Equation of the line $y=mx+q$. This line cuts $y=\frac{1}{x}$, we have to solve $mx+q=\frac{1}{x}$ or $mx^{2}+qx-1=0$. Solutions $x =\frac{-q \pm \sqrt{q^{2}+4m}}{2}=\frac{-q \pm w}{2}$. Points $A(\frac{-q-w}{2},\frac{2}{-q-w}), B(\frac{-q+w}{2},\frac{2}{-q+w})$. This line cuts $y=-\frac{1}{x}$, we have to solve $mx+q=-\frac{1}{x}$ or $mx^{2}+qx+1=0$. Solutions $x =\frac{-q \pm \sqrt{q^{2}-4m}}{2}=\frac{-q \pm v}{2}$. Points $C(\frac{-q-v}{2},\frac{2}{q+v}), D(\frac{-q+v}{2},\frac{2}{q-v})$. Tangent line $t_{A}\ :\ y + \frac{2}{q+w}=-\frac{4}{(q+w)^{2}}(x+\frac{q+w}{2})$. Tangent line $t_{B}\ :\ y + \frac{2}{q-w}=-\frac{4}{(q-w)^{2}}(x+\frac{q-w}{2})$. Both tangent lines cut in $M(\frac{2m}{q},-\frac{2}{q})$. Tangent line $t_{C}\ :\ y - \frac{2}{q+v}=\frac{4}{(q+v)^{2}}(x+\frac{q+v}{2})$. Tangent line $t_{D}\ :\ y - \frac{2}{q-v}=\frac{4}{(q-v)^{2}}(x+\frac{q-v}{2})$. Both tangent lines cut in $N(-\frac{2m}{q},\frac{2}{q})$.