How Many Points Determine a Parabola?

Date: 12/20/2000 at 10:19:02
From: Ramani Jv
Subject: Identification of a parabola

It is well known that a circle is determined uniquely by three points, 
and from this case it is also possible to find its center and radius.

Suppose we are given a plane curve. How can we determine whether or 
not it is a parabola? Under what conditions is a parabola uniquely 
determined?

With regards,
Ramani Jv

Date: 12/20/2000 at 15:55:18
From: Doctor Rob
Subject: Re: Identification of a parabola

Thanks for writing to Ask Dr. Math, Ramani.

Four points determine a parabola up to a choice of two possibilities.

The general equation of a conic section is:

     A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

If you are given five points on this curve, that will give you a 
system of five simultaneous linear equations in the six unknowns A, B, 
C, D, E and F. Then you can solve for five of these in terms of the 
sixth. The sixth can take an arbitrary value, and then the others are 
uniquely determined. That arbitrary value will affect the above 
equation by replacing it with a constant multiple, which is immaterial 
to the curve itself. Thus five points determine a general conic 
section.

In the case of a parabola, you know that B^2 - 4*A*C = 0, so you have 
an additional equation, either A = B^2/4*C or C = B^2/4*A, depending 
on whether C or A is nonzero (both can't be zero, or you would have 
B = 0 too, and the equation would be that of a line, not a conic 
section). That means that only four points and four simultaneous 
linear equations will allow you to find the values of five of the six 
variables A, B, C, D, E and F in terms of the sixth, and that sixth 
can take an arbitrary value that is immaterial to the curve itself, 
except that there will be two solutions corresponding to the solutions 
of a quadratic equation. Thus four points determine a parabola up to 
two choices.

There are some cases in which the parabola will be degenerate, and 
they occur when

             |2*A  B   D |
     Delta = | B  2*C  E | = 0.
             | D   E  2*F|

Then the four points do not determine a parabola, but a pair of lines.

Example: Find the parabola determined by the points (0,0), (4,0), 
(3,5), and (-1,1). Then:

                                     F = 0
                        16*A + 4*D + F = 0
     9*A + 15*B + 25*C + 3*D + 5*E + F = 0
                 A - B + C - D - E + F = 0
                           B^2 - 4*A*C = 0

The solutions to this system are

     B/A = -2 - 4*sqrt(3/5)     or  -2 + 4*sqrt(3/5),
     C/A = 17/5 + 4*sqrt(3/5)   or  17/5 - 4*sqrt(3/5),
     D/A = -4                   or  -4
     E/A = -52/5 - 8*sqrt(3/5)  or  -52/5 + 8*sqrt(3/5),
     F/A = 0                    or  0.

In order to clear denominators, we can choose A = 5, and then the 
equations of the two parabolas are:

     5*x^2 - (10+4*sqrt[15])*x*y + (17+4*sqrt[15])*y^2 - 20*x -
                                               (52+8*sqrt[15])*y = 0

     5*x^2 - (10-4*sqrt[15])*x*y + (17-4*sqrt[15])*y^2 - 20*x -
                                               (52-8*sqrt[15])*y = 0

In these cases Delta = 1920*(-47 +- 12*sqrt[15]), which is nonzero, so 
these two parabolas are not degenerate.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/