# Folium of Descartes - what is this point P?

I came across this example in an old book.

I have this question here. What is this point P, how is it defined? I did some calculations (implicit differentiation) and it seems to me it's the point where $$x = \sqrt[3]{4}, y = \sqrt[6]{4}$$. Why? Because it seems to me that's when the derivative $$\frac{dy}{dx}$$ is not defined. Is that correct?

• sagecell.sagemath.org/… Apr 21 at 12:46
• Another good question can be to calculate the area bounded by this curve....it's a very nice question Apr 21 at 13:57

Yes, that is the correct idea. At the point $$P$$, the derivative $$dy/dx$$ is undefined; or equivalently, the derivative $$dx/dy = 0$$. The function $$f(x,y)$$ satisfies reflection symmetry about $$y = x$$; i.e., $$f(y,x) = f(x,y)$$ for all $$(x,y)$$. Thus the reflection of $$P$$ about $$y = x$$ gives another point $$P'$$ whose derivative $$dy/dx = 0$$.
In any case, either point can be located via implicit differentiation. To locate $$P$$, we differentiate $$f = 0$$ implicitly with respect to $$y$$: $$0 = \frac{df}{dy} = \frac{d}{dy}\left[x^3 + y^3 - 3xy\right] = 3x^2 \frac{dx}{dy} + 3y^2 - 3y \frac{dx}{dy} - 3x,$$ hence $$\frac{dx}{dy} = \frac{x-y^2}{x^2-y}.$$ Then $$P$$ satisfies $$dx/dy = 0$$, i.e. $$x = y^2$$ under the condition $$x^2 \ne y$$ (since if the denominator also equals zero, then $$dx/dy$$ is indeterminate, not zero). So we know $$P$$ lies on a parabola. The intersection of this parabola with $$f = 0$$ is obtained by substitution: $$0 = f(x,y) = f(y^2,y) = (y^2)^3 + y^3 - 3(y^2)y = y^6 - 2y^3 = y^3(y^3 - 2),$$ hence $$y \in \{0, 2^{1/3}\}$$ and we have the candidate solutions $$(x,y) \in \{(0,0), (2^{2/3}, 2^{1/3})\}.$$ But as stated before, we also require $$x^2 \ne y$$, which excludes the candidate $$(0,0)$$. So the unique solution is $$P = (2^{2/3}, 2^{1/3})$$. We also have by symmetry $$P' = (2^{1/3}, 2^{2/3})$$, the unique nontrivial horizontal tangent to $$f = 0$$.
As a further exercise, can you locate the points $$Q$$, $$Q'$$ such that the tangent to $$f = 0$$ is parallel to the line $$y = x$$? What does this say about the "diagonal width" of the loop?