Example 4 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 $N$. (2007, National High School Mathematics League Competition)
Let the intersection point of the tangents to the curve $C$ at points $M$ and $N$ be $P\left(x_{P}, y_{P}\right)$.
By Proposition 4, the equation of the line $M N$ is
$$
x_{P} y + x y_{P} = 2 x_{P} x + 2 \text{. }
$$

Substituting the coordinates of the point $(0,1)$, we get $x_{P} = 2$.
Substituting $y = \frac{4 - y_{P}}{2} x + 1$ into $y = x + \frac{1}{x}$, we get
$$
\left(1 - \frac{y_{P}}{2}\right) x + x - 1 = 0 \text{. }
$$

This equation has two distinct positive roots, solving which yields
$$
2 < y_{P} < \frac{5}{2} \text{, }
$$

which means the trajectory of point $P$ is the line segment between $(2,2)$ and $\left(2, \frac{5}{2}\right)$ (excluding the endpoints).