11. As shown in the figure, in the Cartesian coordinate system, it is known that $F_{1}, F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, respectively. Let point $D(1,0)$ be the midpoint of segment $O F_{2}$. A line $M N$ (not coinciding with the $x$-axis) passes through point $F_{1}$ and intersects the ellipse $C$ at points $M, N$. Extend $M D, N D$ to intersect the ellipse $C$ at points $P, Q$, respectively. Let the slopes of lines $M N, P Q$ exist and be $k_{1}, k_{2}$, respectively. Express $\frac{k_{2}}{k_{1}}$ as a function of $a$, i.e., $f(a)=\frac{k_{2}}{k_{1}}$, and find the range of $f(a)$.
Solution
As shown in the figure, let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), P\left(x_{3}, y_{3}\right)$,
$Q\left(x_{4}, y_{4}\right), M P: x=\frac{x_{1}-1}{y_{1}} y+1$.
Solving the system $\left\{\begin{array}{l}x=\frac{x_{1}-1}{y_{1}} y+1, \\ b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}\end{array}\right.$
$\Rightarrow\left[\frac{b^{2}\left(x_{1}-1\right)^{2}}{y_{1}^{2}}+a^{2}\right] y^{2}+\frac{2 b^{2}\left(x_{1}-1\right)}{y_{1}} y$
$+b^{2}-a^{2} b^{2}=0 \Rightarrow y_{3}=\frac{\left(b^{2}-a^{2} b^{2}\right) y_{1}}{b^{2}\left(x_{1}-1\right)^{2}+a^{2} y_{1}^{2}}=\frac{\left(b^{2}-a^{2} b^{2}\right) y_{1}}{-2 b^{2} x_{1}+b^{2}+b^{2} x_{1}^{2}+a^{2} y_{1}^{2}}$
$=\frac{\left(b^{2}-a^{2} b^{2}\right) y_{1}}{-2 b^{2} x_{1}+b^{2}+a^{2} b^{2}}=\frac{\left(1-a^{2}\right) y_{1}}{-2 x_{1}+1+a^{2}} \Rightarrow x_{3}=\frac{2 a^{2}-\left(1+a^{2}\right) x_{1}}{-2 x_{1}+1+a^{2}}$,
Similarly, $y_{4}=\frac{\left(1-a^{2}\right) y_{2}}{-2 x_{2}+1+a^{2}}, x_{4}=\frac{2 a^{2}-\left(1+a^{2}\right) x_{2}}{-2 x_{2}+1+a^{2}}$.

Thus, $k_{2}=\frac{y_{3}-y_{4}}{x_{3}-x_{4}}=\frac{\frac{\left(1-a^{2}\right) y_{1}}{-2 x_{1}+1+a^{2}}-\frac{\left(1-a^{2}\right) y_{2}}{-2 x_{2}+1+a^{2}}}{\frac{2 a^{2}-\left(1+a^{2}\right) x_{1}}{-2 x_{1}+1+a^{2}}-\frac{2 a^{2}-\left(1+a^{2}\right) x_{2}}{-2 x_{2}+1+a^{2}}}$
$=\frac{k_{1}\left(1-a^{2}\right)\left[\left(x_{1}+2\right)\left(-2 x_{2}+1+a^{2}\right)-\left(x_{2}+2\right)\left(-2 x_{1}+1+a^{2}\right)\right]}{\left[2 a^{2}-\left(1+a^{2}\right) x_{1}\right]\left(-2 x_{2}+1+a^{2}\right)-\left[2 a^{2}-\left(1+a^{2}\right) x_{2}\right]\left(-2 x_{1}+1+a^{2}\right)}$
$=\frac{k_{1}\left(1-a^{2}\right)\left(a^{2}+5\right)\left(x_{1}-x_{2}\right)}{-\left(1-a^{2}\right)^{2}\left(x_{1}-x_{2}\right)}=\frac{k_{1}\left(a^{2}+5\right)}{a^{2}-1} \Rightarrow \frac{k_{2}}{k_{1}}=\frac{a^{2}+5}{a^{2}-1}=1+\frac{6}{a^{2}-1}$.
Therefore, $f(a)=1+\frac{6}{a^{2}-1}$, since $a>2$, we have $f(a) \in(1,3)$.