6. 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 Competition)
(The line equation of the chord $MN$ is $l$: $\left(y_{0}-2 x_{0}\right) x+x_{0} y=2$. Since $l$ passes through the point $(0,1)$, substituting gives $x_{0}=2$. At this point, the line $l:\left(y_{0}-4\right) x+2 y=2$. Substituting into the curve $C: y=x+\frac{1}{x}(x>0)$, we get $\left(y_{0}-2\right) x^{2}-$ $2 x+2=0$. It is easy to see that
$$
\left\{\begin{array}{l}
\Delta=4-8\left(y_{0}-2\right)>0, \\
x_{1}+x_{2}=x_{1} x_{2}=\frac{2}{y_{0}-2}>0,
\end{array}\right.
$$

which means $2<y_{0}<\frac{5}{2}$. Therefore, the locus of the intersection point of the two tangents at points $M$ and $N$ on the curve $C$ is a line segment $x=2\left(2<y<\frac{5}{2}\right)$, excluding the endpoints.)