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### 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.

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.

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.

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/
```
Associated Topics:
High School Equations, Graphs, Translations

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