From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: Parabola through four integer points
Date: 8 Jun 2001 21:55:16 GMT
Newsgroups: sci.math
Summary: When do four points lie on a single parabola?

In article <7ko3byzpozqx@forum.mathforum.com>,
martinton@yahoo.com (Martin Lukarevski) asked
when there was a parabola through 4 given points. Here's
a fairly straightforward algebraic approach.

A line can meet a parabola at most twice, so we must
suppose that no three of the points are on the same line.

Assuming that, we can translate the plane to put the first
point at the origin. We can then do a linear transformation 
to put the next two at (1,0) and (0,1).  The fourth one
will now be some (p,q) not on the lines x=0, y=0, x+y=1
(the lines joining the first three).

Both the translation and the linear map take parabolas to
parabolas, so the original four points are on a parabola
preciely when these images are. (Indeed, we can go from
one equation to the other.) 

Now a general conic will have an equation of the form
  ax^2 + bxy + cy^2 + dx +ey +f = 0.
The condition that it goes through (0,0) is f=0, and
going through (1,0) and (0,1) is then equivalent to
a+d=0 and c+e = 0.  That is, a conic through these
three points has the form
   a(x^2-x) + bxy + c(y^2-y) = 0.

The condition that makes the conic a parabola (or possibly
two parallel lines) is b^2 = 4ac.  If (say) c=0, then b=0, 
and in our curves we have just the two lines a(x^2-x)=0. Since
that is easy to check whether p=1, we may assume c is nonzero
and solve for a, gettting the form
   (b^2/4c) (x^2-x)  + b xy + c (y^2-y) = 0
or
   (b/c)^2 (x^2-x)/4 + b/c xy +  (y^2-y) = 0.

For our given (p,q), now, we are asking for nonzero (real)
b and c satisfying
   (b/c)^2 (p^2-p)/4 + (b/c) pq +  (q^2-q) = 0.
This has a solution precisely when the discriminant
  (pq)^2 - 4 [(p^2-p)/4][q^-q] 
is 0 or positive.  But that comes out to be just
   pq(p+q-1).
Since we have assumed that (p,q) is not on the three lines,
this expression will never be 0. Thus we are through:

Theorem.  Take (p,q) not on the lines x=0, y=0, x+y=1. Then
there is a parabola or pair of parallel lines through
(0,0), (1,0), (0,1), (p,q) precisely when pq(p+q-1) > 0.

Now we can go back to geometry. The three lines divide
the plane into seven regions, and the sign of pq(p+q-1)
depends only on which region you are in.  But that
splitting is preserved by the translation and linear
transformation, so the same conclusion carries back
to the original four points. Thus we have our geometric
result:

Theorem. Take four points, no three on a line. Choose
any three of them and draw the corresponding three lines.
Then there is a parabola or pair of parallel lines through
the four points precisely when the fourth point is in
certain of the regions formed by those lines. 

Checking in the special case, you can see more specifically
that the fourth point must not lie in the interior of the triangle
or the vertical angles at the vertices.

Note finally that the equation for b/c is quadratic. Thus usually
there will be two different parabolas when the fourth point is
in the permissible regions. I think the computation shows in fact
that, even when the four points are on two parallel lines, there will 
also be another, "true" parabola through them unless they are the
vertices of a paallelogram.


William C. Waterhouse
Penn State