How do you get an algebraic equation from a parametrization?

Question

I know the Cycloid is an example of a curve with a parametrization that doesn't have an algebraic equation. But is there a general procedure to find the algebraic equation of parametric curves that do have one? (I am mostly concerned with plane curves for now, so an answer narrowed down to this case will be fine.)

As an example, I'm trying to find the algebraic equation for the Hypocyloid parametrized by:

$$\begin{aligned}t \mapsto & 1/3(2\cos (t) + \cos (2t), 2\sin (t) -\sin (2t)) \\ &= 1/3(2\cos (t) + \cos^2 (t)-\sin^2 (t), 2\sin (t) -2\sin (t)\cos (t))\end{aligned}$$

I know that this Hypocycloid is the vanishing set of a quartic, but the best I can do from the above equation is use the rational parametrization of the circle to get a rational parametrization. Indeed, letting $\tau = \tan (t/2)$ we get

$$(\cos t, \sin t) = \frac 1 {1+\tau^2} (1-\tau^2, 2\tau)$$ so that the Hypocycloid has a parametrization

$$\tau \mapsto \frac 1 {3(1+\tau^2)^2}(3-6\tau^2 -\tau^4,8\tau^3)$$

But I don't know how to get further.

Answer 1

In M2; that is using gröbner bases:

R=QQ[s,t,x,y,z,MonomialOrder=>Lex]
I=ideal(x-(3*s^4-6*t^2*s^2-t^4),y-(8*t^3*s),z-3*(s^2+t^2)^2)
gens gb I -- among other terms 3x4-8x3z+6x2y2+6x2z2+24xy2z+3y4+6y2z2-z4

So your equation after dehomogenisation is: $3x^4-8x^3+6x^2y^2+6x^2+24xy^2+3y^4+6y^2-1$ as a check with plots in Geogebra using Curve and ImplicitCurve seems to verify.