Non-degenerate conic is projectively equivalent to smooth conic

Question

I'm trying to understand the proof of the proposition:

Every non-degenerate conic $C$ in $\mathbb{P}_{\mathbb{C}}^{2}$ is projectively equivalent to the smooth conic $$ C_{0}=\left\{\left[x_{0}, x_{1}, x_{2}\right] \in \mathbb{P}_{\mathbb{C}}^{2} \mid x_{1}^{2}+x_{0} x_{2}=0\right\}. $$

Proof. By a previous result, we may assume that [0,0,1] lies on $C .$ Then $C$ is the zero set of a homogeneous quadratic polynomial of the form $$ Q\left(x_{0}, x_{1}, x_{2}\right)=a x_{0}^{2}+b x_{1}^{2}+c x_{0} x_{1}+d x_{0} x_{2}+e x_{1} x_{2} $$ with $a, b, c, d, e \in \mathbb{C} .$ If $d=e=0$ then $Q$ is a product of two linear factors, contradicting the assumption that $C$ is non-degenerate. Thus $d$ or $e$ is non-zero, which means that [0,0,1] is a smooth point of $C .$ By taking a further projective transformation, we may also assume that the projective tangent line of $C$ at $[0,0,1]$ is defined by $x_{0}=0,$ which means that $e=0$ and $d \neq 0 .$ Then $b \neq 0$ because otherwise, $Q$ would be divisible by $x_{0}$ and $C$ would be degenerate. Multiplying $Q$ with $1 / b,$ we can reduce to the case where $b=1$ This leaves us with $$ Q\left(x_{0}, x_{1}, x_{2}\right)=x_{1}^{2}+x_{0}\left(a x_{0}+c x_{1}+d x_{2}\right) $$ Now we apply the projective transformation $\Phi_{A}$ where $A$ is the invertible matrix $$ \left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & c & d \end{array}\right) $$ This operation transforms $C$ into $\Phi_{A}(C)=C_{0}$.

I had two questions.

1. How do we know that for every non-degenerate conic, we can choose $4$ points in it such that no three points are collinear? (This is relevant in "By a previous result" part)

2. In the above proof, we first start with $Q\left(x_{0}, x_{1}, x_{2}\right)=a x_{0}^{2}+b x_{1}^{2}+b' x_{2}^{2}+c x_{0} x_{1}+d x_{0} x_{2}+e x_{1} x_{2}$, and take projective transformations to make $Q$ pass through $[0,0,1]$. So we let $b'=0$. Taking further projective transformations, we let $e=0$. So my question is, how do we know that the form of $Q$ is 'preserved' through these transformations? To make this question more specific, by the first projective transformation, we can let $b'=0$ thus wiping out the $x_2^2$ from $Q$. By the second projective transformation, we can wipe out the $x_1x_2$ factor, but how do we know that the $x_2^2$ doesn't reappear in the polynomial form of $Q$, as we have taken another projective transformation?

Answer 1

1. Bezout. If three points on your conic were colinear, then that line is an irreducible component of your conic and therefore your conic is degenerate.

2. Any transformation fixing $[0,0,1]$ keeps $b'=0$. So all you need to do is fix $[0,0,1]$ while moving the tangent line to $x_0=0$ - this is easily accomplished by rotating around that point by the rotation that sends the line $dx+ey=0$ to the line $x=0$. The matrix of this projective transformation is exactly the $3\times 3$ matrix $\begin{pmatrix} R & 0 \\ 0 & 1 \end{pmatrix}$ where $R$ is the $2$-d rotation matrix.