Is x+y = -1 an oblique asymptote for the folium of Descartes curve?

[ProgMetal]

Homework Statement

the curve x^3+y^3=3xy is called the folium of descartes.

Show that x+y=3/((x/Y)-1+(y/x)), and hence that x+y = -1 is an oblique asymptote.

Homework Equations

The Attempt at a Solution

I have done the first part of the question, though i am just having trouble showing that it is an oblique asymptote. I am trying to get just x on the denominator so that as x approaches infinity, the fraction will approach 0 and rearrange from there. No luck so far.

 

[nate808]

To show that x+y = -1 is an oblique asymptote, we can use the definition of an oblique asymptote: as x approaches infinity, the difference between the curve and the asymptote approaches 0.

We can rewrite the equation x^3 + y^3 = 3xy as x^3 - 3xy + y^3 = 0. This can be factored to (x+y)(x^2 - xy + y^2) = 0.

Since we are looking at the behavior of the curve as x approaches infinity, we can ignore the second factor and focus on the first one: x+y = 0.

Substituting this into our equation from the first part, x+y = 3/((x/y)-1+(y/x)), we get 0 = 3/(-1+(-1)) = 3/(-2) = -3/2. This shows that as x approaches infinity, the difference between the curve and the asymptote (x+y = -1) approaches 0, making it an oblique asymptote.

Therefore, we can conclude that x+y = -1 is an oblique asymptote for the curve x^3+y^3=3xy, also known as the folium of Descartes.