Fastest way to fit a parabola to set of points?

If the points have no error associated, you may interpolate by three points. Otherwise least squares or any equivalent formulation is the way to go.

I recently needed to find a parabola that passes through 3 points.

suppose you have (x1,y1), (x2,y2) and (x3,y3) and you want the parabola

y-y0 = a*(x-x0)^2

to pass through them: find y0, x0, and a.

You can do some algebra and get this solution (providing the points aren't all on a line) :

let c = (y1-y2) / (y2-y3)
x0    = ( -x1^2 + x2^2 + c*( x2^2 - x3^2 ) )  /  (2.0*( -x1+x2 + c*x2 - c*x3 ))
a     = (y1-y2)  /  ( (x1-x0)^2 - (x2-x0)^2 )
y0    = y1 - a*(x1-x0)^2

Note in the equation for c if y2==y3 then you've got a problem. So in my algorithm I check for this and swap say x1, y1 with x2, y2 and then proceed.

hope that helps!

Paul Probert

A calculated solution is almost always faster than an iterative solution. The "exception" would be for low iteration counts and complex calculations.

I would use the least squares method. I've only every coded it for linear regression fits but it can be used for parabolas (I had reason to look it up recently - sources included an old edition of "Numerical Recipes" Press et al; and "Engineering Mathematics" Kreyzig).