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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.

Y N
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  • Do you get the vertex when t=0? If so, the tangent at t=0 would be normal to the axis of symmetry maybe? (These aren't hints, I'm just talking out of the top of my head.) –  Aug 14 '19 at 07:31
  • @Matthew Daly, yes, the point $t=0$ is included but it's not guaranteed that it would be parabola's vertex. Is that what you were trying to say? – Y N Aug 14 '19 at 07:51
  • Yes, that is what I was trying to ask. Drat. –  Aug 14 '19 at 07:52

1 Answers1

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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.

amd
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    "From a distance", a parabola looks like a double-ray along its axis, and its axis looks like a line through the origin, so you can determine the direction vector of that axis by noticing $(x,y)\propto (a_1,a_2)$ as $t\to\pm\infty$. (That argument is perhaps a bit loosey-goosey. Luckily, the parallelogram approach confirms it. :) – Blue Aug 14 '19 at 12:49
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    @Blue I’m sure that can be formalized somehow. I more often come at it from a projective point of view (the axis direction is the tangent point to the line at infinity) but that’s easier to do with a Cartesian equation. – amd Aug 14 '19 at 19:25