parametric equations of folium of Descartes We know the function of the folium of Descartes is $x^3+y^3=3axy$. The problem is to show that the folium of Descartes has parametric equations $x=\frac{3at}{1+t^3}$, $y=\frac{3at^2}{1+t^3}$ (this part is easy and I got the solution) and use these equations to show that it is tangential to the axes at $0$. what makes me confused is what the second part means? Tangential to the axes at $0$? Does at $0$ mean at $(0,0)$?
Thanks in advance.
 A: Let $a\neq 0$; to simplify the description, let us consider the subset $\mathcal F_a$ of the $folium$ of Descartes 
$$\mathcal F_a:=\operatorname{Im} \gamma_a\subset \mathbb R^2,$$
denoting by $\gamma_a$ the smooth curve
$$\gamma_a: t\in(-1,\infty)\rightarrow \mathcal F$$
defined $via$
$$\gamma_a(t)=\left( \frac{3at}{1+t^3}, \frac{3at^2}{1+t^3}\right). $$
Having a look at the full $folium$ in here, the reduced $folium$ $\mathcal F_a$ corresponds to the left wing ($-1<t<0$) and the central loop ($t\geq 0$) of the full $folium$ of Descartes.


*

*On the parameter $t$


We defined the subset $\mathcal F_a$ of the $folium$ of Descartes avoiding the singularity at $t=-1$. If we consider the full $folium$ of Descartes, i.e. we study $\gamma_a(t)$ for all $t\in\mathbb R$, we are just adding a "branch" to the reduced $folium$ $\mathcal F_a$. This additional branch is the right wing in the full $folium$ of Descartes, as shown in the above link.
As we want to study the behaviour at $(0,0)$, it will follow from the discussion below that the analysis of the reduced $folium$ is sufficient to arrive at the thesis.


*

*On the tangent vector at $(0,0)$: self intersections.


We are interested in showing that $\gamma_a$ is tangential at the $x$-axis  and $y$-axis in $(0,0)$.
As none vector $v\neq (0,0)$ in $\mathbb R^2$ is both tangent to the $x$ and $y$-axis, we expect $(0,0)$ to be a self intersection of the reduced $folium$ $\mathcal F_a$. We need to find those $t'\in(,)$ s.t.
$$\gamma_a(t')=(0,0).~ (*)$$
We solve $(*)$, arriving at the unique finite solution $t=0$. On the other hand,
$$\gamma_a(+\infty):=\lim_{t\rightarrow +\infty}\gamma_a(t)=(0,0);$$
in summary both $t=0$ and $t=\infty$ satisfy $(*)$, and $(0,0)$ is a self intersection (double point) of the reduced $folium$ $\mathcal F_a$.


*

*Computation of the tangent vector at $(0,0)$


By definition of tangent vector to a curve
$$\gamma_a'(t):=\left(\frac{d}{dt}\left(\frac{3at}{1+t^3}\right) ,\frac{d}{dt}\left(\frac{3at^2}{1+t^3}\right) \right)=\left(\frac{3a}{1+t^3}-\frac{9at^3}{(1+t^3)^2} ,\frac{6at}{1+t^3} 
-\frac{9at^4}{(1+t^3)^2}\right)=
\left(\frac{3a-6at^3}{(1+t^3)^2} ,\frac{6at-3at^4}{(1+t^3)^2}\right).
$$
At $\gamma_a(0)=(0,0)$ we have
$$\gamma_a'(0) =\left(3a,0\right),$$
i.e. $\gamma_a'(0)$ is parallel to the $x$-axis. 
Note that
$$\gamma_a'(+\infty) :=\lim_{t\rightarrow+\infty}\gamma_a'(t)=(0,0);$$
in other words, the direct computation of $\gamma_a'(+\infty)$ is not useful: we need to find another way to estimate the tangent vector $\gamma_a'(+\infty)$.
The trick is to observe that, for all $t\in(-1,\infty)$ we can write
$$t=\frac{\frac{3at^2}{1+t^3}}{\frac{3at}{1+t^3}}=\frac{y_a(t)}{x_a(t)},$$
denoting by $x_a(t)$ und $y_a(t)$ the components of $\gamma_a(t)$ at $t$.
Geometrically $t$ is the slope of the line passing through $(0,0)$ and the point $(x_a(t),y_a(t))$ on the reduced $folium$. The choice $t=0$ gives zero slope, which is compatible with our previous computation of $\gamma'_a(0)$. Letting $t\rightarrow +\infty$ one gets infinite slope in $(*)$ at the point $(0,0)$, i.e. a tangent vector parallel to the $y$-axis. We are done.
