Axis of symmetry of a (general) parabola given its parametric equation. Given a parabola's parametric equation I want to find its axis of symmetry. 
The parametric equation is 
$$\begin{cases} x=a_1t^2+b_1t+c_1 \\ y=a_2t^2+b_2t+c_2 \end{cases}, t\in\mathbb{R}.$$
This is parabola as was shown in other question.
I understand that one can start with finding the general equation from this parametric one, proceed with finding the rotation angle, getting rid of $xy$ term and finally finding an axis of symmetry. 
My question is whether there is a simpler way to do this? The resulting line may also be given by a parametric equation.
 A: As I mentioned in this answer, a parabola’s axis is parallel to the diagonal of the parallelogram defined by the tangents at two points. It’s not hard to work out that the tangents at $t=\pm1$ intersect at the point $(c_1-a_1,c_2-a_2)$, so the parabola’s axis is parallel to $$(a_1+b_1+c_1,a_2+b_2+c_2)+(a_1-b_1+c_1,a_2-b_2+c_2)-2(c_1-a_1,c_2-a_2) = (4a_1,4a_2),$$ or simply $(a_1,a_2)$.  
The vertex is the point at which the tangent is orthogonal to this vector: $$(a_1,a_2)\cdot(2a_1t+b_1,2a_2t+b_2) = 0,$$ from which $t=-{a_1b_1+a_2b_2\over 2(a_1^2+a_2^2)}$. I’ll leave working out the equation of the axis from these two bits of information to you.

P.S.: If you do happen to have the equation in general Cartesian form, finding the axis direction is also quite easy: it’s an eigenvector of $0$ of the matrix that corresponds to the quadratic part of the equation. If you work through the calculations, you’ll find that this eigenspace is spanned by $(a_1,a_2)$. The vertex is the point at which the normal is parallel to this vector, which can be expressed as the vanishing of a determinant, giving you a system of two equations to solve for the vertex coordinates.
