Let $K$ be a field with characteristic not equal to $2$. We define a conic $C$ to be a subvariety of $\mathbb{P}^2_K$ defined by a homogeneous equation of degree $2$. I need to show that the equation of $C$ can be written as $$ax^2+by^2+cz^2=0$$ for some $a,b,c\in K $ and some basis $(x,y,z)$. I have two different proof ideas:
Idea 1. I have seen many places (see for example "Elementary Geometry of Algebraic Curves" by Gibson, Chapter 5.2) that the equation of the conic has the general form $$ ax^2+2hxy+by^2+2gx+2fy+c = 0 $$ when viewed as an affine variety in the affine plane (not projective plane). I guess I'll need to show that this is true myself, as this is not our definition of a conic. Furthermore, the equation can be represented in terms of the matrix $$ A= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}. $$ If we write $v=(x,y,1), $ then the equation can be written as $$vAv^T = 0.$$ Now if one can show that $A$ is diagonalizable, then we will have shown that such a basis exists. But I don't know any general result that we can use to say this. Maybe one could use the characteristic polynomial of $A$ and deduce that this polynomial has $3$ distinct roots (and hence $A$ has $3$ distinct eigenvalues, and is hence diagonizable). But this approach doesn't seem very straightforward, as our matrix $A$ is very general.
Idea 2. For the second proof idea I wanted to use the fact that there is a one-to-one correspondence between cones in $\mathbb{A}^{n+1}$ and projective varieties in $\mathbb{P}^n$.
A projective conic $C\subset \mathbb{P}^2$ can be written as the zero locus of a single homogeneous polynomial $f\in K[x,y,z]$ of degree $2$, i.e. $C=V_p(f)$.
We have from the course notes Gathmann 2014 Example 7.6.(d) that a projective conic is isomorphic to $V_p(xz-y^2)$ or $V_p(yz-x^2)$, and every projective conic is isomorphic to $\mathbb{P}^1$. The affine part is isomorphic to $V_a(y-x^2)$ or $V_a(xy-1)$.
I was thinking that if I can show that $$K[x,y]/(y-x^2)\cong K[x,y]/(x^2+y^2+1)\cong K[x,y]/(xy-1),$$ then we can use the one-to-one correspondence and get $\mathbb{P}^1 \cong C\cong V_p(x^2+y^2+z^2)$.
But I'm not sure about this strategy either, or how one would establish such an isomorphism.
Idea 1 should work with a couple of modifications: the equation you have is the affine part in the chart $z = 1$, to get the projective conic you should take the homogenization $$ax^2 + by^2 + cz^2 + 2hxy + 2gxz + 2fyz = uAu^T$$ for $u = (x, y, z)$ and the same $A$. Second, you don't want diagonalizable in the sense of $PAP^{-1}$ being diagonal, but $PAP^T$ being diagonal for some invertible $P$ (and then the required basis is given by $uP$). In fact for any symmetric $n \times n$ matrix $A$, away from characteristic $2$ you can always find a $P$ such that $PAP^T$ is diagonal: find a vector $x$ such that $x^T A x \ne 0$ (otherwise $x^T A y = \frac12((x + y)^T A (x + y) - x^T A x - y^T A y) = 0$ for all $x, y$, so $A = 0$) and inductively "diagonalize" $A$ restricted to $\{y \mid x^T A y = 0\}$.
Since the question is about $n = 3$, which is pretty small, you can also just directly do the algebra on the expression $uAu^T$, this is a form of completing the square, as mentioned by @ancientmathematician.