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Is it possible to convert the parametric curve defined by $$x = t^3 - 3t \text{ and } y = t^2 - 4$$ to an implicit function?

K. Claesson
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    It is a plane curve. Are you asking if it's possible to express it as $y=f(x)$? – rogerl Feb 23 '19 at 21:49
  • @rogerl No, it does not necesserily have to be expressible as an explicit function. An implicit function would be just fine. I just want to eliminate the $t$ and express the curve only in the varibles $x$ and $y$. – K. Claesson Feb 23 '19 at 21:52

4 Answers4

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$x^2 = t^6 - 6t^4+9t^2$, and $t^2 = y+4$; substituting gives $$x^2 = (y+4)^3 - 6(y+4)^2 + 9(y+4) = (y+4)(y+1)^2.$$

rogerl
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$x/t = y + 1$, hence $x^2 / t^2 = (y + 1)^2$, hence $$ \frac{x^2}{y + 4} = (y + 1)^2 $$ However! This is not completely equivalent form ...

Antoine
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If you want it in a more user friendly cartesian form, you can start with $$\frac xt=t^2-3=y+1$$ So $$t=\frac{x}{y+1}$$ $$\implies y=\frac{x^2}{(y+1)^2}-4$$ $$\implies x^2=(y+4)(y+1)^2$$

David Quinn
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If you mean can $y$ be expressed explicitly in term of $x$ or vice versa no, but this is an attempt: $$y=t^2-4$$ $$t=\pm\sqrt{y+4}$$ $$x=t^3-3t=\pm(y+4)^{3/2}\mp3\sqrt{y+4}$$

Henry Lee
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