11. (20 points) Let point $Z$ move on the circle $|z|=3$ in the complex plane, and let $w=\frac{1}{2}\left(z+\frac{1}{z}\right)$. The trajectory of $w$ is the curve $\Gamma$. A line $l$ passes through the point $P(1,0)$ and intersects the curve $\Gamma$ at points $A$ and $B$, and intersects the imaginary axis at point $M$. If $\overrightarrow{M A}=t \overrightarrow{A P}$ and $\overrightarrow{M B}=s \overrightarrow{B P}$, find the value of $t+s$.
11. Given $|z|=3$, let $z=3 e^{\theta \mathrm{i}}$.

Thus, $w=\frac{1}{2}\left(3 \mathrm{e}^{\theta \mathrm{i}}+\frac{1}{3} \mathrm{e}^{-\theta \mathrm{i}}\right)$ $=\frac{5}{3} \cos \theta+\frac{4}{3} \mathrm{i} \sin \theta$.
Eliminating the parameter $\theta$ yields $\Gamma: \frac{x^{2}}{\frac{25}{9}}+\frac{y^{2}}{\frac{16}{9}}=1$.
Therefore, $P(1,0)$ is the right focus of $\Gamma$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(0, y_{0}\right)$. Then
$$
\begin{array}{l}
\left(x_{1}, y_{1}-y_{0}\right)=t\left(1-x_{1},-y_{1}\right) \\
\Rightarrow x_{1}=\frac{t}{1+t}, y_{1}=\frac{y_{0}}{1+t} \\
\Rightarrow \frac{\left(\frac{t}{1+t}\right)^{2}}{\frac{25}{9}}+\frac{\left(\frac{y_{0}}{1+t}\right)^{2}}{\frac{16}{9}}=1 \\
\Rightarrow \frac{16}{25} t^{2}+2 t+1-\frac{9}{16} y_{0}^{2}=0 .
\end{array}
$$

Similarly, $\frac{16}{25} s^{2}+2 s+1-\frac{9}{16} y_{0}^{2}=0$.
Since $s t<0$, $s$ and $t$ are the roots of the equation
$$
\frac{16}{25} x^{2}+2 x+1-\frac{9}{16} y_{0}^{2}=0
$$

By Vieta's formulas, we have
$$
t+s=\frac{-2}{\frac{16}{25}}=-\frac{25}{8} .
$$