15. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the point $M(0,2)$, and the right focus is $F(2,0)$.
(1) Write the equation of the ellipse $C$;
(2) A line $l$ passing through point $F$ intersects the ellipse $C$ at points $A, B$, and intersects the $y$-axis at point $P$; if $\overrightarrow{P A}=m \overrightarrow{A F}, \overrightarrow{P B}=n \overrightarrow{B F}$. Prove that $m+n$ is a constant;
(3) Under the condition in (2), if point $P$ is not inside the ellipse $C$, and point $Q$ is the point symmetric to $P$ with respect to the origin $O$, find the minimum value of the area of triangle $Q A B$.
(1) From the problem: $b=2, c=2, \therefore a^{2}=8$, the equation of the ellipse $C$ is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$;
(2) Let the coordinates of $A, B, P$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),(0, t)$, respectively. From $\overrightarrow{P A}^{8}=m \overrightarrow{A F}$, we have
$$
x_{1}=\frac{2 m}{1+m}, \quad y_{1}=\frac{t}{1+m}
$$

Since point $A$ is on the ellipse $C$, then
$$
\frac{\left(\frac{2 m}{1+m}\right)^{2}}{8}+\frac{\left(\frac{t}{1+m}\right)^{2}}{4}=1
$$

Simplifying gives $2 m^{2}+8 m-t^{2}+4=0$. Similarly, from $\overrightarrow{P B}=n \overrightarrow{B F}$, we get $2 n^{2}+8 n-t^{2}+4=0$.
Since $A, B$ are not coincident, i.e., $m \neq n$, $m, n$ are the two roots of the quadratic equation $2 x^{2}+8 x-t^{2}+4=0$, so $m+n=-4$ is a constant;
(3) According to the problem, the equation of the line $l$ is $\frac{x}{2}+\frac{y}{t}=1$, i.e., $y=-\frac{t}{2}(x-2)$, combined with the equation of the ellipse $C$, eliminating $y$ and rearranging, we get:
$$
\begin{array}{c}
\left(2+t^{2}\right) x^{2}-4 t^{2} x+4 t^{2}-16=0 \\
\Delta=16 t^{4}-4\left(2+t^{2}\right)\left(4 t^{2}-16\right)=32 t^{2}+128>0 \\
\therefore x_{1}+x_{2}=\frac{4 t^{2}}{2+t^{2}}, x_{1} x_{2}=\frac{4 t^{2}-16}{2+t^{2}}
\end{array}
$$

And
$$
\begin{array}{c}
S_{\triangle Q A B}=\frac{1}{2} \cdot|2 t| \cdot\left|x_{1}-x_{2}\right|=|t| \cdot\left|x_{1}-x_{2}\right| \\
S_{\triangle Q A B}^{2}=t^{2}\left(x_{1}-x_{2}\right)^{2}=t^{2}\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]=t^{2}\left[\frac{16 t^{4}}{\left(2+t^{2}\right)^{2}}-\frac{16 t^{2}-64}{2+t^{2}}\right]=t^{2} \cdot \frac{32 t^{2}+128}{\left(2+t^{2}\right)^{2}}=32\left[1-\frac{4}{\left(2+t^{2}\right)^{2}}\right]
\end{array}
$$

From the given, point $P$ is not inside the ellipse $C$, so $|t| \geq 2$, i.e., $t^{2} \geq 4$. Therefore, the minimum value of $S_{\triangle Q A B}^{2}$ is $32 \cdot \frac{8}{9}$, hence the minimum area of triangle $Q A B$ is $\frac{16}{3}$.