Intersecting Parabolas

Date: 09/15/1999 at 08:12:02
From: Sergio
Subject: Parabolas

Prove that the intersections of two parabolas always belong to a 
circle. The proof requested assumes that the relative slopes of the 
parabolas are unknown.

I think the question is incorrect because, when the points of 
intersection are four, they should belong to a quartic.

Thanks in advance for your answer.
Sergio.

Date: 09/15/1999 at 12:22:10
From: Doctor Rob
Subject: Re: Parabolas

Thanks for writing to Ask Dr. Math, Sergio.

Any three of the intersection points determine a circle. What you have 
to prove is that the fourth point also lies on that same circle.

Actually, I don't think this is possible, either. You can find an 
ellipse through these four points, but not a circle, in general.

If you start with two general equations of the two parabolas,

     a*x^2 + b*x*y + c*y^2 + d*x + e*y + f = 0
     A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

where b^2 = 4*a*c and B^2 = 4*A*C, then, by multiplying the first 
equation by B and the second by b, and subtracting one from the other, 
you will get

     (a*B-A*b)*x^2 + (c*B-C*b)*y^2 + (d*B-D*b)*x + (e*B-E*b)*y
                                                    + (f*B-F*b) = 0

which has no x*y-term. This is a circle only if the coefficients
of x^2 and y^2 are equal, that is,

     a*B - A*b = c*B - C*b
       (a-c)*B = (A-C)*b
       B/(A-C) = b/(a-c)

This equation is definitely not satisfied by all pairs of parabolas. 
You can write down a counterexample very easily.

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

Date: 09/20/1999 at 10:40:11
From: Sergio
Subject: Parabolas

Find out when the four points of intersection of two parabolas belong 
to a circle. The solution should be found by evaluating the relative 
slopes of the axes of the parabolas.

Thanks for your answer.
Sergio

Date: 09/20/1999 at 16:16:06
From: Doctor Rob
Subject: Re: Parabolas

Reduce as before. You get the equation

     (a*B-A*b)*x^2 + (c*B-C*b)*y^2 + (d*B-D*b)*x + (e*B-E*b)*y
                                                    + (f*B-F*b) = 0

which has no x*y-term. This is a circle only if the coefficients
of x^2 and y^2 are equal, that is,

   a*B - A*b = c*B - C*b,
   (a-c)*B = (A-C)*b,
   B/(A-C) = b/(a-c).

You also need that the coefficients of x^2 and y^2, in addition to 
being equal, are not zero. If they are both zero, then a/A = b/B = 
c/C, and the first equation above is the equation of either a straight 
line (which implies that there are at most two intersection points, 
which is impossible) or the empty set (which implies that there are no 
intersection points, which is also impossible).

Now if the angles of the axes of the parabolas are t and T, with both 
angles between 0 and 2*Pi, then

     tan(2*t) = b/(a-c)
     tan(2*T) = B/(A-C)

There are special cases to consider when a = c (so t = Pi/4, 3*Pi/4, 
5*Pi/4, or 7*Pi/4) and when A = C (so T = Pi/4, 3*Pi/4, 5*Pi/4, or 
7*Pi/4).

Assuming the equation above is a circle, and using the earlier 
equation in a, b, c, A, B, and C

         tan(2*T) = tan(2*t)
     tan(2*T-2*t) = 0
        2*T - 2*t = n*Pi, for some integer n, 3 >= n >= -3
            T - t = n*(Pi/2)

This implies that the axes of the parabolas are either perpendicular 
(n is odd) or parallel (n is even).

I leave it to you to show that if the axes of two parabolas are 
parallel, there are at most two intersection points. I also leave it 
to you to find the conditions when the axes are perpendicular and 
there are four intersection points, which doesn't always happen.

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