Asymptote to Folium of Descartes

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Theorem

Consider the folium of Descartes $F$, given in parametric form as:

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


The straight line whose equation is given by:

$x + y + a = 0$

is an asymptote to $F$.


Proof

First we note that from Behaviour of Parametric Equations for Folium of Descartes according to Parameter:

when $t = 0$ we have that $x = y = 0$
when $t \to \pm \infty$ we have that $x \to 0$ and $y \to 0$
when $t \to -1^+$ we have that $1 + t^3 \to 0+$, and so:
$x \to -\infty$
$y \to +\infty$
when $t \to -1^-$ we have that $1 + t^3 \to 0-$, and so:
$x \to +\infty$
$y \to -\infty$


We have that:

\(\ds x + y\) \(=\) \(\ds \dfrac {3 a t} {1 + t^3} + \dfrac {3 a t^2} {1 + t^3}\)
\(\ds \) \(=\) \(\ds \dfrac {3 a t \paren {1 + t} } {\paren {1 + t} \paren {1 - t + t^2} }\) Sum of Two Cubes
\(\ds \) \(=\) \(\ds \dfrac {3 a t} {1 - t + t^2}\) simplifying


So setting $t = -1$:

\(\ds x + y\) \(=\) \(\ds \dfrac {3 a \times \paren {-1} } {1 - \paren {-1} + \paren {-1}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {-3 a} {1 + 1 + 1}\)
\(\ds \) \(=\) \(\ds -a\)
\(\ds \leadsto \ \ \) \(\ds x + y + a\) \(=\) \(\ds 0\)

The result follows.

$\blacksquare$


Sources

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