11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$

with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ is the geometric mean of $|F A|$ and $|F B|$. Find the smallest positive integer value of $a$, and the equation of $l$ at this value of $a$.
11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} x_{2}$.
Thus, $\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$, which means
$\left(y_{1}-y_{2}\right)^{2}=\frac{2 c^{2}-a^{2}}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$
$=\frac{a^{2}-40}{a^{2}}\left(x_{1}-x_{2}\right)^{2}$.
Therefore, $a^{2} \geqslant 40$.
When $a \in \mathbf{N}_{+}$, $a \geqslant 7$.
When $a=7$, $\left(y_{1}-y_{2}\right)^{2}=\frac{9}{49}\left(x_{1}-x_{2}\right)^{2}$.
Thus, the slope of line $l$ is $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}= \pm \frac{3}{7}$.
Since point $P(1,1)$ is clearly inside the ellipse, both lines $y-1= \pm \frac{3}{7}(x-1)$ satisfy the condition.

In summary, the smallest positive integer value of $a$ is 7, and at this time, the equations of line $l$ are
$3 x-7 y+4=0$ or $3 x+7 y-10=0$.