A "jeopardy" with common tangents of two hyperbolas

Question

Please do not overlook the pre-degree mathematics, there can be some un-noticed avenues/issues.

For the hyperbola $H=x^2/a^2-y^2/b^2=1$, the equation of tangents is $T:y=mx\pm\sqrt{a^2m^2-b^2}$.

Among conics the hyperbola has four distinctions:

(1) It has two branches, (2) Any tangent at it touches only one branch of the hyperbola. (3) similarly, any chord of the hyperbola cut only one branch. (4) One cannot draw tangent of any slope at a hyperbola unless the slope satisfies: $m^2 > b^2/a^2$.

Like two non-intersecting circles have four common tangents, two (non-conjugate) hyperbolas: $$H_1: x^2/3-y^2/2=1, ~~H_2: -x^2/2+y^2/3=1 ~~~(1)$$ can also have four common tangents.

Here $H_1$ is horizontal with left and right branches, $H_2$ is vertical with up and down branches. Any common tangent to $H_1,H_2$ will touch only one branch (left or right) of $H_1$ and only one branch (up or down) of $H_2$.

Question is to find (and plot) four common tangents of $H_1$ and $H_2$. To re-emphasize any of these tangents are never expected to touch all four branches of the two hyperbolas. These common tangents are often not discussed in the literature, if yes, one may please give a reference.

Answer 1

The duals are $-2X^2+3Y^2=1, 3X^2-2Y^2=1,$ which intersect in $(X,Y)=(1,1),(-1,1),(1,-1),(-1,-1),$ making $xX+yY+1=0$ or $x+y+1=0,-x+y+1=0,x-y+1=0,-x-y+1=0$ the four common tangents.

Answer 2

Let two-non-conjugate hyperbolas as $$H_1: x^2/a^2-y^2/b^2=1, ~~~H_2: -x^2/b^2+y^2/a^2=1~~~(1)$$ The Eq. of their tangents of slope $m$ are $$T_1: y=mx \pm \sqrt{m^2a^2-b^2}, ~~~T_2: y=mx \pm \sqrt{-b^2 m^2+a^2}$$ Equating their y-intercept of $T_1$ and $T_2$, we get $$a^2m^2-b^2=-b^2 m^2 +a^2 \implies (a^2+b^2)m^2=(a^2+b^2) \implies m=\pm 1$$ For this example $a^2=3, b^2=2$, putting $m=\pm 1$ in $T_1$ or $T_2$ we get the four common tangents as $$y=\pm x \pm \sqrt{a^2-b^2} \implies y=\pm x\pm 1.$$ These tangents and along with the two hypperbola are plotted by Jan-Manus in his answer here.

Note: that if $a^2<b^2$ in (1), then $H_1$ and $H_2$ intersect at corners and there is no common tangent to them. Amusingly, the common tangents become non-real having non-real intercepts.