Elliptic curves on a K3 surface 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) ?
 A: your question has a negative answer. 
Consider the following counterexample: Set 
$$Y := \mathbb{P}^1 \times \mathbb{P}^2$$ with canonical projections $p_1$ and $p_2$ and define
$$\mathscr{O}_Y(a,b) := {p_1}^* (\mathscr{O}_{\mathbb{P}^1}(a)) \otimes {p_2}^* (\mathscr{O}_{\mathbb{P}^2}(b))$$
as a sheaf on $Y$. Take a smooth surface $X \subset Y$ defined by a general section 
$$ s \in H^0(Y,\mathscr{O}_Y(2,3)).$$
The defining exact sequence
$$  0 \longrightarrow  \mathscr{O}_Y(-2,-3) \mathop \longrightarrow \limits^{s} \mathscr{O}_Y \longrightarrow \mathscr{O}_X \longrightarrow 0$$
induces a long exact sequence with the following segment
$$  H^1(Y,\mathscr{O}_Y) \longrightarrow H^1(X, \mathscr{O}_X) \longrightarrow H^2(Y, \mathscr{O}_Y(-2,-3)).$$
The chomology of projective spaces, which is well-known, implies 
$$H^1(X, \mathscr{O}_X) = 0 $$
i.e. $X$ has irregularity $q = 0$. On the other hand, X has trivial canonical sheaf 
$$ \kappa_X \cong \kappa_Y \otimes \mathscr{O}_Y(2,3) |X \cong \mathscr{O}_X.$$
Hence X is a K3-surface. One shows, that the restriction of Picard groups $Pic(Y) \longrightarrow Pic(X)$ is bijective, hence $Pic(X)$ has rank = 2. 
Due to $q = 0$ the Picard group $Pic(X)$ injects into  $H^2(X, \mathbb Z)$. Its image, the Neron-Severi group $NS(X) \subset H^2(X, \mathbb Z)$, is a sublattice of rank = 2. The two classes $L_1 := \mathscr{O}_Y(1,0)|X$, $L_2 := \mathscr{O}_Y(0,1)|X \in NS(X)$ form a $\mathbb{Z}$-base of $NS(X)$. The intersection form of $NS(X)$ with respect to that base is the matrix
$$\begin{pmatrix}0&3\\3&2\end{pmatrix}.$$
As a consequence, one easily shows that $X$ does not have any divisors of self-intersection = $-2$. In particular, X has no $-2$ curves.
The adjunction formula implies that any class containing an elliptic curve has self-intersection = $0$. Hence, the extremal rays bounding the convex cone of effective divisors $NE(X) \subset NS(X)_\mathbb{R}$, are exactly the rays $\mathbb{R}^+ L$ generated by the class $L$ of an elliptic curve (Olivier Debarre: Higher-Dimensional Algebraic Geometry. Lemma 6.2). There are only two such rays: The ray generated by $L=L_1$ and the ray generated by $L=-L_1+3L_2$. The class $L_1$ has the elliptic pencil defining the restriction
$$p_1|X: X \longrightarrow \mathbb{P}^1.$$ 
The restriction of the second projection 
$$ p_2|X: X \longrightarrow \mathbb{P}^2$$
is a double cover. Let $\sigma \in Aut(X)$ denote the covering involution. Then the induced map on $NS(X)$ maps $L_1$ to $-L_1 + 3L_2$.
For $n \in \mathbb{N}, n > 0,$ the effective divisor $C_n := nL_2$ has intersection numbers $$(C_n,L_1)=(C_n,-L_1 + 3L_2).$$ Any smooth curve contained in $C_n$ has genus 
$$g = 1 + C_n^2/2 = 1 + n^2.$$ 
The divisor $C_n$ is ample by the Nakai-Moishezon criterion. Hence for $n>>0$ the divisor $C_n$ is even very ample. Any very ample divisor $C_n$ contains a smooth curve $\alpha$ by Bertinis theorem.
But the intersection number of both distinguished classes: 
$$(L_1, -L_1 + 3L_2) = 9 \neq 0.$$ 
This answers your question in the negative, q.e.d.
Note. The example is taken from "Bert van Geemen: Some remarks on Brauer groups of K3 surfaces. Advances in mathematics 197 (2005) 222-247". Chapter 5 of his paper contains more details and a series of further examples with elliptic fibrations.
A: 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.
