Let $X\subset\mathbb{P}^N$ be a rational smooth projective irreducible non degenerated variety of dimension $n=\dim(X)$ and let $$\mathcal{H}=|\mathcal{O}_X(1) \otimes\mathcal{I}_{{p_1}^2,\dots,{p_l}^2}|$$ be the linear system of hyperplane sections singular at $p_1,\dots,p_{l}$.

Assume that $\dim(\mathcal{H})=n$ and that the general divisor $H\in \mathcal{H}$ is singular along a positive dimensional subvariety passing through the points. Moreover assume that the schematic intersection of all of these singularities is singular only at $p_1,\dots,p_l$.

Locally, in a neighborhood of the $p_i$'s, say $p_1$ we can write $H$ as the zero locus of $$H_f=a_{i,j}x_ix_j+h(x_1,\dots,x_n)=0$$ where $x_1,\dots,x_n$ are local coordinates and $h(x_1,\dots,x_n)$ is a polynomial of degree $\geq3$. The quadric $$Q_{H_f}=a_{i,j} x_ix_j=0$$ have in general rank $h\leq n$. Denote with $\mathcal{A}_H$ the vertex of $Q_{H_f}$. Is it true that under these assumptions $\bigcap_{H\in \mathcal{H}} \mathcal{A}_H=p_1$?


I have the feeling this question is not well-posed. Each local system where $H$ is of the form: $$a_{j,i} x_ix_j + h(x_1, \ldots, x_n)$$ depends on $H$. As a consequence, $\mathcal{A}_H$ lives in a local system that depends on $H$. So taking the intersection for all $H \in \mathcal{H}$ is somehow a nonsense, since all these linear spaces live in different neighborhoods.

What could be more meaningful would be to define $\bigcap_{H \in \mathcal{H}} \mathcal{A}_H$ inside $T_{X,x}$, but that would require to write everything in terms of second fundamental forms...

Anyways, even with this reformulation, the question you ask has a negative answer. Let $X = \mathbb{P}^{6} \times \mathbb{P}^{6} \subset \mathbb{P}^{48}$. Let $S^2(X)$ be the variety of trisecant planes to $X$. We have $\dim S^2(X) = \dim S(X) + X + 1 -4 = 32$. Let $x_1,x_2,x_3$ be generic point in $X$ and let $L = \langle x_1,x_2,x_3 \rangle$. Let $y$ be a generic point in $L$ and let $S_y$ be the entry locus in $X$ associated to $y$. This entry locus is a $\mathbb{P}^2 \times \mathbb{P}^2$ containing $x_1,x_2,x_3$.

By Terracini's lemma, we have $T_{X,x} \subset T_{S^2(X),y}$ for all $x \in S_y$. As a consequence, any hyperplane in the linear system $T_{S^2(X),y}^{\perp}$ is tangent to $X$ along $S_y$.

Now, let $\mathcal{H} = T_{S^2(X),y}^{\perp}$, we have $\mathcal{H} = |\mathcal{O}_{X}(1) \otimes \mathcal{I}_{x_1,x_2,x_3}^2|$ and $\dim \mathcal{H} = 15$. The generic $H \in \mathcal{H}$ is tangent to $X$ along $S_y$, hence all $H \in \mathcal{H}$ are tangent to $X$ at least along $S_y$. From the definition of $\mathcal{A}_H$ (which naturally leaves in $T_{X,x}$), we see that $T_{Sy,x} \subset \mathcal{A}_H$, for all $H \in \mathcal{H}$.


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