In Frances Kirwan's book Complex Algebraic Curves she has this as Corollary 3.12: Any irreducible projective conic $C$ in $\mathbb{CP}^2$ is equivalent under a projective transformation to the conic $x^2=yz$.
She begins the proof by saying that applying a suitable projective transformation we may assume that $[0,1,0]$ is a nonsingular point of $C$. This part I do understand because given any nonsingular point on $C$ there does exist a projective transformation that sends it to $[0,1,0]$. This is clear, for example using the General Position Theorem.
However, after that she says that we can also assume that the tangent line to $C$ at $[0,1,0]$ is $z=0$. This is the part which intuitively makes sense (you can kind of move any line to any other line and still fixing $[0,1,0]$). However, my issue is that I don't know how one would justify this properly. How would you show that there does exist such a projective linear transformation?
Thank you!
In affine coordinates $(x, z)$, the tangent line to $C$ at $[0: 1: 0]$ has equation $ax + bz = 0$ for some $a$ and $b$ not both $0$, so you just need an invertible linear transformation of the $(x, z)$ plane mapping this line to $z = 0$.