11. (20 points) Let $A$ be a moving point on the hyperbola $y=\frac{2011}{x}$. From $A$, draw two tangents $A P$ and $A Q$ to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, with points of tangency $P$ and $Q$. If the left focus of the ellipse is $F$, find the minimum value of $\frac{|A F|^{2}}{|P F||Q F|}$.
11. Let $A\left(x^{\prime}, y^{\prime}\right), P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then $x^{\prime} y^{\prime}=2011$.
It is easy to know that the chord of tangents from point $A$ to the ellipse, i.e., the line $P Q$, is given by
$$
\frac{x^{\prime} x}{25}+\frac{y^{\prime} y}{9}=1 .
$$

From $\left\{\begin{array}{l}\frac{x^{\prime} x}{25}+\frac{y^{\prime} y}{9}=1, \\ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1,\end{array}\right.$ eliminating $y$ gives
$$
\left(25 y^{\prime 2}+9 x^{\prime 2}\right) x^{2}-450 x^{\prime} x+625\left(9-y^{\prime 2}\right)=0 \text {. }
$$

Thus, $x_{1}+x_{2}=\frac{450 x^{\prime}}{25 y^{\prime 2}+9 x^{\prime 2}}$,
$$
x_{1} x_{2}=\frac{625\left(9-y^{\prime 2}\right)}{25 y^{\prime 2}+9 x^{\prime 2}} \text {. }
$$

By the second definition of the ellipse, we have
$$
\begin{array}{l}
|P F|=a+e x_{1}=5+\frac{4}{5} x_{1}, \\
|Q F|=a+e x_{2}=5+\frac{4}{5} x_{2} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then }|P F||Q F|=\left(5+\frac{4}{5} x_{1}\right)\left(5+\frac{4}{5} x_{2}\right) \\
=25+4\left(x_{1}+x_{2}\right)+\frac{16}{25} x_{1} x_{2} \\
=25+\frac{1800 x^{\prime}}{25 y^{\prime 2}+9 x^{\prime 2}}+\frac{25 \times 16\left(9-y^{\prime 2}\right)}{25 y^{\prime 2}+9 x^{\prime 2}} \\
=\frac{225\left(y^{\prime 2}+x^{\prime 2}+8 x^{\prime}+16\right)}{25 y^{\prime 2}+9 x^{\prime 2}} \\
=\frac{225|A F|^{2}}{25 y^{\prime 2}+9 x^{\prime 2}} . \\
\text { Hence } \frac{|A F|^{2}}{|P F||Q F|}=\frac{x^{\prime 2}}{25}+\frac{y^{\prime 2}}{9} \\
=\left(\frac{x^{\prime}}{5}\right)^{2}+\left(\frac{y^{\prime}}{3}\right)^{2} \geqslant 2\left|\frac{x^{\prime}}{5}\right|\left|\frac{y^{\prime}}{3}\right| \\
=\frac{2 \times 2011}{15}=\frac{4022}{15} .
\end{array}
$$

Equality holds if and only if $3 x^{\prime}=5 y^{\prime}=\sqrt{30165}$.

Therefore, the minimum value of $\frac{|A F|^{2}}{|P F||Q F|}$ is $\frac{4022}{15}$.