Let $S=\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$ embedded with the Segre embedding given by $\mathcal{O}_S(1,1)$. If we intersect $S$ with a general smooth quadric $Q \subset \mathbb{P}^5$ we get a smooth surface $X \subset S$ of type $(2,2)$. Since $deg(S)=3$ by Bertini we have that $deg(X)=6$.

By construction we have that (up to identifying every $\mathbb{P}^2$ in the ruling by projecting from the $\mathbb{P}^1$ factor) for every point $p \in \mathbb{P}^2$ there exist only $2$ conics $C_1,C_2$ passing through $p$, so this is not a surface of the type $Y=\mathbb{P}^1 \times \mathbb{P}^1$ embedded with bidegree $(2,1)$. In fact we have that $Y \subset \mathbb{P}^5$ but $deg(Y)=4 \neq deg(X)$.

My question now is how to prove that $X$ is rational? (I've the feeling that it must be so but I'm currently not able to figure it out).

Also I had the feeling that $X$ could be viewed as the total space of a conic $\mathcal{C}$ in the $\mathbb{P}^5=\mathbb{P}(\mathbb{C}[x^2,y^2,z^2,xy,xz,yz])$ where $[x,y,z]$ are homogeneous coordinates of $\mathbb{P}^2$, where by total space I mean the space consisting of the union of the conics $C_x$ corresponding to the point $x \in \mathcal{C}$. This would imply that there are $6$ singular conics among the conics in the $\mathbb{P}^2$ fibers of $S$, since $deg(\mathcal{C} \cap \mathbb{S}ec_2(v_2(\mathbb{P}^2)))=2 \cdot 3=6$.

Anyone knows if $6$ is the correct number of singular conics in a smooth quadric section of the Segre cubic scroll in $\mathbb{P}^5$? The fact that $6$ is also the degree is a coincidence?

Thanks in advance.