11. (20 points) Given a line $l$ passing through the point $(0,1)$ intersects the curve $C: y=x+\frac{1}{x}(x>0)$ at two distinct points $M$ and $N$. Find the locus of the intersection of the tangents to the curve $C$ at points $M$ and $N$.
11. Let points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, and the tangents to curve $C$ at points $M$ and $N$ be $l_{1}$ and $l_{2}$, respectively, intersecting at point $P\left(x_{P}, y_{P}\right)$.
If the slope of line $l$ is $k$, then $l: y=k x+1$.
By combining the equations of the curve and line $l$, and eliminating $y$, we get
$$
x+\frac{1}{x}=k x+1 \Rightarrow(k-1) x^{2}+x-1=0 \text {. }
$$

By the problem statement, this equation has two distinct real roots $x_{1}$ and $x_{2}$ in the interval $(0,+\infty)$.
Thus, $k \neq 1$, and
$$
\begin{array}{l}
\Delta=1+4(k-1)>0, \\
x_{1}+x_{2}=\frac{1}{1-k}>0, \\
x_{1} x_{2}=\frac{1}{1-k}>0 .
\end{array}
$$

From this, we solve to get $\frac{3}{4}<k<1$.
Differentiating $y=x+\frac{1}{x}$, we get $y^{\prime}=1-\frac{1}{x^{2}}$.
Thus, $\left.y^{\prime}\right|_{x=x_{1}}=1-\frac{1}{x_{1}^{2}},\left.y^{\prime}\right|_{x=x_{2}}=1-\frac{1}{x_{2}^{2}}$.
Therefore, the line $l_{1}: y-y_{1}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right)$, which is $y-\left(x_{1}+\frac{1}{x_{1}}\right)=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right)$
$\Rightarrow y=\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}}$.
Similarly,
the line $l_{2}: y=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}}$.
Subtracting (2) from (3) gives
$\left(\frac{1}{x_{2}^{2}}-\frac{1}{x_{1}^{2}}\right) x_{P}+\frac{2}{x_{1}}-\frac{2}{x_{2}}=0$.
Since $x_{1} \neq x_{2}$, we have
$x_{P}=\frac{2 x_{1} x_{2}}{x_{1}+x_{2}}=2$.
Adding (2) and (3) gives
$2 y_{P}=\left(2-\left(\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}\right)\right) x_{P}+2\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)$.
Noting that,
$$
\begin{array}{l}
\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=1, \\
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2} x_{2}^{2}} \\
=\left(\frac{x_{1}+x_{2}}{x_{1} x_{2}}\right)^{2}-\frac{2}{x_{1} x_{2}}=1-2(1-k) \\
=2 k-1 .
\end{array}
$$

Substituting into equation (4) gives
$$
2 y_{P}=(3-2 k) x_{P}+2 \text {, }
$$

which simplifies to $y_{P}=4-2 k$.
Given $\frac{3}{4}<k<1$, we have $2<y_{P}<\frac{5}{2}$, meaning the trajectory of point $P$ is the line segment between $(2,2)$ and $(2,2.5)$ (excluding the endpoints).