Rational points on a special class of surfaces 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.
 A: 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.
