# Intersection of a hypersurface $V \subset \mathbb{P}^n$ with a tangent hyperplane $T_PV$ singular

I would like some help with an exercise from Reid's paper of algebraic geometry.

Prove that the intersection of a hypersurface $V \subset \mathbb{P}^n$ (not a hyperplane) with the tangent hyperplane $T_PV$ is singular in P.

Maybe I'm using the wrong approach. I'm trying to prove this by taking partial derivatives and evaluate them in the point P.

Thank you!!

First note that the hypersurface $V$ must be regular (=non-singular) at $P$, else its Zariski tangent space $T_P(V)$ would be $\mathbb P^n$ and not a hyperplane.
Next, since the problem is local at $P$ we can suppose, upon replacing $V$ by $W=V\cap \mathbb A^n$, that we have a hypersurface $W\subset \mathbb A^n_{x_1,...,x_{n}}$ containing the smooth point $P=(0,\cdots,0)$ and that the tangent hyperplane $T_P(W)$ is given by $x_1=0$.
This means that the the polynomial defining $W=Z(f)$ is of the form $$f(x_1,\cdots,x_n)=x_1+f_2(x_1,\cdots,x_n)+\cdots+f_r(x_1,\cdots,x_n)$$ where $r\geq 2$, $f_i(x_1,\cdots,x_n)$ is homogeneous of degree $i$ and $f_r(x_1,\cdots,x_n)\neq 0$.
But then the intersection $W\cap T_P(W)$ is the hypersurface $Z(f_0)\subset T_P(W)=\mathbb A^{n-1}_{0,x_2,...,x_n}$ of zeros of the polynomial $$f_0(x_2,\cdots,x_n)=0+f_2(0,x_2,\cdots,x_n)+\cdots+f_r(0, x_2,\cdots,x_n)$$ and that hypersurface is singular because the displayed polynomial $f_0(x_2,\cdots,x_n)$ has zero as linear term.
A specific example is the parabola $$y=x^2$$ with tangent $$y=0$$ at the origin. Of course one would like to count this intersection point with multiplicity two, and therefore consider it as singular.
For any field, let $$f$$ be the defining function of the hypersurface and $$l$$ the defining function of the tangent plane (in the projective case $$f$$ and $$l$$ are homogeneous polynomials). If we define the intersection by the ideal $$(f,l)$$ with non-reduced structure, then at the intersection point the Jacobian matrix of $$(f,l)$$ has rank one, as the rows are equal. But the zero set $$V(f,l)$$ can be non-singular.