Equivalence of Definitions of Folium of Descartes

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Theorem

The following definitions of the concept of Folium of Descartes are equivalent:

Cartesian Form

The folium of Descartes is the locus of the equation:

$x^3 + y^3 - 3 a x y = 0$

Parametric Form

The folium of Descartes is the locus of the equation given in parametric form as:

$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$


Proof

Let $F$ be a folium of Descartes expressed in parametric form.

Then by definition:

The folium of Descartes is the locus of the equation given in parametric form as:

$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$


Hence:

\(\ds x^3 + y^3\) \(=\) \(\ds \dfrac {\paren {3 a t}^3} {\paren {1 + t^3}^3} + \dfrac {\paren {3 a t^2}^3} {\paren {1 + t^3}^3}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {3 a t}^3 \paren {1 + t^3} } {\paren {1 + t^3}^3}\)
\(\ds \) \(=\) \(\ds \dfrac {3 a \paren {3 a t} \paren {3 a t^2} } {\paren {1 + t^3}^2}\)
\(\ds \) \(=\) \(\ds 3 a \dfrac {3 a t} {1 + t^3} \dfrac {3 a t^2} {1 + t^3}\)
\(\ds \) \(=\) \(\ds 3 a x y\)

Thus $F$ is a folium of Descartes expressed in Cartesian form.

$\blacksquare$