Is there a formula for this parabola?

[Hertz]

Given three arbitrary points on a coordinate system, is there a way to derive an equation that forms the single parabola that passes through all three points?

I guess firstly you would have to prove that given three points, only a single parabola passes through all three, but judging by the fact that a single line passes through two points, and a single horizontal line passes through one point, I would say that a single parabola passes through 3 points.

Anyways, setting that proof aside for now, can anyone think of a way you could do this? I was looking into the standard and general form of a parabola but those forms don't account for a slanted axis of symmetry, which would in general be required to plot most parabolas.

-Edit
Alright, it took me a while, but I see what all of you guys are saying now. There are infinite parabolas that can pass through 3 given points; however, only one parabola of the form ax^2 + bx + c. I finally see that this is the case.

I should be able to find what I'm looking for given this information.

 

[flatmaster]

Given three arbitrary points on a coordinate system, is there a way to derive an equation that forms the single parabola that passes through all three points?

I guess firstly you would have to prove that given three points, only a single parabola passes through all three, but judging by the fact that a single line passes through two points, and a single horizontal line passes through one point, I would say that a single parabola passes through 3 points.

Anyways, setting that proof aside for now, can anyone think of a way you could do this? I was looking into the standard and general form of a parabola but those forms don't account for a slanted axis of symmetry, which would in general be required to plot most parabolas.

You are correct in that three points would would uniquely define a parabola of the type

y = Ax^2 + Bx + C

because there are three constants that need to be defined.

If you have three points that you know are on the parabola (x1,y1) (x2,y2) and(x3,y3) you should be able to solve for your unknown constants A, B, and C in terms of the coordinates of your three known points.

You start with three equations (one for each of your known points) and three unknowns (A,B, and C) and solve for your unknwowns.

I've never seen this done before and I'm not sure why you would do it. Even using a matrix to solve this would get rather messy.

As you said, this would not take into account parabolas that have an axis of symetry that is not parallel to the y axis.

 

[micromass]

I don't think it is even true that every 3 points determine a unique parabola. I think you need 4 points. If you're given that the axis of symmetry is parallel to the y-axis, then you might get away with 3 points in most cases.

Anyway, a general conic section is given by either

[tex]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/tex]

This is a parabola if [itex]B^2=4AC[/itex].

Assume that [itex]A\neq 0[/itex], the A=0 case is very similar. We can assume without loss of generality that A=1. So we have that [itex]C=B^2/4[/itex]. So we have the equation

[tex]x^2+Bxy+\frac{B^2}{4}y^2+Dx+Ey+F=0[/tex]

Assume now that we want a parabola through (0,0), (1,0) and (0,1), then the following system must be satisfied:

[tex]\left\{\begin{array}{l}
F=0\\
1+D+F=0\\
\frac{B^2}{4}+E+F=0
\end{array}\right.[/tex]

This gets us F=0, D=-1 and [itex]E=-\frac{B^2}{4}[/itex]. So the parabola through those three points must have the form

[tex]x^2+Bxy+\frac{B^2}{4}y^2-x-\frac{B^2}{4}y=0[/tex]

As you see, this is not uniquely determined. For every choice of B, there is a parabola.

If we are given a fourth point, for example (3,2), then we get an additional equation

[tex]6+6B+\frac{B^2}{2}=0[/tex]

Solving this gives you two values for B. This gives you two parabolas.

 

[DuncanM]

Yes, it can be done.

One way would be to form two equations in two unknowns.
You know the general form of a parabola, so plug the points in and solve for the values. And if the parabola is symmetric around an axis it simplifies easier.

Although, I don't think you can prove only one parabola would fit the three points. Some points might fit an up-closing parabola and either a right- or left-closing parabola. Same for a down-closing parabola.

 

[Hertz]

Yes, it can be done.

One way would be to form two equations in two unknowns.
You know the general form of a parabola, so plug the points in and solve for the values. And if the parabola is symmetric around an axis it simplifies easier.

Although, I don't think you can prove only one parabola would fit the three points. Some points might fit an up-closing parabola and either a right- or left-closing parabola. Same for a down-closing parabola.

Ah good point. What if we make the restrictions that the vertex of the parabola must be between the first and third points and that the line connecting point one and point three does not include point two? I believe that would then allow us to generalize and say that any three points satisfying condition two will then make one and only one parabola given condition one.

Any objections?

I'm not looking to solve for a specific parabola, I'm looking for a generalized solution to this problem. I want to either find or derive a formula of the form:

f(x) = ... (In terms of x1, x2, x3, y1, y2, and y3.)

that will give the one and only parabola that passes through the three given points. (Assuming the conditions above are met.)

-Edit
Actually, it appears that given any three points meeting the above condition, there will always be three possible parabolas that could connect the points. Hmm..