left|\frac{3 \cos ^{2} \theta-8 \sin \theta}{\sqrt{4+3 \cos ^{2} \theta}}\right|$.
Therefore, $S_{\triangle M A B}=\frac{1}{2}|A B| \cdot d=\frac{1}{2}\left|3 \cos ^{2} \theta-8 \sin \theta\right| \cdot \sqrt{3 \cos ^{2} \theta-8 \sin \theta}$ $=\frac{1}{2}\left(-3 \sin ^{2} \theta-8 \sin \theta+3\right)^{\frac{3}{2}}=\frac{1}{2}\left[-3\left(\sin \theta+\frac{4}{3}\right)^{2}+\frac{25}{3}\right]^{\frac{3}{2}}$.
So when $\sin \theta=-1$, the maximum area of $\triangle M A B$ is $\frac{1}{2} \cdot 8^{\frac{3}{2}}=8 \sqrt{2}$.