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So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(X)$ NE(S)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.

Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way, but that only works for surfaces :

Theorem Let $S$ be a smooth projective surface $H$ an arbitrary ample divisor on $S$ and is only interesting for surfaces for which let $$ Q^+=\{\sigma\in N_1(S) \vert \sigma^2 >0, H\cdot\sigma \geq 0 \} $$ be the "big" positive component" of the interior of the quadric cone theorem does defined by the intersection pairing. Then $$ \overline{NE}(S) = \overline{Q^+} + \sum_{C^2<0} \mathbb R_+[C] $$

There is also one for $K3$'S, using the above notation:

Theorem Let $S$ be a smooth algebraic K3 surface and assume that its Picard number is at least $3$. (If the Picard number is at most $2$, then there are not say anythingtoo many choices for a cone). Then one of the following holds:

(i) $$ \overline{NE}(S) = \overline{Q^+}, or $$

(ii) $$ \overline{NE}(S) = \overline{\sum_{C\simeq \mathbb P^1, say K3'sC^2<0} \mathbb R_+[C]}. $$ The two cases are distinguished by the fact whether there exists a curve in $S$ with negative self-intersection. If the Picard number is at least $12$, then only (ii) is possible.

For proofs and more details, see this paper.
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EDIT: We may assume that the Picard number is at least two, as otherwise the cone is simply a ray generated by any effective curve. In particular, every effective curve is extremal. I will also assume that "curve" means "effective curve". (This edit was prompted by Damiano's comment that is now (sadly) deleted. It was a useful contribution.)

A curve on a surface is simultaneously a curve and a divisor and assuming the surface is smooth or at least $\mathbb Q$-factorial, then the curve, as a divisor, induces a linear functional on $1$-cycles. This works better if the surface is proper, so let's assume that.

So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(X)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.

If $C$ is reducible, then it may have negative self-intersection, but it is not extremal. For an example, blow up two separate points on a smooth surface and take the sum of the exceptional divisors. My guess is that you meant irreducible, so let's assume that.

Now we have $3$ cases:

1) $C^2>0$. In this case $C$ is in the interior of the cone and it cannot be extremal, can't even be on the boundary (Use Riemann-Roch to prove this).

2) $C^2=0$. Since $C$ is irreducible, it follows that it is nef and hence a limit of ample classes, so it is effective, but as Damiano pointed out I have already assumed that. (It is left to the reader to rephrase this if $C$ is assumed to be nef instead of effective). In this case the hyperplane corresponding to $C$ as a linear functional is a supporting hyperplane of the cone, intersecting it at least in the ray generated by $C$. So $C$ is definitely on the boundary, but it may or may not be extremal depending on the surface. For example any curve of self-intersection $0$ on an abelian surface is extremal, but for instance a member of a fibration that also has reducible fibers is not extremal despite being irreducible. For the latter think of a K3 surface with an elliptic fibration that has some $(-2)$-curves contained in some fibers.

3) $C^2<0$. If $C$ is effective, then $C\cdot D>0$ for any irreducible curve $D\neq C$. This means that $C$ and all other irreducible curves lie on different sides of the hyperplane corresponding to $C$ as a linear functional, so the convex cone they generate must have $C$ generating an extremal ray.

Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way, but that only works for surfaces and is only interesting for surfaces for which the "big" cone theorem does not say anything, say K3's.
show/hide this revision's text 2 added 75 characters in body; edited body

EDIT: We may assume that the Picard number is at least two, as otherwise the cone is simply a ray generated by any effective curve. In particular, every effective curve is extremal.

A curve on a surface is simultaneously a curve and a divisor and assuming the surface is smooth or at least $\mathbb Q$-factorial, then the curve, as a divisor, induces a linear functional on $1$-cycles. This works better if the surface is proper, so let's assume that.

So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(X)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.

If $C$ is reducible, then it may have negative self-intersection, but it is not extremal. For an example, blow up two separate points on a smooth surface and take the sum of the exceptional divisors. My guess is that you meant irreducible, so let's assume that.

Now we have $3$ cases(actually for the first two we don't even need to know a priori that $C$ is effective, only that $-C$ is not).:

1) $C^2>0$. In this case $C$ is in the interior of the cone and it cannot be extremal, can't even be on the boundary (Use Riemann-Roch to prove this).

2) $C^2=0$. Since $C$ is irreducible, it follows that it is nef and hence a limit of ample classes, so it is effective. In this case the hyperplane corresponding to $C$ as a linear functional is a supporting hyperplane of the cone, intersecting it at least in the ray generated by $C$. So $C$ is definitely on the boundary, but it may or may not be extremal depending on the surface. For example any curve of self-intersection $0$ on an abelian surface is extremal, but for instance a member of a fibration that also has reducible fibers is not extremal despite being irreducible. For the latter think of a K3 surface with an elliptic fibration that has some $(-2)$-curves contained in some fibers.

3) $C^2<0$. If $C$ is effective, then $C\cdot D>0$ for any irreducible curve $D\neq C$. This means that $C$ and all other irreducible curves lie on different sides of the hyperplane corresponding to $C$ as a linear functional, so the convex cone they generate must have $C$ generating an extremal ray.

Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way, but that only works for surfaces and is only interesting for surfaces for which the "big" cone theorem does not say anything, say K3's.
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