I have the following parametric form of epicycloids:
- $x(t)=\frac{a\cdot\cos t+\cos(a\cdot t)}{1+a}$
- $y(t)=\frac{a\cdot\sin t+\sin(a\cdot t)}{1+a}$
where $a=2,3,4,\ldots$ is a variable that generalized these curves.
I calculated the cartesian forms for $1<a<6$:
- For $a=2$ we have $0=27x^4+54x^2y^2-18x^2-8x+27y^4-18y^2-1$
- For $a=3$ we have $0=64x^6+192x^4y^2-48x^4+192x^2y^4-96x^2y^2 -15x^2+64y^6-48y^4+12y^2-1$
- For $a=4$ we have $0=3125x^8+12500x^6y^2-2500x^6+18750x^4y^4-7500x^4y^2-50x^4-512x^3+12500x^2y^6-7500x^2y^4-100x^2y^2-36x^2+1536xy^2+3125y^8-2500y^6-50y^4-36y^2-27$
- For $a=5$ we have $0=11664x^{10}+58320x^8y^2-9720x^8+116640x^6y^4-38880x^6y^2-135x^6+116640x^4y^6-58320x^4y^4-405x^4y^2-1665x^4+58320x^2y^8-38880x^2y^6-405x^2y^4+9170x^2y^2-80x^2+11664y^{10}-9720y^8-135y^6-1665y^4-80y^2-64$
Is it possible to get a generalized cartesian form that maintains the parameter $a$?
In order to make it more illustrative, what the parameter $a$ does, I plotted these curves for $1<a<6$:
I know that the trick might be expanding $\cos a\cdot t$ and $\sin a\cdot t$ using Chebyshev polynomials. And I see that the coefficients of the cartesian forms (given above for $1<a<6$) posses a certain structure, but how we can generalize these by involving the parameter $a$?




I came across this interesting paper "Implicitization of curves parameterized by generalized trigonometric polynomials" by Hoon Hong (DOI: 10.1007/3-540-60114-7_21), which provides a generalized approach to implicitize parametric curves that are defined as follows:
Hong calculates the resultant $H=res_z(F,G)$ where $F=-(x+iy)+\sum_{k=1}^na_kz^k$ and $G=-(x-iy)z^n+\sum_{k=1}^na_kz^{n-k}$.
The following Mathematica snippet does the job: