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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 11$: Special Plane Curves: Folium of Descartes: $11.27$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): folium of Descartes
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): folium of Descartes
- Weisstein, Eric W. "Folium of Descartes." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/FoliumofDescartes.html