Let $X$ be an elliptic K3 surface. Let $\alpha$ be a smooth curve of genus $\geq3$. Define $$d(\alpha)=\min\lbrace \epsilon\cdot \alpha \ | \ \epsilon \mbox{ is an elliptic curve on } X \rbrace, $$ $$\mathcal{E}^0(\alpha)=\lbrace \mbox{elliptic curves } \epsilon \mbox{ such that } \epsilon\cdot\alpha=d(\alpha) \rbrace.$$
Question: if $\epsilon,\epsilon'\in\mathcal{E}^0(\alpha)$, can we say $\epsilon\cdot\epsilon'=0$ (in other words, they are linearly equivalent) ?
Let me explain why I would not expect this to be true.
Suppose that $X$ is a $K3$ surface with the following properties:
First we ask that the automorphism group $G=\operatorname{Aut}(X)$ is finite. (Remark: by a theorem of Sterk, this is equivalent to requiring that $X$ has fintely many elliptic pencils, though we don't use this.)
Note that this gives us an $G$-invariant line bundle $A$ containing smooth curves of high genus: take any ample line bundle $L$ and form $\bigotimes_{g \in G} g^*L$, or a tensor power of that.
OK, so let's take $A$, and take a smooth curve $\alpha \in A$.
Next we ask that for at least one elliptic pencil $L_1$ in $\mathcal{E}^0(\alpha)$, there is an automorphism $g$ which does not fix $L_1$: say $g^*L_1 = L_2$ for some other bundle $L_2$.
Then $L_2 \cdot \alpha = L_1 \cdot g_* \alpha = L_1 \cdot \alpha$ since the bundle $A$ is $G$-invariant. So $L_2$ is another elliptic pencil in $\mathcal E^0(\alpha)$. That is, sections of the bundles $L_1$ and $L_2$ give you a counterexample.
The gap is that I don't know how to cook up a $K3$ with these properties. But there is a large literature on finite automorphism groups of $K3$ surfaces, by Nikulin, Dolgachev, Kondo, etc. So maybe you could find what you need there.