Finding the common tangent between two arbitrary parabolas Say we have two parabolas:
$$Ax^2 + Bx + C$$
$$Dx^2 + Ex + F$$
where the only constraint is that $A, D\neq0$. Is there an easy way to calculate the common tangent line between them? My attempts thus far have been unsuccessful. I started by saying that at some value of x for each quadratic, the common tangent will intercept the parabola:
$$mx + l = Ax^2 + Bx + C$$
$$mx + l = Dx^2 + Ex + F$$
Therefore, we have the equations:
$$Ax^2 + (B-m)x + (C-l)=0$$
$$Dx^2 + (E-m)x + (F-l)=0$$
Because each of these equations will have a single solution for a given value of $m$ and $l$, we know that the determinant of each quadratic will be equal to $0$:
$$b^2 - 4ac = 0$$
Therefore, we have the system of 2 equations with our 2 unknowns:
$$(B-m)^2 - 4A(C-l) = 0$$
$$(E-m)^2 - 4D(F-l) = 0$$
Any attempt I've made at solving this system of equations has failed. Does anyone have any tips?
 A: The following method applies to any pair of parabolas in any given position/orientation in the two-dimensional $xy$ plane.  It is assumed that the two parabolas are given in parametric form, as follows:
$ \mathbf{r_1}(t) = \mathbf{p_0} + \mathbf{p_1} t + \mathbf{p_2} t^2 $
and
$ \mathbf{r_2}(s) = \mathbf{q_0} + \mathbf{q_1} s + \mathbf{q_2} s^2 $
where $\mathbf{p_0}, \mathbf{p_1}, \mathbf{p_2}, \mathbf{q_0}, \mathbf{q_1}, \mathbf{q_2} $ are constant 2 dimensional vectors, and $t$ and $s$ are scalars, $t , s \in \mathbb{R} $.
The tangent vectors to the two parabolas as functions of the parameters $t$ and $s$, respectively, are given by:
$ \mathbf{T_1} (t) = \mathbf{p_1} + 2 t \mathbf{p_2}  $
$ \mathbf{T_2} (s) = \mathbf{q_1} + 2 s \mathbf{q_2} $
At tangency, we have the following conditions:
$ (\mathbf{r_1}(t) - \mathbf{r_2}(s) ) = \alpha \mathbf{T_1}(t) \hspace{20pt} (1) $
$ (\mathbf{r_1}(t) - \mathbf{r_2}(s) ) = \beta \mathbf{T_2}(s) \hspace{20pt} (2) $
$ \mathbf{T_1}(t) = \gamma \mathbf{T_2}(s) \hspace{20pt} (3)$
Extending all the vectors to three dimensional vectors with their $z$ coordinate equal to zero, we can use the cross product to eliminate $\alpha, \beta $ and $\gamma$.  The last equation becomes,
$ \mathbf{T_1}(t) \times \mathbf{T_2}(s) = 0 \hspace{20pt} (6)$
The first two equations become:
$ (\mathbf{r_1}(t) - \mathbf{r_2}(s) ) \times \mathbf{T_1}(t) = 0 \hspace{20pt} (4) $
$ (\mathbf{r_1}(t) - \mathbf{r_2}(s) ) \times \mathbf{T_2}(s) = 0 \hspace{20pt} (5)$
Expanding equations (4) and (5) and substituting equation (6) in both, results in two quadratic equations in $t$ and $s$.  These two equations can be numerically solved for all solutions using a computer application such as Mathematica.
Once we have the solutions, we have to check for extraneous solutions by checking if equation (6) is satisfied.  Then we're done.  The tangent lines are those connecting $\mathbf{r_1}(t_i)$ with $\mathbf{r_2}(s_i) $ where $(t_i, s_i)$ is the $i$-th solution of (4), (5), (6).
The following figure illustrates the result of this method.

A: HINT:
After differentiation note from this particular Geogebra construction that common tangent slopes form diagonals at respective dropped tangential point abscissas/ordinates.

