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While the number of lines on Del Pezzo surfaces are finite, the number of conics is infinite. More precisely, there are finitely many families $X\to P^1$ whose fibers are plane conics. Let me explain this in more detail.

As you probably know, a degree $d$ Del Pezzo surface $X$ can be realized as the blow-up of $P^2$ in $r=9-d$ points in general position. The Picard group of $X$ has rank $r+1$ and is generated by the classes of the exceptional divisors $E_1,\ldots, E_r$ and $L$ which is the pullback of a general line in $P^2$ via the blow-up morphism $\pi:X\to P^2$. The intersection form on $N^1(X)=\mbox{Pic }X$ is given by $$ E_i\cdot E_j=-\delta_{ij}, \qquad E_i\cdot L=1, \qquad L^2=1. $$Also, the anticanonical class equals $-K=3L-E_1-\ldots-E_r$ in this basis.

If $X$ has degree $\ge 4$, then $-K$ is very ample, and the conics on $X$ correspond precisely to the effective divisor classes such that $$ -K.D=2 \mbox{ and } D^2=0 $$Examples are $L-E_i$ (pullback of a line through the point $p_i$) and $2L-E_1-E_2-E_3-E_4$ (pullback of a conic avoiding $p_5$). Using the AM-GM inequality, one can show that the number of such classes is finite.

In fact it is easy to see that any conic can be written as the sum of two exceptional curves (which form the generators for the effective cone $\overline{NE}(X)$). So $D=E+F$ for some $E,F$ with $E.F=1$. Moreover, using this description, it is not hard to verify that the conic divisors $D$ are even base-point free and so by Riemann-Roch, define morphisms $X\to \mathbb{P}^1$. These morphisms are conic bundles, i.e., every fiber is isomorphic to a plane conic in $X$.

On the other hand, the lines on $X$ correspond to classes satisfying $-K.E=1 \mbox{ and } E^2=-1$ so they don't 'move' in linear systems like the conics do, which explains why their number is finite.

EDIT: There can not be any isolated conics on $X$, since if $D$ is any isolated rational curve, then $D^2<0$ and the adjunction formula $D^2+K.D=2g-2$ implies that $D^2=-1$, so $D$ is an exceptional curve, i.e., a line.
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While the number of lines on Del Pezzo surfaces are finite, the number of conics is infinite. More precisely, there are finitely many families $X\to P^1$ whose fibers are plane conics. Let me explain this in more detail.

As you probably know, a degree $d$ Del Pezzo surface $X$ can be realized as the blow-up of $P^2$ in $r=9-d$ points in general position. The Picard group of $X$ has rank $r+1$ and is generated by the classes of the exceptional divisors $E_1,\ldots, E_r$ and $L$ which is the pullback of a general line in $\PP^2$ P^2$ via the blow-up morphism $\pi:X\to PP^2$P^2$. The intersection form on $N^1(X)=Pic N^1(X)=\mbox{Pic }X$ is given by $$ E_i\cdot E_j=-\delta_{ij}, \qquad E_i\cdot L=1, \qquad L^2=1. $$Also, the anticanonical class equals $-K=3L-E_1-\ldots-E_r$ in this basis.

If $X$ has degree $\ge 4$, then $-K$ is very ample, and the conics on $X$ correspond precisely to the effective divisor classes such that $$ -K.D=2 \mbox{ and } D^2=0 $$Examples are $L-E_i$ (pullback of a line through the point $p_i$) and $2L-E_1-E_2-E_3-E_4$. 2L-E_1-E_2-E_3-E_4$ (pullback of a conic avoiding $p_5$). Using the AM-GM inequality, one can show that the number of such classes is finite.

In fact it is easy to see that any conic can be written as the sum of two exceptional curves (which form the generators for the effective cone $\overline{NE}(X)$). So $D=E+F$ for some $E,F$ with $E.F=1$. Moreover, using this description, it is not hard to verify that the conic divisors $D$ are even base-point free and so by Riemann-Roch, define morphisms $X\to \mathbb{P}^1$. These morphisms are actually conic bundles, i.e., every fiber is isomorphic to a plane conic . in $X$.

On the other hand, the lines on $X$ correspond to classes satisfying $-K.E=1 \mbox{ and } E^2=-1$ so they don't 'move' in linear systems like the conics do, which explains why their number is finite.

EDIT: There can not be any isolated conics on $X$, since if $D$ is any isolated curve, then $D^2<0$ and the adjunction formula $D^2+K.D=2g-2$ implies that $D^2=-1$, so $D$ is an exceptional curve.
show/hide this revision's text 1

While the number of lines on Del Pezzo surfaces are finite, the number of conics is infinite. More precisely, there are finitely many families $X\to P^1$ whose fibers are plane conics. Let me explain this in more detail.

As you probably know, a degree $d$ Del Pezzo surface $X$ can be realized as the blow-up of $P^2$ in $r=9-d$ points in general position. The Picard group of $X$ has rank $r+1$ and is generated by the classes of the exceptional divisors $E_1,\ldots, E_r$ and $L$ which is the pullback of a general line in $\PP^2$ via the blow-up morphism $\pi:X\to PP^2$. The intersection form on $N^1(X)=Pic X$ is given by $$ E_i\cdot E_j=-\delta_{ij}, \qquad E_i\cdot L=1, \qquad L^2=1. $$Also, the anticanonical class equals $-K=3L-E_1-\ldots-E_r$ in this basis.

If $X$ has degree $\ge 4$, then $-K$ is very ample, and the conics correspond precisely to the effective divisor classes such that $$ -K.D=2 \mbox{ and } D^2=0 $$Examples are $L-E_i$ and $2L-E_1-E_2-E_3-E_4$. In fact it is easy to see that any conic can be written as the sum of two exceptional curves (which form the generators for the effective cone $\overline{NE}(X)$). Moreover, using this description, it is not hard to verify that the conic divisors $D$ are base-point free and so by Riemann-Roch, define morphisms $X\to \mathbb{P}^1$. These morphisms are actually conic bundles, i.e., every fiber is isomorphic to a plane conic.

On the other hand, the lines on $X$ correspond to classes satisfying $-K.E=1 \mbox{ and } E^2=-1$ so they don't 'move' in linear systems like the conics do, which explains why their number is finite.