How To Parameterize A Parabola

How to Parametrise a Parabola?

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

Nosotros want to start at $\frac{three}{\sqrt{2}}$ and end at $0$. The above parameterization does the opposite. To reverse it, let's let $x=\frac{iii}{\sqrt{ii}}-t$, and $z=2\left(\frac{3}{\sqrt{2}}-t\correct)^2$. We tin still eliminate $t$ to become $z=2x^ii$, but now, when $t=0$, nosotros're at the $x$-value $\frac{\sqrt{three}}{2}$, and when $t=\frac{\sqrt{three}}{2}$ at the end of its interval, we're at the $10$-value $ten=0$.

There are infinitely many other means 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 tin all the same eliminate $t$ and get $z=ten^2$, we tin do a lot of different things.

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

Fifty-fifty that description doesn't cover all the ways this trouble could be solved. It'southward very open-ended.

Comments

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

Then if I had $two$ points then:

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

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

Is there a general way to solve all parabolic parametric equations? If yep, and then how?
- Ii points aren't plenty to specify a parabola. There are infinitely many parabolas that pass through those two points. To pin down which one you mean, nosotros'd need a third point, too.
- I tried to simplify it lol. Information technology's basically along the intersection of z=2x^ii and y=3 at the points A = (3/sqrt(2) , 3 , 9) and B = ( 0 , 3 , 0)
- So basically information technology'southward the parabola z=2x^ii for x and z. And then what would x(t) and z(t) be. What's the procedure for findind the parametric equations for a gerneral parabola though. Thanks
- It might be useful in the futurity to notice that instead of "the method of find the parametric equations" is more like "a method of find a parametric equation". In that location are many parameterizations and consequently many methods.
- If you take a defining equation, such as $y=2x^two$, in which ane variable $y$ is written every bit a function of the other $x$. Y'all 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 $ten$ in the cease points of the arc. In your case, $t\in[0,\frac{3}{\sqrt{2}}]$.
- @lorde fwiw, I wrote this nigh how to utilise the vertex and a betoken to specify the quadratic; math.stackexchange.com/a/481614/36530 that said, if you are actually request how to parametrize curves in full general and then you lot should realize that while the methods outlined in the comments are proficient, the comment about "a" vs. "the" by ABC is really important to absorb.