Let $X$ be a smooth complex projective variety embedded in $\Bbb P^n$. Fix $p \in \Bbb P^n$ and let $Y = \cup_{x \in X} L_x$ where $L_x$ is the line passing through $p$ and $x$.
Assume that $L_x \cap L_y = \{p\}$ for all $x \neq y$. Then, $Y \backslash \{p\}$ is the total space of a line bundle $L \to X$.
Question : What is $c_1(L)$ ? Is $L$ ample ?
I am a bit embarrased since it is a basic question in projective geometry, I think it might be related to the $\rm{Proj}$ construction but I'm not really familiar enough with it. If $X$ is only a curve or a surface it would be already very nice.
Here is also a guess : maybe there is an hyperplane $H$ so that $H \cap Y = \{p\}$. Then, $L \to X$ becomes an affine cone in the complement of $H$, so $S = k[L]$ is naturally graded and we can take $\text{Proj}(S)$. But I have no ideas what to do after, or if this even makes sense.
I assume $p \not\in X$ (otherwise, the line bundle $L$ is not on $X$, but on the blowup of $X$ at $p$). Then $$ L = O_{\mathbb{P}^n}(1)\vert_X. $$ Indeed, first assume that $X$ sits in a hyperplane and $p$ is away from it. In other words, assume that $\mathbb{P}^n = \mathbb{P}(\Bbbk \oplus V)$ for a vector space $V$ with $X \subset \mathbb{P}(V) \subset \mathbb{P}(\Bbbk \oplus V)$ and $p$ corresponding to $\Bbbk \subset \Bbbk \oplus V$. Then the tautological embedding $$ O_{\mathbb{P}(V)} \oplus O_{\mathbb{P}(V)}(-1) \to (\Bbbk \oplus V) \otimes O_{\mathbb{P}(V)} $$ induces a morphism $$ \mathbb{P}_{X}(O_{\mathbb{P}(V)} \oplus O_{\mathbb{P}(V)}(-1)) \to \mathbb{P}(\Bbbk \oplus V) $$ that contracts the relative hyperplane section $X \cong \mathbb{P}_{X}(O_{\mathbb{P}(V)}) \subset \mathbb{P}_{X}(O_{\mathbb{P}(V)} \oplus O_{\mathbb{P}(V)}(-1))$ to the point $p$ and defines an isomorphism $$ \mathbb{P}_{X}(O_{\mathbb{P}(V)} \oplus O_{\mathbb{P}(V)}(-1)) \setminus \mathbb{P}_{X}(O_{\mathbb{P}(V)}) \cong L. $$ It remains to note that the left side is isomorphic to the total space of $O_{\mathbb{P}(V)}(1)$.
In the case when $X$ does not sit in a hyperplane, your variety $Y$ is just an isomorphic projection (because of the condition $L_x \cap L_y = \{p\}$) of the above construction from $\mathbb{P}^{n+1}$, hence the line bundle is the same.
EDIT. The isomorphism in the last display is a special case of a general result: $$ \mathbb{P}(E \oplus O) \setminus \mathbb{P}(E) = Tot(E) $$ for any vector bundle $E$. Now just note that $$ \mathbb{P}(O \oplus O(-1)) \setminus \mathbb{P}(O) = \mathbb{P}(O(1) \oplus O) \setminus \mathbb{P}(O(1)) $$ (because projectivization does not change under a twist) and apply the above to $E = O(1)$.