Find the asymptote of the folium of Descartes

Find the asymptote of the folium of Descartes $x=\dfrac{3at}{1+t^3}$, $y=\dfrac{3at^2}{1+t^3}$.

Step 1: Locate the singular parameter value

The denominator $1+t^3$ vanishes at $t=-1$; near this value both $x$ and $y$ become infinite, so $t=-1$ generates the (only) asymptote.

Step 2: Slope

\[ k = \lim_{t\to-1}\frac{y}{x} = \lim_{t\to-1}\frac{3at^2/(1+t^3)}{3at/(1+t^3)} = \lim_{t\to-1} t = -1. \tag{1} \]

Step 3: Intercept

Since the slope is $-1$, form $x+y$ and let $t\to-1$: \[x+y = \frac{3at(1+t)}{1+t^3} = \frac{3at(1+t)}{(1+t)(1-t+t^2)} = \frac{3at}{1-t+t^2}. \tag{2}\]

\[ \lim_{t\to-1}(x+y) = \frac{-3a}{1+1+1} = -a. \tag{3} \]

Step 4: Verification

\[a=1,\ t=-0.999\text{: }x\approx-333.7,\ y\approx333.4,\ x+y+1\approx-0.00033\to0\ \checkmark\]