parametrization - Maple Help

algcurves

 parametrization
 Find a parametrization for a curve with genus 0

 Calling Sequence parametrization(f, x, y, t)

Parameters

 f - irreducible polynomial in x and y, with genus 0 x, y, t - variables

Description

 • This procedure computes, if it exists, a parametrization of an algebraic curve f. A parametrization is a birational equivalence from a projective line to the given curve f. Such a parametrization exists if and only if the genus is 0 and the curve is irreducible (which can be checked by AIrreduc).
 • The output of the procedure is a list $\left[X\left(t\right),Y\left(t\right)\right]$ of rational functions in t, such that $\left[X\left(t\right),Y\left(t\right)\right]$ is a point on the curve f for every value of t.
 • For a description of the method used see M. van Hoeij, "Rational Parametrizations of Algebraic Curves using a Canonical Divisor", 23, p. 209-227, JSC 1997.

Examples

 > $\mathrm{with}\left(\mathrm{algcurves}\right):$
 > $f≔{y}^{5}+2x{y}^{2}+2x{y}^{3}+{x}^{2}y-4{x}^{3}y+2{x}^{5}:$
 > $v≔\mathrm{parametrization}\left(f,x,y,t\right)$
 ${v}{≔}\left[\frac{{-}{24192}{}{{t}}^{{5}}{-}{6048}{}{{t}}^{{4}}{+}{2520}{}{{t}}^{{3}}{-}{238}{}{{t}}^{{2}}{+}{7}{}{t}}{{181604}{}{{t}}^{{5}}{-}{103680}{}{{t}}^{{4}}{+}{17280}{}{{t}}^{{3}}{-}{1440}{}{{t}}^{{2}}{+}{60}{}{t}{-}{1}}{,}\frac{{16464}{}{{t}}^{{5}}{+}{6860}{}{{t}}^{{4}}{-}{686}{}{{t}}^{{3}}}{{181604}{}{{t}}^{{5}}{-}{103680}{}{{t}}^{{4}}{+}{17280}{}{{t}}^{{3}}{-}{1440}{}{{t}}^{{2}}{+}{60}{}{t}{-}{1}}\right]$ (1)

Now subs(t=any number,v) should be a point on the curve. Test the result (this should be 0):

 > $\mathrm{normal}\left(\mathrm{subs}\left(x=v\left[1\right],y=v\left[2\right],f\right)\right)$
 ${0}$ (2)
 > $\mathrm{parametrization}\left({x}^{4}+{y}^{4}+a{x}^{2}{y}^{2}+b{y}^{3},x,y,t\right)$
 $\left[{-}\frac{{b}{}{{t}}^{{3}}}{{{t}}^{{4}}{+}{a}{}{{t}}^{{2}}{+}{1}}{,}{-}\frac{{{t}}^{{4}}{}{b}}{{{t}}^{{4}}{+}{a}{}{{t}}^{{2}}{+}{1}}\right]$ (3)