Converting from parametric equation to algebraic (implicit) equation of a parabola

Question

A parabola is given by

$ p(t) = v_0 + v_1 \ t + v_2 \ t^2 $

where $p(t) = (x(t), y(t)) $, and $v_0, v_1, v_2$ are known $2 \times 1 $ vectors, and $v_1$ is not necessarily perpendicular to $v_2$. I want to convert this parametric equation into an implicit equation of the form

$ a x^2 + b xy + c y^2 + d x + e y + f = 0 $

My attempt:

We can write

$ p(t) - v_0 = [ v_1 , v_2 ] [t, t^2]^T $

Let $U = [v_1, v_2]^{-1}$, then

$ U ( p - w_0) = [t, t^2]^T $

The vector $X = [t, t^2]^T$ satisfies

$ X^T Q X + b^T X = 0 $

where

$ Q = \begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix} $

$ b = \begin{bmatrix} 0 && -1 \end{bmatrix} $

Therefore,

$( p - w_0 )^T U^T Q U (p - w_0) + b^T U (p - w_0) = 0 $

And this the quadratic equation in $x, y$ that is the implicit equation of the parabola.

Any alternative solutions, remarks, comments, regarding this problem of conversion are highly appreciated.

Answer 1

I would procede as follow.
Suppose $v_i=(x_i,y_i)$, for $i = 1,2$, then \begin{align} & x = x_0 + x_1 t + x_2 t^2, \\ & y = y_0 + y_1 t + y_2 t^2 \end{align} Then, to eliminate the square of $t$ $$ y_2 x - x_2 y = (y_2 x_0 - x_2 y_0) + (y_2 x_1 - x_2 y_1) t $$ from which, supposing $y_2 x_1 - x_2 y_1 \neq 0$ $$ t = \frac{(y_2 x - x_2 y) - (y_2 x_0 - x_2 y_0)}{y_2 x_1 - x_2 y_1} $$ and substituting in one of the original equation should provide the implicit equation.

Answer 2

I did this problem some year ago for a problem in analytical geometry.
I'm omitting you the steps simply because I don't have them anymore and I only saved the formula.
If you have a curve $$\Gamma(x,y): ax^2+2bxy+cy^2+2dx+2ey+f=0$$ his parametric equation is:
$$\Gamma(t)=\mathbf{V}+\begin{pmatrix}\dfrac{(a+c)^2}{ae-bd}\cdot\dfrac{ab}{2}&a\\\dfrac{(a+c)^2}{cd-be}\cdot\dfrac{ab}{2}&b\end{pmatrix}\cdot\begin{pmatrix}t^2\\t\end{pmatrix}$$ Where $\mathbf{V}$ is the vertex of the parabola. $$\mathbf{V}=\begin{pmatrix}\dfrac{b(ad^2-ce^2)-2ae(be+ad)+bf(a+c)^2}{2(ae-bd)(a+c)^2}\\\dfrac{a(ad^2-ce^2)+2bd(ce+bd)-af(a+c)^2}{2(ae-bd)(a+c)^2}\end{pmatrix}$$