Is it possible to find infinitely many points in a smooth variety such that their dual of corresponding tangent space have nonzero intersection?

Question

Assume $X\subset \mathbb{P}^n$ to be a smooth projective variety defined over an algebraic closed field $k$ (the characteristic of $k$ could be either $0$ or positive). For every point $P\in X$, denote by $T_P X$ its corresponding tangent space. And denote by $$ (T_P X)^{\vee}:=\{\text{hyperplanes } H\supset T_P X\} $$ the set of hyperplanes containing $T_P X$. In fact, if we consider the dual space $(\mathbb{P}^n)^{\vee}$, then $(T_P X)^{\vee}$ could be understood as a linear space in it. In particular, $\dim (T_P X)^{\vee}+\dim X=n-1$.

My question is, is it possible that there are infinitely many distinct points $P_1, P_2, \cdots$ such that $$ \bigcap_{i=1}^{\infty} (T_{P_i} X)^{\vee}\neq \emptyset? $$

So far, according to my best knowledge, when $X$ is a hypersurface, then the answer to the above question is negative. Since in this case, $(T_{P_i} X)^{\vee}$ is exact the Gauss image of $P_i$. And by a result of Zak, the Gauss map for a smooth variety is finite. But what will happen if $X$ is not a hypersurface? For instance, if $X$ is a complete intersection?

Edit: When I first posted the question, I forgot to exclude the obvious examples: the linear subspaces. So maybe I should ask what will happen when the degree of $X$ is strictly greater than $1$.

Answer 1

You are asking whether there exists a hyperplane in $\mathbb{P}^n$ which is tangent to $X$ along a positive-dimensional subvariety. Consider the Veronese surface $V\subset \mathbb{P}^5$; by definition, the hyperplanes in $\mathbb{P}^5$ cut down the full linear system of conics in $\mathbb{P}^2$. In particular, there are hyperplanes which intersect $V$ along a double line, hence which are tangent to $V$ at each point of the line.