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What is the method to find the parametric equations for all types of parabolas (in both directions)?

So if I had $2$ points then:

Parametrise from $A$ to $B$ where $A = \left( \frac{3}{\sqrt2} , 9 \right)$ and $B = ( 0, 0 )$.

What would $x(t)$ and $y(t)$ be?

Is there a general way to solve all parabolic parametric equations? If yes, then how?

braziil
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  • Two points aren't enough to specify a parabola. There are infinitely many parabolas that pass through those two points. To pin down which one you mean, we'd need a third point, too. – G Tony Jacobs Sep 06 '13 at 04:11
  • I tried to simplify it lol. It's basically along the intersection of z=2x^2 and y=3 at the points A = (3/sqrt(2) , 3 , 9) and B = ( 0 , 3 , 0) – braziil Sep 06 '13 at 04:16
  • So basically it's the parabola z=2x^2 for x and z. So what would x(t) and z(t) be.

    What's the process for findind the parametric equations for a gerneral parabola though. Thanks

     – braziil Sep 06 '13 at 04:18
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    It might be useful in the future to notice that instead of "the method of find the parametric equations" is more like "a method of find a parametric equation". There are many parameterizations and consequently many methods. – OR. Sep 06 '13 at 04:48
  • If you have a defining equation, such as $y=2x^2$, in which one variable $y$ is written as a function of the other $x$. You always can get out of it the parameterization $x(t):=t$ and $y(t):=2(x(t))^2=2t^2$. The domain of this parameterization would be the end-points of the values of $x$ in the end points of the arc. In your case, $t\in[0,\frac{3}{\sqrt{2}}]$. – OR. Sep 06 '13 at 04:52
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    @lorde fwiw, I wrote this about how to use the vertex and a point to specify the quadratic; http://math.stackexchange.com/a/481614/36530 that said, if you are really asking how to parametrize curves in general then you should realize that while the methods outlined in the comments are good, the comment about "a" vs. "the" by ABC is really important to absorb. – James S. Cook Sep 06 '13 at 05:03

1 Answers1

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Probably the simplest approach is to let $x=t$ and $z=2t^2$, so we can easily eliminate $t$ and get $z=2x^2$. Now we just let $t$ take values on the interval $\left[0,\frac{3}{\sqrt{2}}\right]$. But... this isn't quite right.

We want to start at $\frac{3}{\sqrt{2}}$ and end at $0$. The above parameterization does the opposite. To reverse it, let's let $x=\frac{3}{\sqrt{2}}-t$, and $z=2\left(\frac{3}{\sqrt{2}}-t\right)^2$. We can still eliminate $t$ to get $z=2x^2$, but now, when $t=0$, we're at the $x$-value $\frac{\sqrt{3}}{2}$, and when $t=\frac{\sqrt{3}}{2}$ at the end of its interval, we're at the $x$-value $x=0$.

There are infinitely many other ways to parameterize this. Sometimes, people like to have $t$ come from the interval $[0,1]$; we could do that. Alternatively $x$ could depend on $t$ in some other, non-linear fashion, as long as we can still eliminate $t$ and get $z=x^2$, we can do a lot of different things.

In fact, let $f(t)$ be any function that decreases monotonically on the interval $[a,b]$, with $f(a)=\frac{\sqrt{3}}{2}$ and $f(b)=0$. Then we can define a parameterization by $x(t)=f(t)$ and $z(t)=2(f(t))^2$, where we get the desired curve by letting $t$ run from $a$ to $b$. In my second paragraph above, I did this, using a linear function for $f$.

Even that description doesn't cover all the ways this problem could be solved. It's very open-ended.

G Tony Jacobs
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