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

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: $$y(x) = \frac{v}{\sqrt a} \sqrt x - x$$

• Well! You've just proven it......... – Mostafa Ayaz Oct 11 '18 at 23:15
• @MostafaAyaz could you tell me how? What I got does not resemble any parabola equations I know – TheRlee Oct 11 '18 at 23:17
• Move the $-x$ term to the other side of the equation and then square both sides. – amd Oct 12 '18 at 1:26

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

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.

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

$$\begin{bmatrix} x&y\end{bmatrix}\begin{bmatrix} 1&1\\1&1\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix} - x = 0$$
• As a remark, squaring an equation can change the solution set. $f(x) = g(x)$ is true when the functions have the same signs, both $+$ or both $-$. $f^2 = g^2$ allows for these solutions, plus any solutions where $f = - g$. I suspect that in this case here, we didn't change the solution set. To check this, starting with $vt = x + y$, squaring won't change the solution set to this equation as long as $x + y > 0$ when $vt > 0$ and $x + y < 0$ when $vt < 0$.To proceed further, I'd have to know if $a,v$ are positive, negative, or mixed. Otherwise I'd have to check each case. Let's assume $a,v$ – DWade64 Oct 12 '18 at 14:34
• both positive numbers, then $x > y$ when $t > v/2a$, and $t = v/2a$ happens to be the time when $y(t)$ reaches its maximum. $x$ is always positive. $y > 0$ between $t = 0$ and $t = v/a$. So, yes, when $vt > 0$ (or since assuming $v$ is positive, when $t > 0$) $x + y$ is indeed positive. And when $t < 0 < v/2a$, then $x + y$ is less than $0$ so we are safe. It's impossible to find solutions where the left hand side is equal and opposite to the right hand side – DWade64 Oct 12 '18 at 14:39
• Otherwise if we didn't do this, since you started with $x = f(t)$ and $y = g(t)$ and found that this implies an equation in terms of $x,y$. If nothing changed, then we should be able to reverse the implication: starting with the $x,y$ equation and letting $x = f(t)$, do we get $y = g(t)$ and for some reason I'm getting that $y = \pm vt - at^2$ which is telling me, since we didn't get the same $y = f(t)$, we changed the solution set when we squared the equation. So I'm getting a little confused now – DWade64 Oct 12 '18 at 14:48
• However using wolfram alpha with random $a,v$ - it's showing me that $x = at$ and $y = vt - at^2$ or $x = at$ and $y = -vt - at^2$ seems to produce the same xy curve. So something weird is going on because of the parametric representation – DWade64 Oct 12 '18 at 14:54
• I think the end of my first comment, and my second comment, are flawed. There are $x,y$ values which make $x + y = - (vt)$, therefore squaring is not valid. We seem to get away with it because of some sort of symmetry in parabola's and parametric functions when $t \to -t$. – DWade64 Oct 12 '18 at 15:07