From: rusin@math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Tangency of Rational Curves
Date: 18 Mar 1998 19:50:37 GMT

In article <350C08D3.3D1C@olemiss.edu>,
Scott Gilbert  <sgilbert@olemiss.edu> wrote:
>Given two CURVES 
>y(x) = P(x)/Q(x) and x(y) = R(y)/S(y), 
>with x and y REAL VECTORS, suppose
>(1) P(x) and R(y) are AFFINE 
>(2) Q(x) and S(y) are QUADRATIC, positive
>    and bounded away from zero. 
>
>Question: Can the curves y(x) and x(y)
>have more than one tangency point?

Sure. Here's how to find them: up to scaling, you're interested in 
pairs of curves
	y=(ax+b)/(x^2+cx+d)   and  x=(y+e)/(y^2+fy+g)
(Scaling would also let you assume  e=0  or  e=1  if you wish.)

In general, any such pair has five points of intersection, as can be
seen by eliminating  x  (say) from the pair of equations. Of course the
intersections need not be real or distinct; the latter is just the
situation you seek. You want the quintic in  y  to be of the form
	(y-y1)*(y^2-Sy+P)^2
for some  S  and  P which give a quadratic with real roots. Well, that
gives 5 equations in {a, ..., g, y1, S, P}. We can easily eliminate
y1, S, and  f, say, using equations which are linear in those variables.
Then we need only choose the other 7 variables to satisfy 2 polynomial
equations. In principle one could eliminate further, but this task
swamps my computer-algebra resources. So I simply selected a few
values of  {a,b,d,g,P} and asked Maple to  fsolve  for c and e.

Actually this isn't quite sufficient; as noted, we need  S^2-4P  positive,
and your positivity condition forces  4d-c^2 and 4g-f^2  to be positive
as well. Thus whenever I solve for  c and  e, I need to check that these
three quantities are all positive. I had to reject a few initial guesses
of  {a,b,d,g,P}  to achieve this, but found
	a=10	b=10	d=10	g=15	P=15
to work. Since  c  was so close to integer, I decided to play a bit and
found
	a=10	b=10	d=10	g=15	P=14.974504
to work, too, and to give integral c=2 (and e=-3.85733, f=7.73370).
The curves are tangent near (-2,4) and (2,4).

Since the solution set is now a nonempty open subset of a 5-dimensional
variety, you have plenty of such pairs of curves to choose from.

dave