Let $G$ be a projective plane degree $2d$ curve with equation $AC-B^2$, with $A$, $B$, $C$ of degrees $\deg A=2d-4$, $\deg B=d$, $\deg C=4$. Then, for $d>2$, a dimension count shows that $G$ is not generic.
Note that $A$ cuts an even divisor on $G$, of degree $2d(2d-4)$.


Is it true that for a smooth projective plane degree $2d$ curve $X$, existence of a degree $2d-4$ curve $A$ of degree $2d-4$ cutting out an even divisor on $X$ implies that $X$ is not generic?


I can show this for $d=3$, by showing that in this case $X$ can be written as $AC-B^2$, as above; in general this isn't feasible, as $A$ can be singular, and it's not clear how to proceed.



showing a plane curve non-generic by exhibiting an even divisor
The condition that there is a degree $2d - 4$ curve cutting out on a plane curve $X$ of degree $2d$ an even divisor is equivalent to the existence of a point
$$
\alpha \in \operatorname{Pic}^0(X)
$$
of order 2 such that the divisor class
$$
(d-2)H + \alpha \in \operatorname{Pic}^{2d(d-2)}(X)\tag{\*}
$$
is effective, where $H$ is the restriction to $G$ of the line class.
Since
$$
2d(d-2) < (2d - 1)(d-1) = g(X) - 1
$$
the plane curve $X$ is not generic.


EDIT. Let me show that for a general smooth plane curve $X$ with a non-trivial point $\alpha \in \operatorname{Pic}^0(X)$ of order 2 the divisor class $(\*)$ is not effective. Since this condition is open (by semicontinuity of cohomology) and the moduli space of pairs $(X,\alpha)$ is connected, it is enough to find one pair $(X,\alpha)$ for which $(\*)$ is not effective.


Consider a general symmetric matrix of size $d \times d$ with entries homogeneous polynomials of degree 2 in three variables. It gives a self dual exact sequence on $\mathbb{P}^2$
$$
0 \to \mathcal{O}(-1)^{\oplus d} \to \mathcal{O}(1)^{\oplus d} \to \mathcal{L} \to 0,
$$
where $\mathcal{L}$ is a line bundle on a smooth plane curve $X \subset \mathbb{P}^2$. The self-duality of the matrix implies that $\mathcal{L}$ has the form
$$
\mathcal{L} \cong \mathcal{O}(dH + \alpha)
$$
where $\alpha$ is a non-trivial point of order 2 on $X$. So, to check that $(\*)$ is not effective in this case, we can just twist the above exact sequence by $\mathcal{O}(-2)$ and use the cohomology exact sequence.


