Let A,B,C be three points on the folium of Descartes

Let $A,B,C$ be three points on the folium of Descartes ($a=1$), $x=\dfrac{3t}{1+t^{3}},\ y=\dfrac{3t^{2}}{1+t^{3}}$, that lie on a single line. The tangent at each of $A,B,C$ meets the curve again at a point $S_{1},S_{2},S_{3}$ respectively. Prove that $S_{1},S_{2},S_{3}$ are collinear.

Step 1: When are three points of the folium collinear?

A line $px+qy=r$ meets the folium where $p\cdot\dfrac{3t}{1+t^3}+q\cdot\dfrac{3t^2}{1+t^3}=r$; clearing the denominator, \[rt^{3} - 3qt^{2} - 3pt + r = 0. \tag{1}\]

By Vieta's formula for the product of the three roots $t_1,t_2,t_3$ of (1) (constant term over leading coefficient, with a sign): \[t_{1}t_{2}t_{3} = -\frac{r}{r} = -1. \tag{2}\]

Step 2: Where the tangent at $t_0$ meets the folium again

The tangent at $t_0$ is a line through the point at $t_0$ with the curve's own slope there, so in the cubic (1) specialized to that line, $t=t_0$ is automatically a double root; write the third root as $s$. By the same Vieta relation applied to this particular cubic,

\[ t_{0}\cdot t_{0}\cdot s = -1 \;\Longrightarrow\; s(t_0) = -\frac{1}{t_{0}^{2}}. \tag{3} \]

Step 3: Collinearity of $S_1,S_2,S_3$

Let $A,B,C$ have parameters $t_1,t_2,t_3$ with $t_1t_2t_3=-1$ by (2). Their tangents meet the folium again at $s_1=-1/t_1^2,\ s_2=-1/t_2^2,\ s_3=-1/t_3^2$. Then

\[ s_{1}s_{2}s_{3} = \frac{(-1)^{3}}{(t_{1}t_{2}t_{3})^{2}} = \frac{-1}{(-1)^{2}} = -1. \tag{4} \]

By the collinearity criterion (2) applied in reverse, $s_1s_2s_3=-1$ is exactly the condition for the three points with parameters $s_1,s_2,s_3$ to lie on a common line. Hence $S_1,S_2,S_3$ are collinear. $\blacksquare$

Step 4: Verification

\[t_1=0.7,\ t_2=-2.3,\ t_3=-1/(t_1t_2)\approx0.62112\text{: the three points are collinear (area residual} \sim10^{-16})\ \checkmark\]

\[\text{Tangent at } t_0\text{, second intersection at } s=-1/t_0^2\text{, checked for } t_0=0.5,-1.3,2.1\text{ (residual} \sim10^{-11})\ \checkmark\]