Without loss of generality we may assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^{\circ}$, and rename the points).

Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d, -1/d)$. Note that all numbers $a$, $b$, $c$, $d$ are pairwise distinct and $c < 0$, $d > a > b > 0$. Since the derivative of the function $y(x) = 1/x$ is equal to $y'(x) = -1/x^2$, the equations of the tangents to $H_1$ at $A$ and $B$ have the forms
$$y = -\frac{1}{a^2}(x-a) + \frac{1}{a} \quad \text{and} \quad y = -\frac{1}{b^2}(x-b) + \frac{1}{b}.$$
Let $M(x_M; y_M)$ be the point of intersection of these tangents, then
$$-\frac{1}{a^2}(x-a) + \frac{1}{a} = -\frac{1}{b^2}(x-b) + \frac{1}{b}, \quad \text{so } x_M = \frac{2ab}{a+b}. \quad \text{Therefore}$$
$$y_M = -\frac{1}{a^2}(x_M - a) + \frac{1}{a} = \frac{2}{a+b}.$$

Similarly, since the derivative of the function $y(x) = -1/x$ is equal to $y'(x) = 1/x^2$, the equations of the tangents to $H_2$ at $C$ and $D$ have the forms
$$y = \frac{1}{c^2}(x-c) - \frac{1}{c} \quad \text{and} \quad y = \frac{1}{d^2}(x-d) - \frac{1}{d}.$$
Let $N(x_N; y_N)$ be the point of intersection of these tangents, then
$$\frac{1}{c^2}(x-c) - \frac{1}{c} = \frac{1}{d^2}(x-d) - \frac{1}{d}, \quad \text{so } x_N = \frac{2cd}{c+d}. \quad \text{Therefore}$$
$$y_N = \frac{1}{c^2}(x_N - c) - \frac{1}{c} = -\frac{2}{c+d}.$$
Since $A$, $B$, $C$, $D$ belong to the same line, we have $a+b = c+d$ and $ab = -cd$ (see the solution of Problem C.1). Hence $x_M = -x_N$ and $y_M = -y_N$, i.e., $M$ and $N$ are symmetric with respect to the origin.