# How to convert the parametric equation into implicit form?

This problem is generated from another Green's theorem related question of mine.

The original equation of the plane curve is not in rational parametric form.

In order to calculate the symbolic solutions of the intersections of the curve itself, I thought its implicit equation form might be helpful or whatever. Nevertheless, such a conversion is interesting itself.

So after simple treatment, I have the following rational parametric equation of the same plane curve as in another problem:

$$\left\{\; \begin{array}{rl} x=& \dfrac{2 t \left(3 t^4+50 t^2-33\right)}{\left(t^2+1\right)^3} \\ y=& \dfrac{2 \left(7 t^6-60 t^4+15 t^2+2\right)}{\left(t^2+1\right)^3} \\ \end{array} \right.$$

How can I convert it into implicit form :$F(x,y)=0$? My difficulty mainly lies in how to implement it via different computer algebra systems, i.e., Mathematica, Maple or whatever.

The answer by Macaulay2 is definitely acceptable, however, there are less documents for beginners on the commands than Maple or Mathematica.

• Try also mathematica.stackexchange.com. – lhf Jun 29 '15 at 13:02
• Maple might be better in this side. So I don't want to limit it to Mathematica only. – LCFactorization Jun 29 '15 at 13:04

In Macaulay2

R=QQ[s,t,x,y,z,MonomialOrder=>Eliminate 2]
I=ideal(x-2*s*t*(3*t^4+50*t^2*s^2-33*s^4),y-2*(7*t^6-60*t^4*s^2+15*t^2*s^4+2*s^6),z-(t^2+s^2)^3)
gens gb I


yields a term free of $s, t$:

$625x^6+1875x^4y^2+1875x^2y^4+625y^6-182250x^4yz+364500x^2y^3z-36450y^5z+585816x^4z^2+1171632x^2y^2z^2+585816y^4z^2-41620992x^2z^4-41620992y^2z^4+550731776z^6$

Dehomogenising $z$ we get:

$625x^6+1875x^4y^2+1875x^2y^4+625y^6-182250x^4y+364500x^2y^3-36450y^5+585816x^4+1171632x^2y^2+585816y^4-41620992x^2-41620992y^2+550731776=0$

Which plots to the same as your parameterization

Edit

I didn't have to use the homogenised versions of everything, because the more straightforward clearing of denominators (see exercises 3.3.13-14 of Ideals, Varieties and Algorithms)

R=QQ[t,x,y]
I=ideal((t^2+1)^3*x-2*t*(3*t^4+50*t^2-33),(t^2+1)^3*y-2*(7*t^6-60*t^4+15*t^2+2))
gens gb I


gives the relation directly.

Interesting (to me at least) is that the original $x=−9\sin(2t)−5\sin(3t), y=9\cos(2t)−5\cos(3t)$ also satisfies this relation.

Edit2

Maple does it too:

> with(Groebner);
> Basis({(t^2+1)^3*x-2*t*(3*t^4+50*t^2-33), (t^2+1)^3*y-2*(7*t^6-60*t^4+15*t^2+2)}, tdeg(t, x, y));

• Thank you! Is there any equivalent commands in Mathematica for the codes you used in Macaulay 2? – LCFactorization Jun 29 '15 at 2:07
• Can you please recommend references for me to learn the principles of such procedure? thanks – LCFactorization Jun 29 '15 at 2:15
• @LCFactorization: I don't have Mathematica, and it's too heavy for wolframalpha, but try GroebnerBasis[{x-2*s*t*(3*t^4+50*t^2*s^2-33*s^4),y-2*(7*t^6-60*t^4*s^2+15*t^2*s^4+2*s^6),z-(t^2+s^2)^3},{s,t,x,y,z},{s,t}]. – Jan-Magnus Økland Jun 29 '15 at 2:17
• @LCFactorization: Ideals, Varieties, and Algorithms An Introduction to Computational Algebraic Geometry and Commutative Algebra by David Cox, John Little and Donal O’Shea – Jan-Magnus Økland Jun 29 '15 at 2:19
• @LCFactorization: Let's illustrate with the circle: $x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$. This is $X=s^2-t^2,Y=2st,Z=s^2+t^2$ where $x=\frac{X}{Z}, y=\frac{Y}{Z}$ and homogenised to be degree two polynomials s.t. the rational functions are degree $0$. That is as a map $k[X,Y,Z]\to k[s,t]$ i.e as a map from $\mathbb{P}^1\to \mathbb{P}^2$. But you can interpret it as a map $\mathbb{A}^2\to\mathbb{A}^3$ as well, and then you can proceed as in the link you provided on implicitization of surfaces. Btw, if you want to save the question, say where you got the problem from, and what you did. – Jan-Magnus Økland Jun 29 '15 at 10:20