Conics and conics of the form $ax^2+by^2+c=0$ The problem of finding rational points on conics is usually discussed (for example in the book of Silverman and Tate) for conics of the form $ax^2+by^2+c=0$.
I assume that those conics are in bijection with all conics, in general of the form $ax^2+by^2+dxy+ex+fy+c=0$, through some natural transformation I haven't been able to think of.
Can someone give me any hints, or point me to a reference?
Note: I'm only interested in conics over the rational.
 A: Alright, I will switch to Lehman's notation; here we have a quadratic form in three variables $x,y,z$ written as
$$ g(x,y,z) = a x^2 + b y^2 + c z^2 + r yz+szx + t xy. $$
At the end, you would want to set $z=1$ and set the mess to zero.
Set
$$ \Delta = 4ab - t^2 $$ and 
$$ \delta = 4 a c - s^2.  $$
Then 
$$ 4 a \Delta g = \Delta \left(2ax + ty + sz \right)^2 + \left( \Delta y + (2ar -st) z \right)^2 + \left( \Delta \delta -  (2ar -st)^2 \right) z^2  $$
Lehman's discriminant for the ternary form is
$$ 4abc + rst - a r^2 - b s^2 - c t^2, $$ 
or
$$
\frac{1}{2} \det
\left(
\begin{array}{rrr}
2a & t & s \\
t & 2b & r \\
s & r & 2c
\end{array}
\right)
$$
Another way to write the conclusion is
$$ 4 a \Delta g = \Delta \left(2ax + ty + sz \right)^2 + \left( \Delta y + (2ar -st) z \right)^2 + 4a \left(4abc + rst - a r^2 - b s^2 - c t^2 \right) z^2  $$
I do not see that there is anything particularly natural about this. It is just repeated completing the square, introducing coefficients so that there is not a giant denominator.
