Polynomial of degree two in two variables and rational points This question of mine and the comments on the answer led me to the following more general problem:
Suppose that we have polynomial $f(x,y)=a_1xy+a_2x^2+a_3y^2+a_4x+a_5y+a_6$:
Consider now the equation $f(x,y)=0$:
I would like to know the following:

What are the most general conditions that you know of on the coefficients $a_1.a_2.a_3,a_4,a_5,a_6$ of the polynomial so that under those conditions there are infinitely many points $(x,y)$ with both rational coordinates such that $f(x,y)=0$?

 A: I am assuming you are working over a number field $K$. Then, $f(x,y)$ describes a conic section. If you homogenize it by adding suitable powers of $z$ to every monomial, you get a ternary quadratic form, which can always be diagonalized by suitable change of variables to obtain something of the form $Q(x,y,z) = ax^2 + by^2 + cz^2$. You are then asking when the quadratic form is isotropic (ie, there is a triple $(x,y,z)\ne (0,0,0)$ such that $ax^2 + by^2 + cz^2 = 0$).
This is the theory of ternary quadratic forms, which is contained in the theory of quaternion algebras. There are many sources on this online (I'd suggest looking up quadratic forms)
Since the Hasse Principle holds for quadratic forms, $Q = 0$ has a solution over a number field $K$ if and only it has solutions over every $K_\nu$, where $\nu$ is a place of $K$ and $K_\nu$ is the completion of $K$ at $\nu$. If $K = \mathbb{Q}$, then $\{K_\nu\}_\nu = \{\mathbb{Q}_p\}_{p\text{ prime}}\cup\{\mathbb{R}\}$. (Here $\mathbb{Q}_p$ is the field of $p$-adic numbers).
Over $\mathbb{R}$, $Q = 0$ has solutions if and only if $a,b,c$ are not all positive and not all negative.
At any finite place $\nu$, you can scale $Q$ by some number so that $a,b,c$ are all $\nu$-adically integral, and one of them is $-1$. WLOG suppose $c = -1$. In this case, $Q = 0$ has a solution iff the Hilbert symbol $(a,b)_p = 1$. If $K = \mathbb{Q}$ you can find formulas for it here: https://en.wikipedia.org/wiki/Hilbert_symbol
This is pretty much as good as good as it gets.
