I'm trying to understand better conic bundles on quartic del Pezzo surfaces *over non-algebraically closed fields*.

Let $k$ be a field. A conic bundle surface is a smooth projective surface $S$ over $k$ equipped with a dominant morphism $S \to C$ to some smooth curve $C$, whose fibres are isomorphic to plane conics.

Let $S$ be a quartic del Pezzo surface over $k$ (i.e. $S$ is a smooth complete intersection of two quadrics in $\mathbb{P}^4$). If $S$ contains a conic $Q$ over $k$, then it admits a conic bundle. Namely, the pencil of hyperplanes through $Q$ gives rise to a conic bundle morphism $S \to \mathbb{P}^1$, determined by the conics residual to $Q$. I want to know whether every conic bundle on $S$ has this form.

Let $S$ be a quartic del Pezzo surface over $k$ equipped with a conic bundle $S \to C$. Then is $C(k) \neq \emptyset$?

Note that $C$ is clearly a smooth curve of genus $0$ in this case.

The answer to the analogue of my question is *no* for del Pezzo surfaces of degree $8$. Take $S = C \times C'$ where $C$ and $C'$ are conics without rational points. Then $S \to C$ is a conic bundle, but there is no conic bundle $S \to \mathbb{P}^1$.

For cubic surfaces, however, the answer is *yes*. A cubic surface with a conic bundle contains a line and thus a rational point. Hence the base of the conic bundle is always $\mathbb{P}^1$.

I'm guessing that the answer to my question is *no*, but I don't see how to construct explicit counter-examples. Note that a similar argument to the cubic surface case shows that such a counter-example must have no $0$-cycle of odd degree.