Showing that $x(t)=at^2$, $y(t)=vt-at^2$ parameterizes a parabola

Question

Solving a physics problem I obtained following motion equations $$ x(t) = at^2 $$ $$ y(t) = vt - a t^2 $$ And I want to determine what type of curve is it on the interval of $\left<0,v\right>$. Wolfram says it's a parabola but I can't prove it. All I got is what is obvious: $$ y(x) = \frac{v}{\sqrt a} \sqrt x - x $$

Answer 1

$x = at^2\\ y = vt - at^2\\ y = vt - x\\ vt = x+y\\ v^2t^2 = (x+y)^2\\ \frac {v^2}{a} x = (x+y)^2$

That is a parabola that has been rotated 45 degrees from standard.

If that is not obvious, you could rotate the frame..

$x' = \frac {\sqrt 2}{2} x + \frac {\sqrt 2}{2} y\\ y' = \frac {\sqrt 2}{2} y - \frac {\sqrt 2}{2} x$

Which will rotate your parabola into standard form.

$\frac {v^2}{a} \frac {\sqrt 2}{2}(x' - y') = 2x'^2\\ y'= x' - 2\sqrt 2(\frac a{v^2}) x'^2$

Or you could work with the quadradic form:

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

And since the matrix is singular, one of the eigenvalues is 0, and it is a parabola.