2. Given a line $l$ passing through the point $(0,2)$ intersects the curve $C$: $y=x+\frac{1}{x}(x>0)$ at two distinct points $M$ and $N$. Then the locus of the intersection of the tangents to the curve $C$ at $M$ and $N$ is $\qquad$.


2. $x=1,10 \Rightarrow k>0 \text {. } \\
\text { and } x_{1}+x_{2}=\frac{2}{1-k}>0, x_{1} x_{2}=\frac{1}{1-k}>0 \\
\Rightarrow 0<k<1 \text {. } \\
\end{array}
$$

It is easy to know that the equation of $l_{1}$ is
$$
\begin{aligned}
& y-y_{1}=\left(1-\frac{1}{x_{1}^{2}}\right)\left(x-x_{1}\right) \\
\Rightarrow & y=\left(1-\frac{1}{x_{1}^{2}}\right) x+\frac{2}{x_{1}} .
\end{aligned}
$$

Similarly, $l_{2}: y=\left(1-\frac{1}{x_{2}^{2}}\right) x+\frac{2}{x_{2}}$.
Thus, $x_{1} \neq x_{2}, x_{P}=\frac{2 x_{1} x_{2}}{x_{1}+x_{2}}=1$.
$$
\begin{array}{l}
\text { Also, } y_{P}=\frac{1}{2}\left\{\left[2-\left(\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}\right)\right] x_{P}+2\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)\right\} \\
=-k x_{P}+2=2-k \\
\Rightarrow 1<y_{P}<2 .
\end{array}
$$

Therefore, the trajectory of the intersection point is $x=1,1<y<2$.