Consider a smooth surface of the following form
$$
S = \{f(x,y,t) = p\_0(t)x^2+p\_1(t)xy+p\_2(t)x+p\_3(t)y^2+p\_4(t)y+p\_5(t) = 0\}\subset\mathbb{A}^3
$$
over $\mathbb{Q}$, and set
$$
U\_S = \{t' \in \mathbb{Q} : |\: f(x,y,t') = 0 \text{ for some } (x,y)\in\mathbb{Q}^2\}\subset\mathbb{Q}.
$$
Is there any example of such a smooth and irreducible surface $S$ such that the projection $S\rightarrow\mathbb{A}^1\_t$ is dominant, and $U\_S$ is non empty and non Zariski dense in $\mathbb{Q}$?


Thank you.



Rational points on a special class of surfaces
I am just posting my comments as an answer. Without a hypothesis that the geometric generic fiber of $\pi:S\to \mathbb{A}^1\_t$ is irreducible, the result is false. For a smooth compactification of $S$ on which $\pi$ extends to a morphism to $\mathbb{P}^1\_t$, the finite part of the Stein factorization of $\pi$ is either an isomorphism to $\mathbb{P}^1\_t$ (precisely when the geometric generic fiber is irreducible) or it is a degree-$2$ cover, i.e., a hyperelliptic curve. For appropriate choices of the coefficient polynomials $p\_i(t)$, this can be any hyperelliptic curve. If the genus is $\geq 2$, then this curve has only finitely many $\mathbb{Q}$-points (by Mordell's Conjecture / Falting's Theorem), so the image in $\mathbb{P}^1\_t$ is also a finite set.


However, if the geometric generic fiber is irreducible, then the compactification over $S$ is a conic bundle over $\mathbb{P}^1\_t$. After base change from $\mathbb{Q}$ to some number field, this surface is rational, i.e., the surface is geometrically rational. There is a conjecture (perhaps due to Colliot-Thélène) that the set of rational points on a geometrically rational variety over a number field is dense in the Brauer subset of the set of adelic points. Assuming this conjecture, once there is a single rational point (so that the Brauer subset is nonempty), the set of rational points is Zariski dense. Thus, the image in $\mathbb{P}^1\_t$ is also Zariski dense.


