Folium of Descartes

Graphing the curve

The folium of Descartes can be expressed in polar coordinates as$${\displaystyle r=\frac{3a\sin \theta \cos \theta }{{\sin }^3\theta +{\cos }^3\theta },}$$which is plotted on the left. This is equivalent to

${\displaystyle r=\frac{3a\sec \theta \tan \theta }{1+{\tan }^3\theta }.}$

Another technique is to write ${\displaystyle y=px}$ and solve for ${\displaystyle x}$ and ${\displaystyle y}$ in terms of ${\displaystyle p}$. This yields the rational parametric equations:

${\displaystyle x=\frac{3ap}{1+p^3},y=\frac{3a{p}^2}{1+p^3}.}$

We can see that the parameter is related to the position on the curve as follows:

• ${\displaystyle p<-1}$ corresponds to ${\displaystyle x>0}$, ${\displaystyle y<0}$: the right, lower, "wing".
• ${\displaystyle -1<p<0}$ corresponds to ${\displaystyle x<0}$, ${\displaystyle y>0}$: the left, upper "wing".
• ${\displaystyle p>0}$ corresponds to ${\displaystyle x>0}$, ${\displaystyle y>0}$: the loop of the curve.

Another way of plotting the function can be derived from symmetry over ${\displaystyle y=x}$. The symmetry can be seen directly from its equation (x and y can be interchanged). By applying rotation of 45° clockwise for example, one can plot the function symmetric over rotated x axis.

This operation is equivalent to a substitution:$${\displaystyle x=\frac{u+v}{\sqrt{2}},y=\frac{u-v}{\sqrt{2}}}$$and yields$${\displaystyle v=\pm u\sqrt{\frac{3a\sqrt{2}-2u}{6u+3a\sqrt{2}}},u<3a/\sqrt{2}.}$$Plotting in the Cartesian system of ${\displaystyle (u,v)}$ gives the folium rotated by 45° and therefore symmetric by ${\displaystyle u}$-axis.

Properties

It forms a loop in the first quadrant with a double point at the origin and has asymptote$${\displaystyle x+y=-a.}$$It is symmetrical about the line ${\displaystyle y=x}$. As such, the curve and this line intersect at the origin and at the point $(3a/2,3a/2).$

Implicit differentiation gives the formula for the slope of the tangent line to this curve to be

${\displaystyle \frac{dy}{dx}=\frac{ay-x^2}{y^2-ax},}$

with poles $x=y^2/a$ and value 0 or ${\displaystyle \pm \mathrm{\infty }}$ at the origin.

Using either one of the polar representations above, the area of the interior of the loop is found to be $3a^2/2$. Moreover, the area between the "wings" of the curve and its slanted asymptote is also $3a^2/2.$

Relationship to the trisectrix of Maclaurin

The folium of Descartes is related to the trisectrix of Maclaurin by affine transformation. To see this, start with the equation$${\displaystyle x^3+y^3=3a\cdot xy,}$$and change variables to find the equation in a coordinate system rotated 45 degrees. This amounts to setting

$${\displaystyle x=\frac{X+Y}{\sqrt{2}},y=\frac{X-Y}{\sqrt{2}}.}$$In the ${\displaystyle X,Y}$ plane the equation is$${\displaystyle 2X(X^2+3Y^2)=3\sqrt{2}a(X^2-Y^2).}$$

If we stretch the curve in the ${\displaystyle Y}$ direction by a factor of ${\displaystyle \sqrt{3}}$ this becomes$${\displaystyle 2X(X^2+Y^2)=a\sqrt{2}(3X^2-Y^2),}$$which is the equation of the trisectrix of Maclaurin.