Let $\Bbb{P}^2$ be the projective space over $\Bbb{C}$.
It is a well known fact that the blowup of $\epsilon:S\to\Bbb{P}^2$ at $9$ points $P_1,...,P_9$ in general position generates infinitely many exceptional curves, i.e. infinitely many smooth, irreducible curves $C\subset S$ of genus zero with $C^2=-1$.
The simplest examples are:
a) the exceptional divisors $E_i:=\epsilon^{-1}(P_i)$.
b) the strict transforms of the lines $P_iP_j$ with $i\neq j$.
c) the strict transforms of the conics $C_{i_1,...,i_5}$ passing through five points $P_{i_1},...P_{i_5}$ with $1\leq i_1<...<i_5\leq 9$.
I'm now interested in generating an infinite set of exceptional curves distinct from the ones above.
Here's my idea: let $\mathcal{C}$ be the set of cubics with multiplicity 2 in $P_1$, multiplicity 1 in $P_2,...,P_7$ and not passing through $P_8,P_9$. If $C\in\mathcal{C}$, then its strict transform $\widetilde{C}$ is smooth of genus zero with $\widetilde{C}^2=C^2-2^2-6\cdot 1^2=9-10=-1$.
I'm trying to show that $\#\mathcal{C}=\infty$. I think it is possible to argue that the cubics with an order 2 singularity at $P_1$ and smooth at $P_2,...,P_7$ form a pencil $\mathcal{P}$ and there are only finitely many cubics in $\mathcal{P}$ passing through either $P_8$ or $P_9$.
I'm having difficulty to formalize the argument for the dimension of $\mathcal{P}$. How do I prove that the multiplicity 2 at $P_1$ makes the dimension drop by two?
(if I'm talking nonsense, please let me know)
Thank you!
In fact, a rational $(-1)$-curve $C$ on a surface $S$ never deforms. In particular, it is not a member of a movable linear systems. This follows from the exact sequence $$ 0 \to \mathcal{O}_S \to \mathcal{O}_S(C) \to \mathcal{O}_C(-1) \to 0 $$ which shows that $$ h^0(S,\mathcal{O}_S(C)) = h^0(S,\mathcal{O}_S) = 1. $$ So, to get infinitely many curves you must find (or show the existence of) infinitely many appropriate divisor classes.