Let $C \subseteq \mathbb{P}^2$ be defined by $A x^2 + B y^2 + C z^2 = 0$ and let $P$ be a point on $C$ with $\mathbb{Q}$ coordinates.
I was wondering what assumptions do I need to establish that there is a bijection between lines defined over $\mathbb{Q}$ going through $P$ and the rational points of $C \backslash \{P\}$?
Thank you!
The correct statement here is that for a smooth conic $C\subset \Bbb P^2_k$ with a $k$-rational point $P$, there's a bijection between $k$-rational lines through $P$ and $k$-rational points on $C$. This is a consequence of Bezout's theorem: the intersection of any $k$-rational line $L$ with $C$ is two points counted with multiplicity. As $C$ is smooth, it has a unique tangent at each point - if $L$ is the tangent at $P$, this means it intersects $C$ with multiplicity two at $P$, and this gives you $P$. If $L$ is not the tangent, then it intersects $C$ with multiplicity one at $P$ and has a unique other $k$-rational point of intersection with $C$.
In order to get the correct statement from your post, we should require that $C$ is smooth, which means that $A,B,C\neq 0$, and then also make one of the following changes: either replace $C\setminus \{P\}$ with $C$, or replace "lines through $P$" with "non-tangent lines to $C$ through $P$".