# Steps to solve parametric equation of Folium of Descartes

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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 $$y=\frac{3t^2}{1+t^3}$$

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### Sue

Updated on September 10, 2020

• Sue almost 2 years

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?

• Donald Splutterwit almost 5 years
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 $(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2})$.
• Nosrati almost 5 years
@sue with $t=\frac{y}{x}$ obtain $y$.