Fields of characteristics and affine conics Let $k$ be any field of characteristic not equal to $2$, and $V$ a $3$-dimensional $k$-vector space. Let $Q: V \to k$, be a non degenerate quadratic form on $V$. How can we show that if $\,0\neq e_1\in V\,$ satisfies $Q(e_1) = 0$, then $V$ has a basis $e_1,e_2,e_3$ such that $Q(x_1 e_1 + x_2 e_2 + x_3 e_3) = x_1 x_3 + ax_2^2$.
And how to deduce that a nonempty nondenerate conic $C \subseteq \mathbb{P}_k^2$, is projectively equivalent to $XZ = Y^2$?
What I am able to note is I need to work the symmetric bilinear form $\phi$ associated to $Q$; and since $\phi$ is non-degenerate, there is a vector $e_3$ such that $\phi(e_1,e_3) = 1$.
 A: You are off to a good start with considering $\phi$ and noting that you can choose $e_3$ so that $\phi(e_1, e_3) = 1$.
Now let $c = \phi(e_3, e_3)$, which may or may not be $0$.  But we can replace $e_3$ by $e'_3 = \frac{c}{2} e_1 - e_3$ and compute that $\phi(e'_3, e'_3) = 
\frac{c^2}{4}\phi(e_1,e_1) - 2\frac{c}{2}\phi(e_1,e_3) + \phi(e_3,e_3) = 0$.  So, replacing $e_3$ with $e'_3$, we may assume $\phi(e_1,e_1) = \phi(e_3,e_3) = 0$ while $\phi(e_1,e_3) = \phi(e_3,e_1) = 1$.
Let $e_2$ be a third vector to complete the basis.  Set $b = \phi(e_1,e_2)$.  Setting $e'_2 = e_2 - b e_3$, we have that $\phi(e_1,e'_2) = 0$, so replacing $e_2$ by $e'_2$, we may assume $\phi(e_1,e_2) = \phi(e_2, e_1) = 0$.  Now let $c = \phi(e_2,e_3)$ and set $e'_2 = e_2 - c e_1$.  The $\phi(e'_2,e_3) = 0$ and $\phi(e'_2,e_1) = 0$, so replacing $e_2$ by $e'_2$ again, we get that the matrix for $\phi$ has the form
$$
\begin{pmatrix}
0 & 0 & 1 \\ 0 & a & 0 \\ 1 & 0 & 0
\end{pmatrix}
$$
for some $a$.  Because $\phi$ is non-degenerate, $a \neq 0$.  This yields the desired $Q$.
If $-a$ has a square root in $k$ (always the case if $k$ is algebraically closed), then replacing $e_2$ by $e'_2 = \frac{1}{\sqrt{-a}} e_2$, we may assume that $Q(e_2) = \phi(e_2,e_2) = -1$.
For the second part, let $C \subset \mathbb{P}^2_k$ be a non-degenerate conic defined by a non-degenerate quadratic form $Q$.  Note that the existence of the vector $e_1 \neq 0$ such that $Q(e_1) \neq 0$ is equivalent to the fact that the conic $C$ is not empty.
Applying a projective equivalence corresponds to making a change of basis in $V$.  And we just showed that we could change our basis so that $Q$ has the form
$$
\begin{pmatrix}
0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0
\end{pmatrix}.
$$
But this $Q$ defines the conic $C$ given by the equation $XZ - Y^2 = 0$, or $XZ = Y^2$.  (Edit: Actually, depending on conventions, you are more likely to interpret this $Q$ as defining the conic $2 XZ = Y^2$, but see the remark below.)
Edit: As Georges Elencwajg pointed out in the comments, you can use $Z \mapsto Z/a$ to get $XZ = aY^2$ projectively equivalent to $XZ = Y^2$ for any $a \neq 0$.
