11. (20 points) Given the hyperbola $C: \frac{x^{2}}{4}-y^{2}=1$, a line $l$ passing through the point $M(1,-1)$ intersects the right branch of the hyperbola $C$ at points $A$ and $B$, and intersects the $x$-axis at point $N$. Let
$$
\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N}, \overrightarrow{M B}=\lambda_{2} \overrightarrow{B N} \text {. }
$$

Find the range of $\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}$.
11. Let the line $l: y+1=k(x-1)$. Then
$$
N\left(\frac{1}{k}+1,0\right) \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Substituting the equation of line $l$ into the equation of the hyperbola $C$ yields
$$
\begin{array}{l}
\left(1-4 k^{2}\right) x^{2}+8 k(k+1) x-4(k+1)^{2}-4 \\
=0 .
\end{array}
$$

Since equation (1) has two distinct positive roots $x_{1} 、 x_{2}$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\Delta=64 k^{2}(k+1)^{2}+16\left(1-4 k^{2}\right)\left[(k+1)^{2}+1\right]>0, \\
x_{1}+x_{2}=\frac{-8 k(k+1)}{1-4 k^{2}}>0, \\
x_{1} x_{2}=\frac{-4(k+1)^{2}-4}{1-4 k^{2}}>0
\end{array}\right. \\
\Rightarrow \frac{1}{2}<k<\frac{1+\sqrt{7}}{3} \text {. } \\
x_{1}=\frac{1+\lambda_{1}\left(\frac{1}{k}+1\right)}{1+\lambda_{1}}, y_{1}=\frac{-1}{1+\lambda_{1}} \text {. } \\
\end{array}
$$

Since $\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N}$, we get

Substituting it into the equation of the hyperbola yields
$$
\left[\left(\frac{1}{k}+1\right)^{2}-4\right] \lambda_{1}^{2}+\left[2\left(\frac{1}{k}+1\right)-8\right] \lambda_{1}-7=0 \text {. }
$$

Similarly,
$$
\left[\left(\frac{1}{k}+1\right)^{2}-4\right] \lambda_{2}^{2}+\left[2\left(\frac{1}{k}+1\right)-8\right] \lambda_{2}-7=0 \text {. }
$$

Thus, $\lambda_{1} 、 \lambda_{2}$ are the two roots of the equation
$$
\left(t^{2}-4\right) \lambda^{2}+(2 t-8) \lambda-7=0
$$

where $t=\frac{1}{k}+1 \in\left(\frac{\sqrt{7}+1}{2}, 3\right)$.
$$
\begin{array}{l}
\text { Then } \lambda_{1}+\lambda_{2}=\frac{2 t-8}{4-t^{2}}, \lambda_{1} \lambda_{2}=\frac{7}{4-t^{2}} . \\
\text { Hence } \frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(\lambda_{1}+\lambda_{2}\right)^{2}}{\lambda_{1} \lambda_{2}}-2 \\
=\frac{(2 t-8)^{2}}{7\left(4-t^{2}\right)}-2 .
\end{array}
$$

Let $z=t-4$. Then $z \in\left(\frac{\sqrt{7}-7}{2},-1\right)$,
$$
\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=-\frac{4}{7\left(\frac{12}{z^{2}}+\frac{8}{z}+1\right)}-2 \text {. }
$$

Therefore, $\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}} \in\left(-\frac{74}{35}, 2\right)$.