Let $\mathcal{D} \approx \mathbb{P}^{\delta\_d}$ be the space of homogeneous degree $d$ 
polynomials in three variables (up to scaling), where $\delta\_d = \frac{d(d+3)}{2}$. 
Define $\mathcal{A}$ to be space of degree $d$ curves with a strict node at 
the point $[1,0,0]$, ie 


$$ \mathcal{A} := \{ f \in \mathcal{D}: f([1,0,0]) =0, ~~ \nabla f |\_{[1,0,0]} =0, 
 \quad det \nabla^2 f ([1,0,0]) \neq 0 \}$$


Is it true that a generic element of $\mathcal{A}$ has finitely many singular points, provided 
$d$ is sufficiently large? It seems to me that a much stronger statement should be 
true, ie a generic element of $\mathcal{A}$ has only one singular point. Is there some reference 
for this result? It seems to me this is true, but I am not sure how to prove this 
rigorously. 


By generic, I mean that the set of curves that has only finitely many nodes 
forms a dense open subset of $\mathcal{A}$. 



Does a generic curve inside the space of curves with a node at a specific point have only finitely many nodes?
Ritwik: I believe we have discussed this before. Consider the blowing up at $[1,0,0]$,
$$
\nu:X\to \mathbb{P}^2
$$
with exceptional divisor $E$. Now for the family $\overline{\mathcal{A}}$ of plane curves $C$ of degree $d$ having a singularity at $[1,0,0]$, consider the family of transform curves $\nu^\*C - 2E$. By considering curves such as $x^{d-2}y^2$ and $x^{d-2}z^2$, you see that this family of curves on $X$ is base point free ($\textbf{edit}$: in a neighborhood of $E$; curves such as $y^d$ and $z^d$ show the linear system is base point free everywhere). Therefore, by Bertini's theorem, a general member is smooth on $X$. Since $\nu$ is an isomorphism away from $[1,0,0]$, it follows that a general member $C$ of $\overline{\mathcal{A}}$ is smooth away from $[1,0,0]$. $\textbf{Edit}$: Also the intersection number of these curves with $E$ is $2$, so a general member on $X$ will intersect $E$ in precisely $2$ distinct points, proving that a general member $C$ of $\overline{\mathcal{A}}$ has an ordinary double point at $[1,0,0]$.


