Rex Given that the 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 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 the curve $C$ at points $M$ and $N$ are $l_{1}$ and $l_{2}$, respectively, and their intersection point $P$ has coordinates $\left(x_{P}, y_{P}\right)$. If the slope of the line $l$ is $k$, then the equation of $l$ is $y = kx + 1$. From the system of equations $\left\{\begin{array}{l}y = x + \frac{1}{x}, \\ y = kx + 1\end{array}\right.$, eliminating $y$ gives $x + \frac{1}{x} = kx + 1$, which simplifies to $(k-1)x^2 + x - 1 = 0$. By the problem's conditions, this equation has two distinct real roots $x_{1}$ and $x_{2}$ in $(0, +\infty)$, so $k \neq 1$, and $\Delta = 1 + 4(k-1) > 0$,

(1) $x_{1} + x_{2} = \frac{1}{1-k} > 0$,

(2) $x_{1} x_{2} = \frac{1}{1-k} > 0$, thus $\frac{3}{4} < k < 1$.

Differentiating $y = x + \frac{1}{x}$, we get $y' = 1 - \frac{1}{x^2}$. Therefore, $\left.y'\right|_{x=x_{1}} = 1 - \frac{1}{x_{1}^2}$ and $\left.y'\right|_{x=x_{2}} = 1 - \frac{1}{x_{2}^2}$. Thus, the equation of line $l_{1}$ is $y - y_{1} = \left(1 - \frac{1}{x_{1}^2}\right)(x - x_{1})$, which simplifies to $y - \left(x_{1} + \frac{1}{x_{1}}\right) = \left(1 - \frac{1}{x_{1}^2}\right)(x - x_{1})$, and further simplifies to the equation of line $l_{1}$: $y = \left(1 - \frac{1}{x_{1}^2}\right)x + \frac{2}{x_{1}}$. (4) Similarly, the equation of line $l_{2}$ is $y = \left(1 - \frac{1}{x_{2}^2}\right)x + \frac{2}{x_{2}}$. (5) Subtracting (5) from (4) 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 $x_{1} + x_{2}$, we have $x_{P} = \frac{2 x_{1} x_{2}}{x_{1} + x_{2}}$, (6) Substituting (2) and (3) into (6) gives $x_{P} = 2$. (4) + (5) gives
$$
2 y_{P} = \left[2 - \left(\frac{1}{x_{1}^2} + \frac{1}{x_{2}^2}\right)\right] x_{P} + 2\left(\frac{1}{x_{1}} + \frac{1}{x_{2}}\right),
$$

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

Substituting into (7) gives: $2 y_{P} = (3 - 2k) x_{P} + 2$. Since $x_{P} = 2$, we get $y_{P} = 4 - 2k$. From $\frac{3}{4} < k < 1$, we have $2 < y_{P} < \frac{5}{2}$, so the trajectory of point $P$ is the line segment between $(2, 2)$ and $\left(2, \frac{5}{2}\right)$ (excluding the endpoints).