Equivalence of Definitions of Folium of Descartes
Jump to navigation
Jump to search
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.
$\blacksquare$