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Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - Derivation of Descartes' Method: Cartesian Parabola

Gary Rubinstein (Stuyvesant High School)

Descartes did not explain his derivations, writing (Smith 192):

I have also omitted here the demonstration of most of my statements, because they seem to me so easy that if you take the trouble to examine them systematically the demonstrations will present themselves to you and it will be of much more value to you to learn them in that way than by reading them.

And later, as the final sentence of his book, he wrote (Smith 240):

I hope that posterity will judge me kindly, not only as to the things which I have explained, but also as to those which I have intentionally omitted so as to leave to others the pleasure of discovery.

Descartes did not explain how he derived the complex rules that convert the coefficients \(p, q, r, s, t,\) and \(u\) of the given sextic equation into the six parameters \(n, a, b, h, k,\) and \(d\) for defining the Cartesian parabola and the circle. All he did was verify that these complex rules are correct by algebraically finding the intersection of the Cartesian parabola and the circle and showing that it leads to the proper sextic equation.

Before taking on the challenge of rediscovering Descartes’ daunting construction, I searched for previously published analyses of his work. In the College Mathematics Journal (18, 1987), I found a nice derivation of the formula for the Cartesian parabola in an article about Isaac Newton by V. Frederick Rickey. Newton dubbed this curve the ‘trident’ as it resembles the pitchfork wielded by the sea god.

Deriving the formula for the Cartesian parabola

In Figure 13, below, point B is the origin, lengths \({\rm{BA}}=b\) and \({\rm{DE}}=a\) are fixed, and variable lengths \({\rm{CG}}=y\) and \({\rm{CF}}=x\) are coordinates of the point C on both the Cartesian parabola and the ordinary parabola. Therefore, the equation of the (ordinary) parabola is \[y=\frac{x^{2}}{n}+{\rm{BD}}\] (with BD changing as the parabola slides up and down) and \({\rm{GA}}=b-x.\) By similar triangles, \(\frac{b-x}{y}=\frac{x}{\rm{FE}},\) so \({\rm{FE}}=\frac{xy}{b-x}.\) Also, when D coincides with B, FD is the \(y\)-value for each point on the parabola, so \[{\rm{FD}}=\frac{{\rm{FC}}^{2}}{n}=\frac{x^{2}}{n}.\]

Figure 13. Deriving the equation for the Cartesian parabola. Instructions: Move the vertex of the parabola up and down.  The parameters \(n, a,\) and \(b\) can be adjusted with the sliders.

But \[{\rm{FD}}={\rm{DE}}-{\rm{FE}}=a-{\rm{FE}}=a-\frac{xy}{b-x}.\] Equating two expressions for FD, we can say \[\frac{x^{2}}{n}=a-\frac{xy}{b-x}=\frac{ab-ax-xy}{b-x}.\] Clear the denominators to get \[bx^2-x^{3}=nab-nax-nxy.\] Now isolate \(nxy\): \[nxy=nab-nax-bx^{2}+x^{3}\] and solve for \(y\) to get: \[y=\frac{x^{2}}{n}-\frac{bx}{n}-a+\frac{ab}{x}\] as the equation of the Cartesian parabola (Rickey 372).

From the equation of the Cartesian parabola, and knowing already what Descartes’ results were, I next set out to re-derive Descartes’ parameters for his construction – that is, starting with the coefficients of a sextic equation, to come up with the center and radius of the circle and also the two parameters that define the Cartesian parabola so that the \(x\)-intercepts of the intersections of the circle and Cartesian parabola would equal the real roots of the sextic equation.

Gary Rubinstein (Stuyvesant High School), "Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - Derivation of Descartes' Method: Cartesian Parabola," Convergence (March 2016)