The common tangent of two tilted parabolas

Question

I very recently asked a question here about finding the tangent to a tilted parabola but this was only part of a larger question I had. I figured that I would be able to solve it myself with answers to this question but I am still completely stumped.

The equation of a parabola with directrix $l_1x+m_1y+n_1=0$ and focus $P(x_P,y_P)$ is $$\frac{(l_1x+m_1y+n_1)^2}{{l_1}^2+{m_1}^2}=(x-x_P)^2+(y-y_P)^2$$

This parabola is then $S_1\equiv a_1x^2+2h_1xy+b_1y^2+2g_1x+2f_1y+c_1=0$ where \begin{align} a_1 &= -{m_1}^2 \\ h_1 &= l_1m_1 \\ b_1 &= -{l_1}^2 \\ g_1 &= n_1l_1+\left({l_1}^2+{m_1}^2\right)x_P \\ f_1 &= m_1n_1+\left({l_1}^2+{m_1}^2\right)y_P \\ c_1 &= {n_1}^2-\left({l_1}^2+{m_1}^2\right)\left({x_P}^2+{y_P}^2\right) \end{align}

Likewise the second parabola has the directrix $l_2x+m_2y+n_2=0$ and focus $Q(x_Q,y_Q)$. The equation is the same as the first except all the 1's become 2's and $P$'s become $Q$'s in the subscripts.

I know that it is possible for there to be up to three tangent lines common to both of these parabolas but I am unable to find any of them.

I believe that I need to go one of two routes. The first is using calculus to find the derivative $\mathrm{d}y/\mathrm{d}x$ of each parabola and, knowing that the slopes of the parabolas must be equal at some point $A(x_A,y_A)$ (actually three points) set the values equal and somehow come up with the tangent. The second is not using calculus. I know that the tangent at $A$ to $S_1=0$ has the equation $$T_A\equiv (a_1x_A+h_1y_A+g_1)x+(h_1x_A+b_1y_A+f_1)y+g_1x_A+f_1y_A+c_1=0$$ So we must have $T_A=T_B$ where $T_B$ is the tangent line through $B$ on $S_2=0$. However I don't think this way will yield all the tangents.

While writing this question I found this. This is my exact question asked over two years ago except significantly lacking detail and without answers. I am unable to add a bounty to that question and I believe that this question is more likely to receive useful answers.

Answer 1

There is @MvG's approach in this answer. Let's take the two parabolas $(x+1)^2+(y-1)^2-\frac{(x+y+1)^2}{2}$ and $(x-1)^2+(y+1)^2-\frac{(x-y+1)^2}{2}$.

Then the dual conics (whose points correspond to tangent lines of the original conics) are defined by the adjugate $3\times 3$ matrices of those defining the original parabolas. In this case $x^2+4xy+y^2-2x+2y=0$ and $3x^2-y^2-2x-2y=0$ (up to scalar multiple) which intersect in $(0:0:1),(0:-2:1),(\frac{-1-\sqrt{7}}{2}:\frac{3+\sqrt{7}}{2}:1),(\frac{-1+\sqrt{7}}{2}:\frac{3-\sqrt{7}}{2}:1)$.

That is the lines $z=0,-2y+z=0,\frac{-1-\sqrt{7}}{2}x+\frac{3+\sqrt{7}}{2}y+z=0$ and $\frac{-1+\sqrt{7}}{2}x+\frac{3-\sqrt{7}}{2}y+z=0$, which corresponds to the line at infinity and the lines $-2y+1=0,\frac{-1-\sqrt{7}}{2}x+\frac{3+\sqrt{7}}{2}y+1=0$ and $\frac{-1+\sqrt{7}}{2}x+\frac{3-\sqrt{7}}{2}y+1=0$.

Added: The general case has always dual conics with vanishing constant terms, since those corresponds to the minors that have vanishing determinants because the original conics are parabolas. This makes the line at infinity always a common tangent.

Your equations are represented by the matrices $\begin{pmatrix} a_i&h_i&g_i\\h_i&b_i&f_i\\g_i&f_i&c_i\end{pmatrix}$, $i=1,2$, and the dual conics are represented by $$\begin{pmatrix} +\begin{vmatrix}b_i&f_i\\f_i&c_i\end{vmatrix}&-\begin{vmatrix}h_i&f_i\\g_i&c_i\end{vmatrix}&+\begin{vmatrix}h_i&b_i\\g_i&f_i\end{vmatrix}\\-\begin{vmatrix}h_i&g_i\\f_i&c_i\end{vmatrix}&+\begin{vmatrix}a_i&g_i\\g_i&c_i\end{vmatrix}&-\begin{vmatrix}a_i&h_i\\g_i&f_i\end{vmatrix}\\+\begin{vmatrix}h_i&g_i\\b_i&f_i\end{vmatrix}&-\begin{vmatrix}a_i&g_i\\h_i&f_i\end{vmatrix}&+\begin{vmatrix}a_i&h_i\\h_i&b_i\end{vmatrix}\end{pmatrix}.$$ And those two new equations has tangents to the two original parabolas as points. Which means that the common points, that is their intersection of four points, correspond to common tangents of the original equation. Now the duality goes as $lx+my+nz=0$ is both seen as a line and the point $(l:m:n)$ on the dual curve. To get an equation in the plane (as opposed to the projective plane) set $z=1$. Also $(p:q:1)$ is $(p,q)$.

As for how to find intersection points between two conics see this answer. I used a Gröbner basis, see WA.