14. 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 $N$.
14. Let points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$, and the tangents to curve $C$ at points $M$ and $N$ be $l_{1}$ and $l_{2}$, respectively, with their intersection point being $P\left(x_{P}, y_{P}\right)$. If the slope of line $l$ is $k$, then the equation of $l$ is $y=k x+1$.
From the system of equations $\left\{\begin{array}{l}y=x+\frac{1}{x}, \\ y=k x+1\end{array}\right.$, eliminating $y$ gives
$$
x+\frac{1}{x}=k x+1 \Rightarrow(k-1) x^{2}+x-1=0 \text {. }
$$

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

Solving this, we get $\frac{3}{4}<k<1$.
Differentiating $y=x+\frac{1}{x}$, we get $y^{\prime}=1-\frac{1}{x^{2}}$.
Thus, the equation of line $l_{1}$ is
$$
y-y_{1}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right) \text {, }
$$

which simplifies to $y=\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}}$.
Similarly, the equation of line $l_{2}$ is
$$
y=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}} \text {. }
$$
Subtracting (2) from (1) 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
$$
x_{p}=\frac{2 x_{1} x_{2}}{x_{1}+x_{2}}=2 \text {. }
$$
Adding (1) and (2) gives
$$
2 y_{P}=\left[2-\left(\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}\right)\right] x_{P}+2\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right) \text {, }
$$

where $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=1$,
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2} x_{2}^{2}} \\
=\left(\frac{x_{1}+x_{2}}{x_{1} x_{2}}\right)^{2}-\frac{2}{x_{1} x_{2}}=1-2(1-k)=2 k-1 .
\end{array}
$$

Substituting into equation (3) gives $2 y_{P}=(3-2 k) x_{P}+2$.
Since $x_{P}=2$, we get $y_{P}=4-2 k$.
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 $\left(2, \frac{5}{2}\right)$ (excluding the endpoints).