Parabola through three points

by David Radcliffe

Chris Harrow asked the following question on Twitter:

My interpretation is that we fix three non-collinear points in the plane, and we wish to describe all parabolas passing through these points. (There are infinitely many of these, since the axis of symmetry need not be vertical.)

For simplicity, I assumed that the three fixed points are (0,0), (1,0), and (0,1). This is reasonable because any three non-collinear points can be moved to (0,0), (1,0), and (0,1) by an affine transformation (linear transformation + translation), and because the image of a parabola under an affine transformation is again a parabola.

The general equation of a conic section is Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A, B, and C are not all zero. A conic is a parabola (possibly degenerate) if and only if B^2 = 4AC. Since the parabola is assumed to contain (0,0), (1,0), and (0,1), we find that F = 0, A + D = 0, and C + E = 0. Furthermore, A cannot be zero, since we cannot draw a horizontal parabola through these points. So we may assume, after dividing through by A, that A = 1.

Combining these equations leaves us with a one-parameter family of parabolas:

x^2 + Bxy + \frac{B^2}{4} y^2 - x - \frac{B^2}{4} y = 0.

I created a GeoGebra animation to explore how changing the value of B affects the parabola. It would be interesting to explore the curves traced out by the vertex and the focus as B varies. Wolfram Alpha produced some formulas, but they are not easy to interpret.

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