Example 9 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 Competition Problem)
Let the coordinates of points $M$ and $N$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. The tangents to curve $C$ at points $M$ and $N$ are $l_{1}$ and $l_{2}$, respectively, and the coordinates of their intersection point $P$ are $\left(x_{P}, y_{P}\right)$.
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 $(k-1) x^{2}+x-1=0$.
By the problem statement, this equation has two distinct real roots $x_{1}, x_{2}$ in $(0,+\infty)$, so $k \neq 1$, and
$\Delta=1+4(k-1)>0, x_{1}+x_{2}=\frac{1}{1-k}>0, x_{1} x_{2}=\frac{1}{1-k}>0$.
Solving this, we get $\frac{3}{4}<k<1$.
Differentiating $y=x+\frac{1}{x}$, we get $y^{\prime}=1-\frac{1}{x^{2}}$, so $\left.y^{\prime}\right|_{x=x_{1}}=1-\frac{1}{x_{1}^{2}},\left.y^{\prime}\right|_{x=x_{2}}=1-\frac{1}{x_{2}^{2}}$. Therefore, the equation of line $l_{1}$ is $y-y_{1}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right)$, which simplifies to $y-\left(x_{1}+\frac{1}{x_{1}}\right)=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right)$, or $y=\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}}$.
Similarly, the equation of line $l_{2}$ is $y=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}}$.
Subtracting the two equations, we get $\left(\frac{1}{x_{2}^{2}}-\frac{1}{x_{1}^{2}}\right) x_{P}+\frac{2}{x_{1}}-\frac{2}{x_{2}}=0$. Since $x_{1} \neq x_{2}$, we have $x_{P}=\frac{2 x_{1} x_{2}}{x_{1}+x_{2}}$. From (*) we know $x_{P}=2$.
Adding the two equations, we get $2 y_{P}=\left[2-\left(\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}\right)\right] x_{P}+2\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)$.
Here, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=1, \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2}-x_{2}^{2}}=1-2(1-k)=$ $2 k-1$, so $2 y_{P}=(3-2 k) x_{P}+2$. Since $x_{P}=2$, we get $y_{P}=4-2 k$.
Given $\frac{3}{4}<x<1$, we have $2<y_{P}<\frac{5}{2}$.
Thus, the trajectory of point $P$ is the line segment between $(2,2)$ and $\left(2, \frac{5}{2}\right)$ (excluding the endpoints).