Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous 
degree $d$ polynomials in three vriables, where 
$\delta_d = \frac{d(d+3)}{2}$. Let 
$$ X \subset \mathcal{D} \times \mathbb{P}^2$$ 

be a smooth embedded complex submanifold, not necessarily closed. 
Given a point $p\in \mathbb{P}^2$, we get a hyperplane 
$$\tilde{H}_p \in  \mathcal{D} \times \mathbb{P}^2.$$
Note that a point $p$ first of all gives a hyperplane $H_p$ 
in $\mathcal{D}$ (which is the space of degree $d$ 
polynomials passing through the point $p$). This gives us a hyperplane 
$$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$
Is it true that 
if $\tilde{H}_p$ is not transverse to $X$, then for every open  neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists a 
point $q\in U$ such that the 
hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This 
seems to intuitively say that if something is not transverse
then you can ``perturb'' it so that it is transverse. 
More precisely, I want to claim that the set of points $q$ 
where $\tilde{H}_q$ intersects $X$ transversally is an open, 
dense subset of $\mathbb{P}^2$. It is certainly open. 
Please note that apriori it is possible that there does 
not exist any point $q$ such that $\tilde{H}_q$ intersects 
$X$ transversally. This is the part I don't see how to prove, 
although intuitively it seems obvious.