Let $\{F_t\}$, $t\in \Delta$ be an algebraic family of projective hypersurfaces, where $\Delta=\{z\in\mathbb C||z|<1\}$ is the unit disk. Assume $F_0$ is reducible and $Z$ is one of the components of $\{F_0=0\}$. In particular, $Z$ corresponds to a homogeneous polynomial $G$ with $G^k|F_0$ for some $k\ge 1$ and the $\textit{multiplicity}$ of $Z$ is defined to be the maximal integer $k$ satisfying such condition.
I'm trying to understand the multiplicity through a complex analytic point of view, via the following way.
Consider the total space of the family $$X=\{(x,t)\in \mathbb P^n\times \Delta|F_t(x)=0\}$$ with projection $\pi:X\to \Delta$.
Let $(p,0)\in X$ be a general point on the component $Z$ (in particular, $p$ is a smooth point of $Z$ and does not meet other components of $\{F_0=0\}$). Let $\Delta_p$ is a holomorphic disk in $X$ which is transversal to $Z$ at $p$. Then the restriction
$$\pi|_{\Delta_p}:\Delta_p\to \Delta$$
is identified with $z\mapsto z^m$ for some $m\ge 1$.
Question: Is it true that $m$ coincides with multiplicity of $Z$ in general?
My motivating example is the family of plane curves $\{y^2-txz=0\}\subset \mathbb P^2_{[x,y,z]}\times \Delta$, take $p=[1,0,1]$ and $\Delta_p=(1,t,1;t^2)$ which is 2-to-1 to $\Delta$, so the degree coincides with the multiplicity of $\{y^2=0\}$. Also the degree is not dependent on the choice of $p$ (except at $p=[0,0,1]$, where the total space is singular) and choice of normal disk.
This leads to my question above. However, I cannot prove this result after some trials. I appreciate it if anyone can show me proof or providing a counterexample.