[proofplan] We pass from the projective hypersurface to its affine cone and compute the Zariski tangent space of the cone at an affine representative $a$ of $p$. The tangent space to the cone is the kernel of the differential of $F$, hence is cut out by the linear form whose coefficients are the partial derivatives of $F$ at $a$. [Euler's formula](/theorems/2014) for homogeneous polynomials shows that the radial line $ka$ lies in this affine tangent space, so projectivizing gives the projective tangent hyperplane. Finally, smoothness of $p$ ensures that this linear form is nonzero, so the displayed equation defines a genuine hyperplane. [/proofplan]

[step:Compute the tangent space to the affine cone]Let \begin{align*} \widehat X:=V(F)\subset \mathbb A_k^{n+1} \end{align*} be the affine cone over $X$. Define the $k$-[linear map](/page/Linear%20Map) \begin{align*} L_a:k^{n+1}&\to k \end{align*} by \begin{align*} L_a(v)=\sum_{i=0}^n \frac{\partial F}{\partial X_i}(a)v_i \end{align*} for $v=(v_0,\dots,v_n)\in k^{n+1}$. Since $\widehat X$ is the affine hypersurface cut out by the single polynomial $F$, the [Jacobian matrix](/page/Jacobian%20Matrix) at $a$ is the $1\times(n+1)$ matrix \begin{align*} J_a=\left(\frac{\partial F}{\partial X_0}(a),\dots,\frac{\partial F}{\partial X_n}(a)\right). \end{align*} By the affine Jacobian tangent-space computation in [citetheorem:9456], applied to the affine variety $\widehat X\subset \mathbb A_k^{n+1}$ and the generator $F$ of its defining ideal near $a$, the Zariski tangent space to the cone at $a$ is \begin{align*} T_a\widehat X=\ker L_a \end{align*} as a $k$-linear subspace of $k^{n+1}$.[/step]

[guided]The point of passing to the affine cone is that tangent spaces to affine hypersurfaces are computed by first-order linearization. We define \begin{align*} \widehat X:=V(F)\subset \mathbb A_k^{n+1} \end{align*} and use the affine representative \begin{align*} a=(a_0,\dots,a_n)\in k^{n+1}\setminus\{0\} \end{align*} of $p=[a_0:\cdots:a_n]$. The first-order part of $F$ at $a$ is encoded by the $k$-linear map \begin{align*} L_a:k^{n+1}&\to k \end{align*} given by \begin{align*} L_a(v)=\sum_{i=0}^n \frac{\partial F}{\partial X_i}(a)v_i \end{align*} for $v=(v_0,\dots,v_n)\in k^{n+1}$. In matrix language, this is multiplication by the Jacobian row \begin{align*} J_a=\left(\frac{\partial F}{\partial X_0}(a),\dots,\frac{\partial F}{\partial X_n}(a)\right). \end{align*} We now apply the affine Jacobian tangent criterion in [citetheorem:9456]. Its hypotheses are satisfied because $\widehat X$ is an affine variety in $\mathbb A_k^{n+1}$ whose defining ideal is generated locally at $a$ by the single polynomial $F$. The theorem identifies the Zariski tangent space with the kernel of the Jacobian matrix at the point. Therefore \begin{align*} T_a\widehat X=\ker J_a=\ker L_a. \end{align*} This is the linear subspace of tangent vectors $v\in k^{n+1}$ satisfying \begin{align*} \sum_{i=0}^n \frac{\partial F}{\partial X_i}(a)v_i=0. \end{align*}[/guided]

[step:Show that the radial line lies in the cone tangent space] Because $F$ is homogeneous of degree $d$, each monomial of $F$ has total degree $d$. Writing $F$ as a finite sum of monomials and differentiating term by term gives Euler's homogeneous identity \begin{align*} \sum_{i=0}^n X_i\frac{\partial F}{\partial X_i}=dF. \end{align*} Evaluating at $a$ gives \begin{align*} L_a(a)=\sum_{i=0}^n a_i\frac{\partial F}{\partial X_i}(a)=dF(a). \end{align*} Since $p\in X$, the homogeneous equation defining $X$ gives $F(a)=0$. Hence $L_a(a)=0$, so \begin{align*} ka\subseteq T_a\widehat X. \end{align*} [/step]

[step:Projectivize the affine tangent space] By definition of the projective tangent space to a projective subvariety through its affine cone, the projective tangent space at $p=[a]$ is the projectivization of the affine tangent space to the cone at $a$: \begin{align*} T_pX=\mathbb P(T_a\widehat X)\subset \mathbb P_k^n. \end{align*} The previous step shows that the radial line $ka$ is contained in $T_a\widehat X$, so this projectivization is well-defined as a projective linear subspace passing through $p$. Since $T_a\widehat X=\ker L_a$, its projectivization is exactly the zero locus in projective space of the corresponding homogeneous linear form: \begin{align*} \mathbb P(T_a\widehat X)=V_+\left(\sum_{i=0}^n \frac{\partial F}{\partial X_i}(a)X_i\right). \end{align*} Thus \begin{align*} T_pX=V_+\left(\sum_{i=0}^n \frac{\partial F}{\partial X_i}(a)X_i\right). \end{align*} [/step]

[step:Verify that the equation defines a genuine hyperplane and is independent of the representative] Since $p$ is smooth on the projective hypersurface $X$, the hypersurface Jacobian criterion gives that not all partial derivatives \begin{align*} \frac{\partial F}{\partial X_0}(a),\dots,\frac{\partial F}{\partial X_n}(a) \end{align*} vanish. Therefore the displayed linear form is nonzero, and its projective zero locus is a hyperplane in $\mathbb P_k^n$. It remains to check that the formula does not depend on the chosen affine representative. If $b=\lambda a$ for some $\lambda\in k^\times$, then, because $\frac{\partial F}{\partial X_i}$ is homogeneous of degree $d-1$, we have \begin{align*} \frac{\partial F}{\partial X_i}(b)=\lambda^{d-1}\frac{\partial F}{\partial X_i}(a) \end{align*} for every $0\le i\le n$. Hence the linear form obtained from $b$ is $\lambda^{d-1}$ times the linear form obtained from $a$. Multiplying a nonzero homogeneous linear equation by a nonzero scalar does not change its projective zero locus. Therefore the hyperplane is well-defined, and the stated formula gives the projective tangent hyperplane to $X$ at $p$. [/step]