TITLE: Symplectic orthogonality and projective duality: how do they work together?
QUESTION [6 upvotes]: Let $(V,\omega)$ be a $2n$-dimensional linear symplectic space, and $(\mathbb{P}V,\theta_\omega)$ the corresponding $(2n-1)$-dimensional contact manifold.
Given a smooth $(n-1)$-dimensional smooth projective variety $X\subset \mathbb{P}V$, I can define its orthogonal
$$
X^\perp:=\{ \pi\in \mathbb{P}V^*\mid\pi=v^\perp\textrm{ for some }[v]\in X  \}\, ,
$$
as well as its dual
$$
X^\ast:=\{ \pi\in \mathbb{P}V^*\mid\pi\supset T_v\widehat{X}\textrm{ for some }[v]\in X  \}\, ,
$$
where $\widehat{X}$ is the cone over $X$.
Here's the point: I'm trying to define a "companion" $X_\bullet$ of $X$ via the equation
$$
X_\bullet^\ast=X^\perp\, .\quad\quad (1)
$$
Example. If $D<V$ is a linear $n$-dimensional subspace, and $X=\mathbb{P}D$, then $(1)$ unambiguously defines $X_\bullet$, which turns out to be $\mathbb{P}D^\perp$. In particular, if $D$ is $\omega$-Lagrangian, then $X_\bullet=X$ is also $\theta_\omega$-Legendrian.

QUESTION: how to characterise the $X$'s possessing a well-defined (by means of $(1)$) "companion" $X_\bullet$? In particular, does this property hold for arbitrary (i.e., not necessarily linear) $\theta_\omega$-Legendrian $X$'s? Is it true that in this case $X=X_\bullet$?

I'm sure that this problem has been already addressed somewhere: any reference/keyword will be most welcome.
EDITED after Libli's and Robert Bryant's comments
Of course, by using the reflexivity theorem (see, e.g., this paper by Tevlev, Theorem 1.7a), one can solve $(1)$ and obtain
$$
X_\bullet=(X^\perp)^*\, .
$$
So, the answer to my question is "for all $X$". Nevertheless (recall that I'm assuming $X$ to be smooth), I can still ask, how to characterise those $X$ whose $X_\bullet$ enjoys some extra properties, like being smooth and (which this is what really I'm interested in) having the same dimension as $X$ - or even being $X$ itself (the only purpose of the symplectic orthogonality is to have $X$ and its dual sitting in the same space, and compare them).

REPLY [3 votes]: Varieties having the same dimension as their duals have been classified by Ein (see http://link.springer.com/article/10.1007%2FBF01391495). They are:
_hypersurfaces,
_$\mathbb{P}^1 \times \mathbb{P}^{n-1} \subset \mathbb{P}^{2n-1}$;
_$G(2,5) \subset \mathbb{P}^9$,
_$\mathbb{S}_{10} \subset \mathbb{P}^{15}$. 
EDIT : this results holds if one assumes the variety non-degenerate (that is non-included in a hyperplane).