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Folium of Descartes

In Geometry, the Folium of Descartes is an algebraic curve defined by the equation

$x^3 + y^3 - 3 a x y = 0 ,$.
It forms a loop in the first quadrant with a double point at the origin and asymptote
$x + y + a = 0 ,$.
It is symmetrical about $y = x$.

Then name comes from the Latin word folium which means "leaf".

The curve was featured, along with a portrait of Descartes, on an Albanian stamp in 1966.

History

The curve was first proposed by Descartes in 1638. Its claim to fame lies in an incident in the development of calculus. Descartes challenged Fermat to find the tangent line to the curve at an arbitrary point since Fermat had recently discovered a method for finding tangent lines. Fermat solved the problem easily, something the Descartes was unable to do. Since the invention of calculus, the slope of the tangent line can be found easily using implicit differentiation.

Graphing the curve

Since the equation is degree 3 in both x and y, and does not factor, it is difficult to find solve for one of the variables. However, the equation in polar coordinates is:
$r = frac\left\{3 a sin theta cos theta\right\}\left\{sin^3 theta + cos^3 theta \right\}.$
which can be plotted easily. Another technique is to write y = px and solve for x and y in terms of p. This yields the parametric equations:

$x = \left\{\left\{3ap\right\} over \left\{1 + p^3\right\}\right\},, y = \left\{\left\{3ap^2\right\} over \left\{1 + p^3\right\}\right\}$.

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
$x^3 + y^3 = 3 a x y ,$,
and change variables to find the equation in a coordinate system rotated 45 degrees. This amounts to setting $x = \left\{\left\{X+Y\right\} over sqrt\left\{2\right\}\right\}, y = \left\{\left\{X-Y\right\} over sqrt\left\{2\right\}\right\}$. In the $X,Y$ plane the equation is
$2X\left(X^2 + 3Y^2\right) = 3 sqrt\left\{2\right\}a\left(X^2-Y^2\right)$.
If we stretch the curve in the $Y$ direction by a factor of $sqrt\left\{3\right\}$ this becomes
$2X\left(X^2 + Y^2\right) = a sqrt\left\{2\right\}\left(3X^2-Y^2\right)$
which is the equation of the trisectrix of Maclaurin.

References

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