Any irreducible ipersursurface in $X \subset \mathbb{P}^n$ of degree $d$ with an isolated singular point $p \in X$ of multiplicity $d$ is a cone over a hypersurface $Y$ of degree $d$ in $\mathbb{P}^{n-1}$.

In fact, take any point $q \in X$ different from $p$. Then the line $\overline{pq}$ has at least $d+1$ intersections with $X$ (counted with the right multiplicities), so by Bézout theorem it must be contained in $X$. But then $X$ is a cone of vertex $p$.

Coversely, if $Y \subset \mathbb{P}^{n-1} \subset \mathbb{P}^n$ is a hypersurface, then the cone $X$ vith vertex a point $p \notin \mathbb{P}^{n-1}$ is a hypersurface $X \subset \mathbb P^n$ with a point of multiplicity $d$.

Any irreducible hypersursurface in $X \subset \mathbb{P}^n$ of degree $d$ with an isolated singular point $p \in X$ of multiplicity $d$ is a cone over a hypersurface $Y$ of degree $d$ in $\mathbb{P}^{n-1}$.

In fact, take any point $q \in X$ different from $p$. Then the line $\overline{pq}$ has at least $d+1$ intersections with $X$ (counted with the right multiplicities), so by Bézout theorem it must be contained in $X$. But then $X$ is a cone of vertex $p$.

Coversely, if $Y \subset \mathbb{P}^{n-1} \subset \mathbb{P}^n$ is a hypersurface of degree $d$, then the cone $X$ of vertex a point $p \notin \mathbb{P}^{n-1}$ is a hypersurface $X \subset \mathbb P^n$ with a singular point of multiplicity $d$.