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

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

• Try putting the equation in polar form and solving for $r$. Jan 9 at 7:43
• 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? Jan 9 at 8:08
• 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. Jan 9 at 8:14
• 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. Jan 9 at 8:26
• @JyrkiLahtonen So this means, that this lemma cannot be used to solve this problem, right?... Jan 9 at 8:29

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

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 $$y=-x-1$$.