Find the asymptotes, if any of the Folium of Descartes: $x^3+y^3=3xy$.

We are only taking rectilinear asymptotes into consideration

My solution goes as follows:

We know that, in order to find parallel asymptotes i.e horizontal and vertical asymptotes, from an equation of a curve, we equate, the real linear factors of the coefficient of the highes power of $x$ and $y$ in the given equation. If the coefficients of the highest power of $x$ and $y$ have is a constant or not factorizable into real linear factors, then it has no horizontal and vertical asymptotes respectively. In the equation of $f(x)=x^3+y^3+3xy$, both the coefficients of the highes power of $x$ and $y$ are constants and hence, no parallel asymptotes are possible.

However, I am not been able to calcklate the oblique asymptotes using the following lemma stated as follows:

$y=mx+c$, is an oblique asymptote of $y=f(x)$, iff $\exists $ a finite $m=\lim_{|x|\to\infty}\frac{y}{x}$ and $c=\lim_{|x|\to\infty} y-mx$.

Now, I couldn't evaluate $m$, or rather, obtain $\frac{y}{x}$ as a function of $x$, in order to evaluate the limit $m=\lim_{|x|\to\infty}\frac{y}{x}$ and, $c=\lim_{|x|\to\infty} y-mx$. Is there any way, we can actually, find the oblique asymptotes using the above mentioned lemma stated above? Is it at all possible?

I know that there are numerous posts on this site relating to the same topic. But, I want to calculate the oblique asymptotes only using the above lemma

  • $\begingroup$ Try putting the equation in polar form and solving for $r$. $\endgroup$ Jan 9 at 7:43
  • 2
    $\begingroup$ Divide the equation by $x^3$. You get $$1+\left(\frac y x\right)^3=\frac{3y}x\cdot\frac1{x}.$$ If $|x|\to\infty$ and $y/x\to m$, what do you get? $\endgroup$ Jan 9 at 8:08
  • $\begingroup$ Observe that because $y$ is only an implicit function of $x$, you cannot use any method relying on having an explicit function $f$ such that $y=f(x)$. "Technically" the lemma cannot be applied, as it is written, at all. $\endgroup$ Jan 9 at 8:14
  • $\begingroup$ Some useful references in general can be found in my comments to Why this method gives an asymptote while being like nonsense, is this a coincidence or some logic below is not discovered. $\endgroup$ Jan 9 at 8:26
  • $\begingroup$ @JyrkiLahtonen So this means, that this lemma cannot be used to solve this problem, right?... $\endgroup$
    – Arthur
    Jan 9 at 8:29

1 Answer 1


\begin{eqnarray} x^3+y^3&=&3xy\\ x\left(1+\left(\frac{y}{x}\right)^3\right)&=&3\cdot\frac{y}{x}\\ x(1+u^3)&=&3u\\ x&=&\frac{3u}{1+u^3}\\ &=&\frac{3u}{(1+u)(u^2-u+1)}\\ &=&\frac{3}{(1+u)(u-1+\frac{1}{u})}\tag{1}\\ |x|&=&\frac{3}{\left|(1+u)(u-1+\frac{1}{u})\right|} \end{eqnarray}

As $|x|\to\infty,\quad(1+u)(u-1+\frac{1}{u})\to0$

But for all real values of $u$, either $(u-1+\frac{1}{u})\ge1$ or $(u-1+\frac{1}{u})\le-3$. So it must be the case that as $|x|\to\infty,\quad u\to-1$. So $m=-1$ and $y-mx=x+y$.

From equation (1) we get

$$x(1+u)= x+y=\frac{3}{u-1+\frac{1}{u}}$$

So as $|x|\to\infty,\quad u\to-1$ and $x+y\to-1$. Thus $c=-1$

So the equation of the asymptote is $y=-x-1$.


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