11. (20 points) Given the standard equation of the ellipse $\Gamma$ as $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a=2 b>0)$, and $\Gamma$ passes through the point $P(4,1)$, $A$ and $B$ are two moving points on $\Gamma$. If the external angle bisector of $\angle A P B$ is the tangent line to $\Gamma$ at point $P$, find the locus of the midpoint $M$ of segment $A B$.
11. From the problem, we know $\frac{4^{2}}{a^{2}}+\frac{1^{2}}{b^{2}}=1$.

Combining $a=2 b$, we get the standard equation of the ellipse $\Gamma$ as $\frac{x^{2}}{20}+\frac{y^{2}}{5}=1$.
It is easy to see that the slope of the tangent line to $\Gamma$ at point $P$ is -1, thus, the slope of the angle bisector of $\angle A P B$ is 1.
(1) When the slopes of lines $P A$ and $P B$ exist, $k_{P A} k_{P B}=1$.
Below, we prove that line $A B$ passes through a fixed point.
$$
\text { Let }\left\{\begin{array}{l}
y^{\prime}=y-1, \\
x^{\prime}=x-4 .
\end{array}\right.
$$

Assume the equation of line $A B$ is
$m x^{\prime}+n y^{\prime}=1(m 、 n$ are not both 0$)$.
Then $\Gamma: 4 y^{\prime 2}+x^{\prime 2}+8 x^{\prime}+8 y^{\prime}=0$.
Thus, $4 y^{\prime 2}+x^{\prime 2}+8\left(x^{\prime}+y^{\prime}\right)\left(m x^{\prime}+n y^{\prime}\right)=0$.
Rearranging, we get
$$
(8 n+4) y^{\prime 2}+8(m+n) x^{\prime} y^{\prime}+(8 m+1) x^{\prime 2}=0 \text {, }
$$

which is $(8 n+4) k^{2}+8(m+n) k+8 m+1=0$, where $k=\frac{y^{\prime}}{x^{\prime}}=\frac{y-1}{x-4}$ represents the slope of the line connecting point $(x, y)$ and point $P$.
From $k_{P A} k_{P B}=1$, we know $8 n+4=8 m+1$.
Thus, line $A B$ passes through the point $\left(x^{\prime}, y^{\prime}\right)=\left(\frac{8}{3},-\frac{8}{3}\right)$.
Hence $(x, y)=\left(\frac{20}{3},-\frac{5}{3}\right)$, i.e., through point $Q\left(\frac{20}{3},-\frac{5}{3}\right)$.
(2) When lines $P A$ and $P B$ are perpendicular to the coordinate axes, it is easy to see that the equation of $A B$ is $y=-\frac{1}{4} x$, which also passes through point $Q\left(\frac{20}{3},-\frac{5}{3}\right)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M(x, y)$. Then $\left\{\begin{array}{l}\frac{x_{1}^{2}}{20}+\frac{y_{1}^{2}}{5}=1, \\ \frac{x_{2}^{2}}{20}+\frac{y_{2}^{2}}{5}=1 .\end{array}\right.$
Subtracting the two equations, we get
$$
x+4 y \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0 \text {. }
$$

Combining $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{y+\frac{5}{3}}{x-\frac{20}{3}}$, rearranging, we get
$$
\frac{\left(x-\frac{10}{3}\right)^{2}}{\frac{125}{9}}+\frac{\left(y+\frac{5}{6}\right)^{2}}{\frac{125}{36}}=1 .
$$

Noting that the equation of the line of the angle bisector of $\angle A P B$ is $y=x-3$, its second intersection point with $\Gamma$ is $N\left(\frac{4}{5},-\frac{11}{5}\right)$.
Thus, the trajectory of the midpoint $M$ of segment $A B$ is the ellipse
$$
\frac{\left(x-\frac{10}{3}\right)^{2}}{\frac{125}{9}}+\frac{\left(y+\frac{5}{6}\right)^{2}}{\frac{125}{36}}=1
$$

within $\Gamma$, i.e., the elliptical arc segment between points $P$ and $N$.