Does a generic curve inside the space of curves with a node at a specific point have only finitely many nodes? 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 $\mathcal{A}$.  
 A: Yes, a generic curve with one node at a fixed point of $\mathbb P^2$ over a field of characteristic $0$ has only that singularity. This is true in any degree $d$ at least $2$. By looking at a generic union of $d$ lines, two of which pass through a point $p$, we see that a generic element of the linear system of curves or degree $d$ with a singularity at $p$ has a node at $p$, and no other fixed points. The result follows from Bertini's Theorem for linear systems.
[Edit:] for a more elementary approach, it is easy to show that the curve with equation $xyz^{d-2} + x^d + y^d$ has no singularities outside of the obvious one.
A: 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 $[1,0,0]$.
