system
Given the following post, respond as if you're commenting on Reddit:
human
In Euclidean space, is there any set of three points that can't all be interested by a parabola?

Now that I think about it, if they were in a straight line they couldn't be parabolic.

Can a set of three points all be intersected by two different parabolas?
gpt
It depends on what you mean by a parabola.

If you mean a function y=ax^2 + bx + c with a not equal to zero, then three points in a straight line can't all be intersected by a parabola. If you let a=0, that is, let a line be a "degenerate" parabola, then it's fine.

The other question is: do you allow rotated parabolas? For example, curves of the form x = ay^2 + by + c? If not, then if two points have the same x coordinate, it won't work.

So if you only allow functions y=ax^2 + bx + c, regardless of what you allow a to be, it won't work. However, if you allow everything I've mentioned above (rotations and a=0), then you can always find such a parabola.

More detail if you know a little bit about matrices and if you're interested:

You can actually write the problem as follows. Let's assume for simplicity that the x component of all the points are distinct. If not, rotate your coordinates so that they are; this can always be accomplished. This step is why we have to allow rotations of "standard" parabolas.

Now, what you want is numbers a, b, and c so that for each of your points (x1, y1), (x2, y2), (x3, y3), you have y = ax^2 + bx + c. You can write this as a matrix equation: [y1, y2, y3]^T = M [a b c]^T , where the T means transpose, and M is a 3x3 matrix with ith row [xi^2, xi, 1]. You should check that this is in fact the condition we need to satisfy. Now, a solution for [a,b,c] definitely exists if M is invertible, because then we can just multiply the equation above on the left by M^-1 . You can check by computing the determinant of M and finding that it's not zero that M is always invertible because we assumed x1, x2, and x3 were all different.