Is it possible for a smooth, transcendental plane curve to be such that for every point $(a,b)$ on the curve, there is a polynomial with rational coefficients $p(x,y)\in\mathbb{Q}[x,y]$ where $p(a,b)=0$? The curve $y=e^x$ is not such a curve, because $\pi$ and $e^\pi$ are algebraically independent over $\mathbb{Q}$, so there is no polynomial in $Q[x,y]$ of which $(\pi, e^\pi)$ is a zero.
Equivalently: is there a smooth, transcendental plane curve $C$ such that, given any point $(a,b)$ on the curve, there is an algebraic curve which intersects $C$ at $(a,b)$?
