I know that the parametric equations of the Folium of Descartes are $\displaystyle x = \frac{3t}{(1+t)^3} $and $\displaystyle y = \frac{3t^2}{(1+t)^3}$. What are the steps to achieve this parametric equation from the given equation $x^3 + y^3 = 3xy$, given that $\displaystyle t= \frac yx$? How do I substitute the t value to get this parametric equation?
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You do appreciate that there is something arbitary about a prarmeterisation. Eg $x^2+y^2=1$ could be parameterised by $( \cos \theta, \sin \theta)$ or $(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2})$. – Donald Splutterwit Sep 09 '17 at 13:03
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@sue with $t=\frac{y}{x}$ obtain $y$. – Nosrati Sep 09 '17 at 13:07
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with $y=tx$ we get $$x^3+t^3x^3=3x^2t$$ for $x\neq 0$ we get $$x(1+t^3)=3t$$ therefore $$x=\frac{3t}{1+t^3}$$ from here we get easy $$y=\frac{3t^2}{1+t^3}$$
Dr. Sonnhard Graubner
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Substituting $t=\frac{y}{x}$ into the first equation, we get: \begin{align*} x=\frac{3t}{1+t^3}&\implies x=\frac{3\left(\frac{y}{x}\right)}{1+\frac{y^3}{x^3}}\\ &\implies x=\frac{3x^2y}{x^3+y^3}\\ &\implies x^3+y^3=3xy\\ \end{align*}
Roman Chokler
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