Let $X$ be a projective variety in projective space $P_n$. Let $H$ be a hyperplane, we say $H$ is tangent to $X$ at $p$ if $T_p(X)+T_p(H)\neq T_p(\mathbb{P}_N).$ My question is to prove a hyperplane $H$ doesnot vanish on $X$ then $H$ is tangent to a projective variety $X$ at $p$ iff $X\cap H$ is singular at $p$.
Any help would be appreciated. Also I am a beginner in algebraic geometry and do not know the language of schemes, so I would prefer an answer in the language of variety.
Up to a linear change of coordinates $(X_0: \dots : X_N)$ we can assume that $p=(1:0: \dots :0)$ and $H = \{ X_N =0 \} $ Since the problem is local we can restrict to the affine chart $\{X_0\neq 0 \}$. Suppose that $X\cap\{X_0\neq 0 \}$ is given by the zeros of $f_1, \dots, f_k \in \mathbb C[x_1, \dots , x_N]$ then $$ T_pX = \ker df_{1p} \cap \dots \cap\ker df_{kp} $$ By dimension reasons, $H$ is tangent to $X$ at $P$ iff $T_pX \subset T_pH = \ker dx_{Np}$ This translates to $\frac{\partial f_i}{\partial x_j}(p)= 0$ for every $i$ and $j< N$.
We have that $H\cap X\cap\{X_0\neq 0 \}$ is given by the vanishing of $g_i = f_i(x_1, \dots, x_{N-1}, 0)$, for $1\leq i \leq k$. Therefore $dg_{ip} = 0$ for every $i$ hence $H\cap X$ is singular at $p$.