Linear and Quadratic Diophantine Equations

Consider the following question:

Let $F(x,y)$ be an irreducible polynomial in two variables over  $\mathbb{Q}$. Find all rational numbers $x_0,y_0$ such that $F(x_0, y_0) = 0.$

When $F$ is linear, this problem is easy. The equation

$$F(x,y) = ax + by + c = 0$$

defines a line, and letting $y=t$, the solutions are

$$\left\{ \left( -\frac{b}{a} t - \frac{c}{a}, t\right) \, : \, t \in \mathbb{Q}\right\}.$$

When $F$ is quadratic, the solution is not completely trivial, but it is well understood. In this case, the equation $F=0$ has infinitely many rational solutions if and only if it has at least one solution. Moreover, it is easy to describe all solutions when there is one. If $(x_0, y_0)$ is a solution and $L$ is a non-tangent line through $(x_0, y_0)$, then $L$ will intersect the curve $F=0$ in exactly one other point $(x_1, x_1)$. Also $x_1, y_1\in\mathbb{Q}$ since a quadratic polynomial over  $\mathbb{Q}$ with $1$ rational root has both roots rational. Thus the rational points on $F=0$ are in bijection with the slopes of lines through $(x_0, y_0)$.

Chapter 2 of [Kato et al.] is about how to decide whether or not an $F$ of degree $2$ has a rational point. The answer is that $F=0$ has a rational solution if and only if $F=0$ has a solution with $x_0,y_0\in\mathbb{R}$ and a solution with $x_0,y_0\in\mathbb{Q}_p$ for every ``$p$-adic field'' $\mathbb{Q}_p$. This condition, though it might sound foreboding, is easy to check in practice. I encourage you to flip through chapter 2 of loc. cit.