I have a smooth projective surface $X$, and two family of elliptic curves on it: $E_{1,t}$ and $E_{2,t}$, (I don't know what either $t$ runs through!) such that

(1), for any i={1，2}, the closed points of $X$ are the disjoint union of closed points of all $E_{i,t}$.

(2), the intersection number of $E_{1,t}$ and $E_{2,t}$ is always 1.

Can we conclude that $X$ is an abelian surface?

I have a smooth projective surface $X$, and two flat family of elliptic curves on it: $E_{1,t}$ and $E_{2,t}$, (I don't know what either $t$ runs through!) such that

(1), for any i={1，2}, the closed points of $X$ are the disjoint union of closed points of all $E_{i,t}$.

(2), the intersection number of $E_{1,t}$ and $E_{2,t}$ is always 1.

Can we conclude that $X$ is an abelian surface?

I have a smooth projective surface $X$, and two flat family of elliptic curves on it: $E_{1,t}$ and $E_{2,t}$, (I don't know what either $t$ runs through!) such that

(1), for any i={1，2}, the closed points of $X$ are the disjoint union of closed points of all $E_{i,t}$.

(2), the intersection number of $E_{1,t}$ and $E_{2,t}$ is always 1.

Can we conclude that $X$ is an abelian surface?

I have a smooth projective surface $X$, and two family of elliptic curves on it: $E_{1,t}$ and $E_{2,t}$, (I don't know what either $t$ runs through!) such that

(1), for any i={1，2}, the closed points of $X$ are the disjoint union of closed points of all $E_{i,t}$.

(2), the intersection number of $E_{1,t}$ and $E_{2,t}$ is always 1.

Can we conclude that $X$ is an abelian surface?