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The title really says it all: I want to know how to find a parametric equation for a curve defined by an implicit Cartesian equation.

I would imagine that this is impossible in general, but I'm curious about when it is possible.

I've search for a little while online, but the few resources I've turned up have been old and incomplete. (See this reddit link, for example.)

For context, I was trying to convert the equation $x^4+y^4=1$ into a parametric form. (I am attempting to construct a library of "nice" convex parametric curves.)

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I've reread the reddit post I linked in my question a few times, and I've at least figured out how to construct a parametrization of $x^4+y^4=1$. Recall the trigonometric identity $$\cos^2t=\frac{1+\cos 2t}{2}.$$ We then set $x=\cos t$, so $x^2=\cos^2t$. We know from the original Cartesian equation that $x^2=\sqrt{1-y^4}$, so then $$\sqrt{1-y^4}=\cos^2t=\frac{1+\cos 2t}{2}.$$ Solving this for $y$ yields $$y=\pm\sqrt[4]{1-\left(\frac{1+\cos 2t}{2}\right)^2},$$ so our desired parametrization is $$\left(\cos t,\pm\sqrt[4]{1-\left(\frac{1+\cos 2t}{2}\right)^2}\right).$$ Using this approach, it can also be shown that a parametrization of $x^n+y^n=1$, where $n$ is even, is given by $$\left(\cos t,\pm\sqrt[n]{1-\left(\frac{1+\cos 2t}{2}\right)^{\frac{n}{2}}}\right).$$

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