A projective change of coordinates on $\Bbb P^n_K$ between coordinates $x_0,\cdots,x_n$ and coordinates $y_0,\cdots,y_n$ is a way of writing $y_i = \sum_{j=0}^n c_{ij}x_j$ for $c_{ij}\in K$, such that the matrix $(c_{ij})$ is invertible. If you are thinking of the projective space $\Bbb P^n_K$ as the space of lines in $\Bbb A^{n+1}_K$, this is equivalent to a linear automorphism of $\Bbb A^{n+1}_K$.
There is a reasonably simple way to put every irreducible projective conic in the form $xy+z^2$ over any algebraically closed field $K$ of any characteristic. We'll need one lemma first:
Lemma. Suppose $K$ is an algebraically closed field and $p(x,y)\in K[x,y]$ is a nonzero polynomial which is homogeneous of degree $d$. Then $p(x,y)$ splits completely in to $d$ homogeneous degree-1 factors.
Proof. By writing $p(x,y)=x^ap'(x,y)$ for a polynomial $p'(x,y)\in K[x,y]$ which is not divisible by $x$, we may assume that $p(x,y)$ is not divisible by $x$. Now consider $p(x,y)/x^d = p(1,y/x)$, a nonzero degree-$d$ polynomial in the single variable $y/x$: since $K$ is algebraically closed, this splits completely in to linear factors, or $p(1,y/x)=c\prod_{i=1}^d (y/x - r_i)$. But then $p(x,y)=x^dp(1,y/x)=c\prod(y-r_ix)$, and we've proven the claim. $\blacksquare$
Now let's get down to business. Start with the irreducible conic cut out by the irreducible equation $$a_{00}x^2+a_{01}xy+a_{02}xz+a_{11}y^2+a_{12}yz+a_{22}z^2$$ with $a_{ij}\in K$ and examine $a_{00}x^2+a_{01}xy+a_{11}y^2$: by the lemma, this factors as $L_1L_2$ for two degree-one homogeneous polynomials $L_1,L_2\in K[x,y]$. Now we have two cases: either $L_1$ and $L_2$ are proportional, or they aren't.
If $L_1$ and $L_2$ are proportional, are, then up to scaling our equation by a constant factor, we may assume $L_1=L_2$, and then either $L_1 \mapsto x$, $y\mapsto y$, $z\mapsto z$, or $x\mapsto x$, $L_1\mapsto y$, $z\mapsto z$ is a linear automorphism of the vector space of homogeneous degree-one polynomials. WLOG, suppose it's the first - then our conic becomes $x^2+a_{02}'xz+a_{12}'yz+a_{22}'z^2$. Since $a_{02}'xz+a_{12}'yz+a_{22}'z^2 = z(a_{02}'x+a_{12}'y+a_{22}'z)$ and $x$, $z$, and $a_{02}'x+a_{12}'y+a_{22}'z$ are linearly independent by irreducibility (if not, then our equation is a homogeneous polyonmial of degree 2 in two variables, which is reducible by the lemma), there is a linear automorphism of the vector space of homogeneous degree-one polynomials which sends $x\mapsto x$, $a_{02}'x+a_{12}'y+a_{22}'z\mapsto y$, and $z\mapsto z$, so our conic becomes $x^2+yz$, and permuting variables, we wind up at $xy+z^2$.
If $L_1$ and $L_2$ aren't proportional, then there is a linear automorphism of the vector space of homogeneous degree-one polynomials which sends $L_1 \mapsto x$, $L_2\mapsto y$, and $z\mapsto z$. Using this as our change of coordinates, our equation becomes $xy+a_{02}'xz+a_{12}'yz+a_{22}'z^2$, then the change of coordinates $x\mapsto x-a_{12}'z$, $y\mapsto y-a_{02}'z$, $z\mapsto z$ removes the $a_{02}'$ and $a_{12}'$ terms, leaving us with $xy+a_{22}''z^2$. Since irreducibility is unaffected by coordinate changes, we have $a_{22}''\neq 0$, and therefore since $K$ is algebraically closed we have some $c\in K$ so that $c^2=1/a_{22}''$; the change of coordinates fixing $x$ and $y$ and scaling $z$ by $c$ gives us $xy+z^2$.
If you'd like a different form of your conic, then you can modify and rerun this proof, or transform $xy+z^2$ in to the desired form.
