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 $N$.
Let $l: y=k x+1, M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$,
Combining $\left\{\begin{array}{l}y=x+\frac{1}{x}, \\ y=k x+1\end{array} \Rightarrow(k-1) x^{2}+x-1=0\right.$
$\Rightarrow\left\{\begin{array}{l}\Delta=1+4(k-1)>0, \\ x_{1}+x_{2}=-\frac{1}{k-1}>0, \\ x_{1} x_{2}=-\frac{1}{k-1}>0\end{array}\right.$
Since $y^{\prime}=1-\frac{1}{x^{2}}$, the tangent line $M B: y-x_{1}-\frac{1}{x_{1}}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right)$
$\Rightarrow y=\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}}$ (1), similarly $N B: y=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}}$
(1) - (2) gives $\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}}=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}}$
$\Rightarrow\left(\frac{1}{x_{1}^{2}}-\frac{1}{x_{2}^{2}}\right) x=\frac{2}{x_{1}}-\frac{2}{x_{2}} \Rightarrow x=\frac{2 x_{1} x_{2}}{x_{1}+x_{2}}=2 \Rightarrow x_{B}=2$.
(1) + (2) gives $2 y=\left(2-\frac{1}{x_{1}^{2}}-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{1}}+\frac{2}{x_{2}}$, substituting $x=2$ we get
$y_{B}=2-\frac{1}{x_{1}^{2}}-\frac{1}{x_{2}^{2}}+1=3-\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=3-\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2} x_{2}^{2}}$
$=3-\frac{\left(-\frac{1}{k-1}\right)^{2}+\frac{2}{k-1}}{\left(-\frac{1}{k-1}\right)^{2}}=3-[1+2(k-1)]=4-2 k$.
Since $\frac{3}{4}<k<1 \Rightarrow y_{B}=4-2 k \in\left(2, \frac{5}{2}\right)$.
Therefore, the trajectory equation of the intersection point $B$ of the tangents at points $M$ and $N$ on the curve $C$ is $x=2, y \in\left(2, \frac{5}{2}\right)$.
Its trajectory is a line segment connecting $(2,2)$ and $\left(2, \frac{5}{2}\right)$, excluding the two endpoints.