Roman surface is the image of unit sphere under the map $f(x,y,z)=(yz,xz,xy).$ This gives an implicit formula of $x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}-xyz=0$.
The tangent plane of Roman surface at any point intersects the surface in a quartic curve $\cal C$, how to prove $\cal C$ is reducible over $\mathbb{C}$?
![Block[{x0=54/121,y0=18/121,z0=12/121},ContourPlot3D[{x^2y^2+y^2z^2+z^2x^2-x y z==0,(z-z0) (-x0 y0+2 x0^2 z0+2 y0^2 z0)+(x-x0) (2 x0 y0^2-y0 z0+2 x0 z0^2)+(y-y0) (2 x0^2 y0-x0 z0+2 y0 z0^2)==0},{x,-1/2,1/2},{y,-1/2,1/2},{z,-1/2,1/2},MaxRecursion->4]]](https://i.sstatic.net/lGd1g8A9.png)
![Block[{x0=250/729,y0=50/729,z0=20/729},ContourPlot3D[{x^2y^2+y^2z^2+z^2x^2-x y z==0,(z-z0) (-x0 y0+2 x0^2 z0+2 y0^2 z0)+(x-x0) (2 x0 y0^2-y0 z0+2 x0 z0^2)+(y-y0) (2 x0^2 y0-x0 z0+2 y0 z0^2)==0},{x,-1/2,1/2},{y,-1/2,1/2},{z,-1/2,1/2},MaxRecursion->6]]](https://i.sstatic.net/TDAUfdJj.png)
Similar post: Show that the tangent plane of the saddle surface $z=xy$ at any point intersects the surface in a pair of lines.
My attempt:
The tangent plane at $(x_0,y_0,z_0)$ is (WA) $$(x - x_0) (2 x_0 y_0^2 + 2 x_0 z_0^2 - y_0 z_0) + (y - y_0) (2 x_0^2 y_0 - x_0 z_0 + 2 y_0 z_0^2) + (z - z_0) (2 x_0^2 z_0 - x_0 y_0 + 2 y_0^2 z_0)=0$$ $\cal C$ is the intersection of this plane with Roman surface.
By the way, the formula here gives rational points on the unit sphere, mapped by $f$ to rational points on Roman surface, so that they can be substituted in to check your calculations. Here are some rational points on the unit sphere: $\left(\frac{2}{3},\frac{2}{3},\frac{1}{3}\right),\left(\frac{2}{11},\frac{6}{11},\frac{9}{11}\right),\left(\frac{1}{9},\frac{4}{9},\frac{8}{9}\right),\left(\frac{2}{27},\frac{10}{27},\frac{25}{27}\right)$ mapped by $f$ to rational points on Roman surface: $\left(\frac{2}{9},\frac{2}{9},\frac{4}{9}\right),\left(\frac{54}{121},\frac{18}{121},\frac{12}{121}\right),\left(\frac{32}{81},\frac{8}{81},\frac{4}{81}\right),\left(\frac{250}{729},\frac{50}{729},\frac{20}{729}\right)$.
