China Mathematical Competition

Denote the coordinates of $M$ and $N$ as $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Denote the tangent lines of $C$ at $M$ and $N$ by $l_1$ and $l_2$ respectively, with their intersection point being $P(x_p,y_p)$. Suppose the slope ratio of line $l$ is $k$. Then we can write the equation of $l$ as $y=kx+1$.

Eliminating $y$ from $$\{\begin{array}{l}y=x+\frac{1}{x},\\ y=kx+1,\end{array}$$ we get $x+\frac{1}{x}=kx+1$, i.e. $(k-1)x^2+x-1=0$. By the assumption, we know that the equation has two distinctive real roots, $x_1$ and $x_2$, on $(0,+\infty )$. Then $k\ne 1$, and $$\Delta =1+4(k-1)>0,$$ $$x_1+x_2=\frac{1}{1-k}>0,$$ $$x_1{x}_2=\frac{1}{1-k}>0.$$ From the above we get $\frac{3}{4}<k<1$.

We find the derivative of $y=x+\frac{1}{x}$ as $y^{\prime }=1-\frac{1}{x^2}$. Then $y^{\prime }{|}_{x=x_1}=1-\frac{1}{x_1^2}$ and $y^{\prime }{|}_{x=x_2}=1-\frac{1}{x_2^2}$. Therefore, the equation of line $l_1$ is $$y-y_1=\left(1-\frac{1}{x_1^2}\right)(x-x_1)$$ or $$y-\left(x_1+\frac{1}{x_1}\right)=\left(1-\frac{1}{x_1^2}\right)(x-x_1).$$ After simplification, we get $$y=\left(1-\frac{1}{x_1^2}\right)x+\frac{2}{x_1}$$ In the same way, we get the equation of $l_2$, $$y=\left(1-\frac{1}{x_2^2}\right)x+\frac{2}{x_2}$$ By subtracting, 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 $x_1\ne x_2$, we have $$x_p=\frac{2x_1{x}_2}{x_1+x_2}$$ Substituting $x_1+x_2$ and $x_1{x}_2$ from above, we obtain $x_p=2$.

By adding, we obtain $$2y_p=\left(2-\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\right)x_p+2\left(\frac{1}{x_1}+\frac{1}{x_2}\right)$$ where $$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1{x}_2}=1,$$ $$\begin{array}{rl}\frac{1}{x_1^2}+\frac{1}{x_2^2}& =\frac{x_1^2+x_2^2}{x_1^2{x}_2^2}=\frac{(x_1+x_2{)}^2-2x_1{x}_2}{x_1^2{x}_2^2}\\ & ={\left(\frac{x_1+x_2}{x_1{x}_2}\right)}^2-\frac{2}{x_1{x}_2}=2k-1.\end{array}$$ Substituting it into the previous equation, we have $2y_p=(3-2k)x_p+2$. Since $x_p=2$, then $y_p=4-2k$. As $\frac{3}{4}<k<1$, we get $2<y_p<\frac{5}{2}$.

Therefore, the locus of point $P$ is the segment between $(2,2)$ and $(2,2.5)$ (not including the endpoints).