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.
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.