Parametric Equations that don't follow the rule:

Question

If you have a pair of parametric equations for a parabola, say $x=-2+t^2$, and $y=1+4t$. From what I have learnt parametric equations for a parabola can be generalised to $x=g+at^2$ and $y=h+2at$ where the equation of the parabola was $(y-h)^2=4a(x-g)$. But clearly, these $x=-2+t^2$, and $y=1+4t$ do not follow that structure so I found the equation of the parabola by making 't' the subject and getting $(y-1)^2=16(x+2)$ meaning that for that parabola 'a' was 4 as $4a=16$. And then I made parametric equations seeing that $a=4$ and got $x=-2+4t^2$, and $y=1+8t$ is there any correlation between the two sets of parametric equations? And why do they graph the same parabola?

One thing I did find is that the ratio between the coefficient of the '$t$' in the first $x$ equation so $1$ and the coefficient of the '$t$' in the second $x$ equation so $4$ it is equal to the ratio of the coefficient of the '$t$' in the first $y$ equation and the coefficient in the second $y$ equation. As in $\dfrac{1}{4}=\left(\dfrac{4}{8}\right)^2$, is this some relationship? And this relationship was the same for two other different examples where $\dfrac{x_1}{x_2}=\left(\dfrac{y_1}{y_2}\right)^2$.

Thanks

Answer 1

A parabola is a geometric shape, while a parametric equation describes a path on a curve. But for a given shape the path is not unique, you can shift $t$, change speed and even direction while still lying on the same parabola.

For instance, $x=t,y=t^2$ is a parametrization of the parabola with cartesian equation $y=x^2$. Another valid parametrization is $x=\sinh t, y=\frac{1}{2}(\cosh(2t)-1)$. Still another parametrization is $x=t\sin(t),y=\frac12t^2(1-\cos(2t))$.

It's nonsense to say that the parametric equation of a parabola must have a given restricted form, while there are infinitely many ways, without necessarily an easy way to convert the parameter from one parametrization to another: it may involve an equation with no closed-form solution.

That being said, you have two parametric equations (I change the name of the parameter of the second one on purpose):

• $x=-2+t^2, y=1+4t$
• $x=g+au^2, y=h+2au$

Is there a simple relation between $t$ and $u$? Let's first try to see if they describe the same parabola. It's easier to check the cartesian equation, that describes the coordinates of points on the shape, rather than an arbitrary path:

$$t^2=x+2=\left(\frac{y-1}{4}\right)^2$$

Hence the cartesian equation $y^2-2y-16x-31=0$

The other leads to

$$t^2=\frac{x-g}{a}=\left(\frac{y-h}{2a}\right)^2$$

Hence the cartesian equation $y^2-2hy-4ax+h^2+4ag$

It's the same equation if $a=4,g=-2,h=1$. Then the two parametric equations describe the same curve. But not necessarily in the same way!

We now have the two sets of equations:

• $x=-2+t^2, y=1+4t$
• $x=-2+4u^2, y=1+8u$

Now, what happens if $t=2u$?