I'm trying to find the asymptotes of the Folium of Descartes, which has the equation $$x^3+y^3-3xy=0$$
I was also told to find the curve length in the first quadrant, and to do so I parametrized it by finding the intersection between the curve and the line $y=tx$. The parametric equation is $$(x, y) = (\frac{3t}{1+t^3},\frac{3t^2}{1+t^3})$$
There is an invalid $t$ value, which is $t=-1$, which looks to be about the slope of the asymptotes on the graph. However, I'm not sure how to obtain this value with calculus instead of guessing. I have one way, but it feels really hacky. Differentiating implicitly, $$\frac{dy}{dx} = \frac{x^2-y}{x-y^2}$$
Assuming that $x$ and $y$ approach $\infty$ at the same rate, the equation simplifies to $-1$.
$$\frac{dy}{dx} = \frac{x^2-y}{x-y^2} \implies \lim_{x,y\to\infty}\frac{dy}{dx} \approx \frac{x}{-y}=-1$$
But how can I be sure that that assumption is correct? Am I allowed to reason that, since the equation of the folium appears symmetrical for both $y$ and $x$ (i.e. if I replace $x$ with $y$ and vice versa I get the same equation), $x$ and $y$ approach $\infty$ at the same speed? This assumption also feels gimmicky—is there a better way?
Even after I find the slope, I'm not sure how to find the line, in form $y=mx+b$.
Best Answer
I notice that in your parametrization $$ x+y+1 = \frac{3t}{1+t^3} + \frac{3t^2}{1+t^3}+1 = \frac{(1+t)^2}{1-t+t^2}$$
so $x+y+1=0$ is the asymptote.