Equation for the projection of a cubic surface from a point

Question

Suppose $f \in k[x,y,z,w]$ gives a degree 3 curve $Z(f)$ in $\mathbb{P}^3$ with $Q = [0 : 0 : 0 : 1]$ in $Z(f)$, and we project points in $Z(f) \setminus Q$ to the plane $\{ w = 0\} \subset \mathbb{P}^3$ by sending a point $P$ to the intersection $QP \cap \{ w=0\}$ where $QP$ is the line through $Q$ and $P$, is there a way to calculate an equation for the image in terms of $x,y,z$ such that the image is the zero locus of this equation? My objective is to understand why the image should be a degree 4 curve.

Answer 1

After a change of coordinates you can write the equation of the surface as $$f(x,y,z,u)=u^{2}\cdot L+ u\cdot Q+ C=0,$$ where $L(x,y,z), Q(x,y,z), C(x,y,z)$ are homogenous polynomials of degree $1,2,3,$ respectively. The branch locus is given by $$Q^2-4 LC=0.$$