Answer
The parametrization is
$\left( {\frac{{a{t^n}}}{{1 + {t^{2n + 1}}}},\frac{{a{t^{n + 1}}}}{{1 + {t^{2n + 1}}}}} \right)$ for $t \ne - 1$.
Work Step by Step
Similar to the method of Exercise 80, we find the intersection of $y=t\ x$ and the curve by substituting $y=t\ x$ into ${x^{2n + 1}} + {y^{2n + 1}} = a{x^n}{y^n}$, which gives
${x^{2n + 1}} + {t^{2n + 1}}{x^{2n + 1}} = a{x^n}\left( {{t^n}{x^n}} \right)$,
${x^{2n + 1}}\left( {1 + {t^{2n + 1}}} \right) - a{t^n}{x^{2n}} = 0$,
${x^{2n}}\left( {x\left( {1 + {t^{2n + 1}}} \right) - a{t^n}} \right) = 0$.
The solutions are $x=0$ or $x = \frac{{a{t^n}}}{{1 + {t^{2n + 1}}}}$ for $t \ne - 1$
Since $y=t x$, for non trivial solution we have $y = \frac{{a{t^{n + 1}}}}{{1 + {t^{2n + 1}}}}$.
Thus, the parametrization is
$\left( {\frac{{a{t^n}}}{{1 + {t^{2n + 1}}}},\frac{{a{t^{n + 1}}}}{{1 + {t^{2n + 1}}}}} \right)$ for $t \ne - 1$.
