Parabola Through Four Points

Any four points in the plane determine two conjugate parabolas, each 
of which passes through all four points.  Of course, the axis of 
these parabolas aren't necessarily parallel to the y axis.  Given 
the points

            P1         P2         P3        P4
          (x1,y1)    (x2,y2)    (x3,y3)   (x4,y4)

we are free to translate the coordinate system so that x3=y3=0.  In
terms of this coordinate system we can fit a second degree polynomial
to the three points P1,P3,P4.  Since P3 is at (0,0) this polynomial
has the form
                     y = c1 x + c2 x^2

Similarly we can fit a second degree polynomial to the three points
P2,P3,P4.  This polynomial will be of the form

                     y = d1 x + d2 x^2

The coefficients c2 and d2 are given by

             x4 y1 - x1 y4                 x4 y2 - x2 y4
        c2 = -------------            d2 = -------------
             x1 x4 (x1-x4)                 x2 x4 (x2-x4)

The numerators of these coefficients are invariant under rotations but
the denominators are not.  Thus, if we rotate the coordinate system
relative to the configuration of points, we can vary the values of c2
and d2.  There exists a particular orientation in which c2 = d2, and
this is the orientation for which a single polynomial of degree 2
passes through all four points.

We rotate the coordinate system by applying the transformation

           xi'  =  xi cos(theta) + yi sin(theta)
           yi'  = -xi sin(theta) + yi cos(theta)

Substituting these transformed point coordinates into the expressions
for c2 and d2, and setting c2=d2, we arrive at a quadratic equation of
the form
             A tan^2(theta) + B tan(theta) + C  =  0

If we choose our (arbitrary) initial orientation and scale factor such
that x4=1 and y4=0 then the coordinates of this quadratic are simply

   A = y2-y1      B = 2(x2-x1)      C = x2(x2-1)/y2  -  x1(x1-1)/y1

The two roots of the quadratic give the two orientations in which the
four points lie on a single second degree polynomial.  (Incidentally, 
this shows that if a parabola with vertical axis intersects a 
horizontal line at points P,Q, and a second parabola intersects the 
first at the points P,Q,R,S, then the slope of the second parabola's 
axis is half the slope of the line through R,S.)

More generally, suppose we are given the configuration of four points
only in a relative sense, e.g., as the lengths of the three segments

        r1 = P1-P3      r2 = P2-P3      r4 = P4-P3

and the angles

     theta_1 = angle between the rays P3-P1 and P3-P4

     theta_2 = angle between the rays P3-P2 and P3-P4


If we let sj and cj denote the sine and cosine of theta_j, then the 
angle theta that the ray P3-P4 must make with the positive x axis 
in order for the four given points to lie on a single parabola must 
satisfy the quadratic

       a tan^2(theta)  +  b tan(theta)  +  c   =   0

where

        a = s1 s2 (s2 r2 - s1 r1)

        b = 2 s2 s1 (c1 r1 - c2 r2)

        c = s2 c1 (r4 - c1 r1) - s1 c2 (r4 - c2 r2)

As an example, consider the four points (-4,4), (-3,2), (0,0), and
(3,1).  This relative configuration of points can be represented by

               r1    = 5.6568542
               r2    = 3.6055512
               r4    = 3.1622776
             theta_1 = 2.0344439
             theta_2 = 2.2318394

In order to make a single parabola pass through all four points we 
must place P3 at the origin and orient the set of points such that the 
segment P3-P4 makes an angle of either 0.5741332 or -0.7504994 radians 
with the positive x axis.  In our original coordinate system we had P4
at (3,1), so the angle between P3-P4 and the x axis was atan(1/3), 
which equals 0.321750554, so this means we have to rotate those four 
points from that configuration by either 0.2523827 radians or 
-1.0722500 radians.  In the first case the coordinates of the four 
points become
                     x           y
                ----------  ----------
          P1    -4.8721282   2.8744332
          P2    -3.4043843   1.1875047
          P3     0.0000000   0.0000000
          P4     2.6552487   1.7174558

and the parabola that passes through those points is

            y  =  0.1643056 x^2  +  0.2105432 x

In the second case the coordinates of the four points become

                     x           y
                ----------  ----------
          P1    -4.8721282   2.8744332
          P2    -3.4043843   1.1875047
          P3     0.0000000   0.0000000
          P4     2.6552487   1.7174558

and the parabola that passes through those points is

            y  =  -6.0691452 x^2  +  13.1037433 x

We can transform both of these parabolas back to the original 
coordinate system, and show how both general parabolas pass through 
all four points.  This is illustrated in the figure below.



We've shown how to find the parabola passing through ANY four points, 
but you may wonder how it's possible for a single parabola to pass 
through each of the four points in a configuration such as shown below:

                                 *


                                 *

                          *             *

In general, if any of the points falls inside the triangle formed by 
the other three, then we can still find the parabola that contains all 
four points, but the parabola's coefficients are complex.

As an example, consider the four points with the following coordinates

                              x    y

                      P1     -1    4
                      P2     -2   -2
                      P3      0    0
                      P4      4    0

Here the point P3 falls inside the triangle formed by P1,P2,P4.  Using
the formula given above, this leads to the equation

            24*tan^2(theta) + 8*tan(theta) + 29  =  0

where theta is the angle through which the xy coordinate axes must be
rotated (counter clockwise) to give a coordinate system in which all
four points lie on a single polynomial of degree 2.  Solving this for
theta gives the two complex solutions

       tan(theta)  =  (-0.1666666) +(-) (1.086533734)i

Both of these roots give the following complex value for theta:

              theta  =  2.076894833  +  (1.205552068)i

In the rotated coordinate system x'y', the coordinates of the four
points are
                    x'                             y'

  P1       (7.2459) - (1.6174)i          (-1.9362) - (6.0528)i
  P2       (-1.4184) + (4.1313)i         (4.9457) + (1.1848)i
  P3             0 + 0i                         0 + 0i
  P4       (-3.5273) - (5.3162)i         (-6.3641) + (2.9464)i

The second degree polynomial that passes through all four points is

                   y'  =  A*(x')^2 + B*(x')

where
                A  =  (-0.0135548593) + (0.0263763418)i
                B  =  (-0.0213675212) - (1.0655569140)i

In the original xy coordinate system we have the following equation
for this conic

            a*x^2 + b*x*y + c*y^2 + d*x + e*y  =  0

where

    a = AC^2     b = 2ACS     c = AS^2     d = BC+S     e = BS-C

with
          C = cos(theta) = (-0.8818250591) - (1.3290520970)i
          S = sin(theta) = ( 1.5910307810) - (0.7366239911)i

Clearly this conic is a parabola, because b^2 = 4ac.  For our example,
the real and imaginary parts of the conic coefficients are as shown
below.
                       real              imaginary

          a        -0.048423136        -0.057852337
          b         0.141858076        -0.085942630
          c         0.034868277         0.084228678
          d         0.193692546         0.231409346
          e         0.062913888        -0.350541923

It's very interesting (to me) that the real and imaginary parts of
these parabola coefficients, taken separately, give two real conics
(hyperbolas), each of which passes through all four points as
shown below.