Here is an outline of the solution.
First we consider a related problem of finding common points of two conics: if they are described by equations $F(x,y)=0,G(x,y)=0$, then the $x$ coordinates of their common points are the roots of the resultant of $F,G$ with respect to $y$. The resultant is a polynomial in $x$ of degree at most $4$. So while can't immediately say $x$ is a solution of a cubic, we can use the standard methods to find $x$ using square, cube and fourth roots.
To get from common points to common tangents, we apply a kind of duality: every line not passing through the origin is uniquely described by an equation of the form $ax+by=1$, so we can parametrize lines using a nonzero point $(a,b)$. It turns out that tangents to a conic describe a conic themselves: $ax+by=1$ is tangent to $F(x,y)$ iff the resultant has a double root iff its discriminant, which works out to a quadratic in $a,b$, vanishes.
Therefore finding a common tangent to two conics is equivalent to finding a common point of two conics, which reduces to solving a quartic, which can be reduced to solving cubics.
Note: there are some minor technical issues, for example we had to ignore tangents passing through the origin, and also the last resultant mentioned might reduce to a linear equation. The "correct" way of dealing with such issues is by passing to the projective plane, but if we instead decide to ignore the "pathological" cases (intersections at zero and infinity), the above works just fine.
Edit: in case of parabolas, one of the common tangents is the line at infinity, so I believe that in this case we would, indeed, get cubic equations which the coefficients satisfy.
