Let $C_1=V(F),C_2=V(G)$ be smooth curves in $\Bbb{P}_{\Bbb{C}}^2$ and consider the pencil of curves $C_{(s:t)}=sF+tG$ for $(s:t)\in\Bbb{P}^1$.
I'm trying to prove/disprove that $C_{(s:t)}$ is smooth for all but finitely many $(s:t)\in\Bbb{P}^1$.
If $F,G\in \Bbb{C}[X,Y,Z]$, the conditions for $P\in C_{(s:t)}$ to be singular is that $(sF+tG)_X$, $(sF+tG)_Y$, $(sF+tG)_Z$ and $sF+tG$ are all zero at $P$.
This looks like a very strong condition, so it really looks like there can't be many $(s:t)$ that makes the four terms above be zero for all $P$.
But I'm having real difficulty proving this. I've tried simple cases, for example when $F$ is quadratic and $G$ is linear, and the result was clear, but more general examples looked really complicated.
I know I'm doing a brute-force approach, and if there's a better on, I'd be happy to know.
Hint: Consider the set
$$Problem:=\{((x:y:z),(s:t)) \in \mathbb{P}^2\times \mathbb{P}^1\;|\; (x: y: z) \text{ is a singular point of }V(sF+tG) \}.$$
This is a closed subset of $\mathbb{P}^2\times \mathbb{P}^1$, and so it is a projective variety. It has an obvious projection $Problem \rightarrow \mathbb{P}^1, ((x:y:z),(s:t)) \mapsto (s: t).$ Try to show that the image of the projection consists of finitely many points only (use projectivity, and topology of $\mathbb{P}^1$).