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A unibranch singular point of an algebraic variety on a plane algebraic curve. The ordinary cusp has local equation $y^2=x^3$ and parametrization $(x,y)=(t^2,t^3)$; its branch has a repeated tangent and a vanishing first derivative at the singular point. Higher-order cusps may have different local equations. A cuspidal cubic gives the basic projective example.

Take $D=\{y^{2}z=x^3\}$, $L=\{z=0\}$ and $C=3L$. This projective cuspidal cubic is irreducible and meets $L$ only at $p=[0:1:0]$. At $p$, the derivative with respect to $z$ is $y^2=1$, so $D$ is smooth. In the affine chart $y=1$, it is $z=x^3$, and $L$ is $z=0$, so $I_p(L,D)=3$. Multiplicity is linear in the divisor $C=3L$, giving

Apply a projective transformation to put the cuspidal cubic in the form $D=\{F=y^{2}z-x^3=0\}$. Since $L$ meets $D$ at only one smooth point $p$, the Bézout theorem forces intersection multiplicity three there. Hence $L$ is the tangent line at a flex.

The homogeneous Hessian matrix has determinantBy the Hessian criterion for a flex, every smooth flex lies in $D\cap \{x{y}^2=0\}$. These intersections are $[0:0:1]$, the cusp, and $[0:1:0]$, which is smooth. Thus the unique smooth flex of a cuspidal cubic is $p=[0:1:0]$, with tangent line $L=\{z=0\}$. Conversely the pair in part (b) has exactly the required properties. Therefore there is one projective-equivalence class, represented by

WriteThe equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the cuspidal cubicIt is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.

If $y=0$, the other equations becomeTheir solutions are $(0,0,0)$, which already lies on $C$, and the two pointsThus the irreducible components areEquivalently, the radical ideal of the cusp with two isolated-point components is