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 $y'(x) = \frac{1}{x^2}$, 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
$$b^2 + 2ab - a^2 = 0. \quad (1)$$
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 $b_1 \neq b_2$. Therefore, the equation of the line $M_1 M_2$ has the form
$$y = \frac{1}{b_1 b_2} (x - b_1) - \frac{1}{b_1} \iff y = \frac{1}{b_1 b_2} x - \frac{1}{b_2} - \frac{1}{b_1} \iff y = \frac{x - (b_1 + b_2)}{b_1 b_2}.$$
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 $y'(-a) = -\frac{1}{(-a)^2}$, 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 $N$).