FINAL ROUND

Without loss of generality we may assume that the point $M(a;1/a)$ lies on the upper-right half-hyperbola $H_1$, $y=1/x$. $M_1(b_1;-1/b_1)$ and $M_2(b_2;-1/b_2)$ be the points of tangency with $H_2$, mentioned in the problem condition (see the Fig.).



Since the derivative of the function $y(x)=-1/x$ is equal to $y^{\prime }(x)=\frac{1}{x^2}$, the equation of the line tangent to $H_2$ at $(b;-1/b)$ has the form $$y=\frac{1}{b^2}(x-b)-\frac{1}{b}.$$ Since $M(a;1/a)$ belongs to this tangent, we have $\frac{1}{a}=\frac{1}{b^2}(a-b)-\frac{1}{b}$. Therefore, the abscissae $b_1$ and $b_2$ of $M_1$ and $M_2$ satisfy the equation $$b^2+2ab-a^2=0.(1)$$ The equation of the straight line passing through $M_1$ can be written as $y=k(x-b_1)-\frac{1}{b_1}$, and if this line passes through $M_2$, then

$\frac{1}{b_2}=k(b_2-b_1)-\frac{1}{b_1}$, so $k=\frac{1}{b_1{b}_2}$ since $b_1\ne b_2$. Therefore, the equation of the line $M_1{M}_2$ has the form $$y=\frac{1}{b_1{b}_2}(x-b_1)-\frac{1}{b_1}⟺y=\frac{1}{b_1{b}_2}x-\frac{1}{b_2}-\frac{1}{b_1}⟺y=\frac{x-(b_1+b_2)}{b_1{b}_2}.$$ Since $b_1$ and $b_2$ are the roots of (1), by Vieta's theorem, we have $b_1{b}_2=-a^2$, $b_1+b_2=-2a$. Then the equation of the line $M_1{M}_2$ has the form $$y=\frac{x+2a}{-a^2}.$$ Note that $y=-1/a$, for $x=-a$, i.e., the point $N(-a;-1/a)$ lying on the hyperbola $H_1$ belongs to the line $M_1{M}_2$. It remains to note that the slope of the line tangent to $H_1$ at $N(-a;-1/a)$ is equal to $y^{\prime }(-a)=-\frac{1}{(-a{)}^2}$, i.e., coincides with the slope of the line $M_1{M}_2$. Therefore, the line $M_1{M}_2$ touches the hyperbola $H_1$ (at the point $N$).