Different singularities can share a tangent cone

The field $k$ is a commutative ring and an integral domain by [F16], so $k[x]$ and then $P=k[x,y]=k[x][y]$ are integral domains by [F17]; the five displayed polynomials are elements of $P$ by [F1, F2], each generates the principal ideal $I_i=(f_i)$ of [F12], and each quotient $A_i=P\mathrm{/}{I}_i$ defines the closed subscheme $X_i=\mathrm{Spec}{A}_i$ of ${\mathbb{A}}_k^2$.

Homogeneous decomposition of the first three equations: by the unique monomial expansion [F2] and the notion of homogeneous part [F3] one has $f_1=y^2+(-3x^{2}y-2y^3)+(2x^4+y^4)$, $f_2=-y^2+(-2x^{2}y-x{y}^2)+(x^4+x^2{y}^2)$ and $f_3=(3x^{2}y-y^3)+(x^4+2x^2{y}^2+y^4)$, the last expansion using the binomial theorem for $(x^2+y^2{)}^2$ by [F19]; hence the lowest nonzero homogeneous parts are $y^2$, $-y^2$ and $3x^{2}y-y^3=y(3x^2-y^2)$, and none of $f_1,f_2,f_3$ has a term of degree $0$ or $1$.

Homogeneous decomposition of the remaining two equations: $f_4=(x^6+3x^4{y}^2+3x^2{y}^4+y^6)+(-4x^2{y}^2)$ and $f_5=x^6+(-x^2{y}^3-y^5)$ by [F2, F3] with $(x^2+y^2{)}^3$ expanded by the binomial theorem [F19], so the lowest nonzero homogeneous parts are $-4x^2{y}^2$ and $-y^3(x^2+y^2)=-x^2{y}^3-y^5$, and $f_4$ has no term of degree at most $3$ while $f_5$ has no term of degree at most $4$.

The origin is a $k$-rational point of every $X_i$: evaluation $\epsilon$ at $(0,0)$ has the maximal kernel $(x,y)$ by [F9], each $f_i$ has zero constant term by steps 1.2 and 1.3 so $f_i\in (x,y)=\ker \epsilon$, hence [F10] factors $\epsilon$ through a $k$-algebra homomorphism $A_i\to k$, which [F11] exhibits as a $k$-rational point of $X_i$, namely the origin $0$.

The multiplicities of the five equations at the origin are $2,2,3,4,5$: at $a=0$ the translated expansion $f_i(0+t)=f_i(t)$ is exactly the homogeneous decomposition of steps 1.2 and 1.3, so the least $j$ with $f_{i,j}\mathrm{\ne }0$ is $2,2,3,4,5$ for $i=1,\dots ,5$ by [F4], equivalently the $\mathfrak{m}$-adic orders of the $f_i$ in the local ring of ${\mathbb{A}}_k^2$ at the origin.

Factorization of the five lowest nonzero homogeneous parts over $k$: by [F14] the nonconstant polynomials $X^2-3$ and $X^2+1$ in $k[X]$ have roots $\sqrt{3}$ and $i$ in $k$ with $(\sqrt{3}{)}^2=3$ and $i^2=-1$; characteristic $0$ gives $2\cdot 1\mathrm{\ne }0$, $3\cdot 1\mathrm{\ne }0$ and $4\cdot 1\mathrm{\ne }0$ by [F15], so $3\mathrm{\ne }0$, $2\mathrm{\ne }0$ and $-1\mathrm{\ne }0$, and since $k$ is a field [F16] the elements $\sqrt{3}$ and $i$ are nonzero, because $3=(\sqrt{3}{)}^2$ and $-1=i^2$ would otherwise vanish; consequently $3x^{2}y-y^3=y(\sqrt{3}x-y)(\sqrt{3}x+y)$ and $x^2+y^2=(x+iy)(x-iy)$, while $x^2{y}^2=x\cdot x\cdot y\cdot y$ and $y^3(x^2+y^2)=y\cdot y\cdot y\cdot (x+iy)(x-iy)$ are immediate from commutativity.

The first two curves are genuinely different: if the ideals agreed, $(f_1)=(f_2)$, then $f_1\in (f_2)$ and $f_2\in (f_1)$, so by [F13] there are $h,g\in k[x,y]$ with $f_1=h{f}_2$ and $f_2=g{f}_1$; substituting gives $(1-hg)f_1=0$, and since $f_1\mathrm{\ne }0$ by step 1.2 and $k[x,y]$ is a domain by [F17] we get $hg=1$, so $h$ is a unit of $k[x,y]$; two applications of [F18], first with the domain $R=k[x]$ (a domain by [F17]) and then with the field $R=k$, show that each unit of $k[x,y]$ is a nonzero constant $c\in k^{\times }$, and comparing the coefficient of $x^2{y}^2$ in $f_1=c{f}_2$ by the unique expansion [F2] gives $0=c$ on the left against $c\mathrm{\ne }0$ on the right, a contradiction; hence $(f_1)\mathrm{\ne }(f_2)$ and $X_1\mathrm{\ne }{X}_2$ as closed subschemes of ${\mathbb{A}}_k^2$.

The tangent cones: each $f_i$ lies in $\mathfrak{q}=(x,y)$ because its value at the origin is zero, and $I_i=(f_i)$ is principal, so the principal case of [F5] gives ${\mathrm{in}}_{\mathfrak{q}}(I_i)=(f_{i,\min })$ and [F5, F6] identify the scheme-theoretic tangent cone as ${\mathrm{Cone}}_0(X_i)\cong \mathrm{Spec}(P\mathrm{/}(f_{i,\min }))$; explicitly these are $P\mathrm{/}(y^2)$ for $i=1$; $P\mathrm{/}(y^2)$ for $i=2$, because the scalars $\pm 1$ are nonzero and invertible in $k$ by [F15, F16] so $(-y^2)=(y^2)$ by [F13]; $P\mathrm{/}(y(3x^2-y^2))$ for $i=3$, because $3x^{2}y-y^3=y(3x^2-y^2)$; $P\mathrm{/}(x^2{y}^2)$ for $i=4$, because $-4$ is nonzero in characteristic $0$ by [F15] and invertible in $k$ by [F16], so $(-4x^2{y}^2)=(x^2{y}^2)$; and $P\mathrm{/}(y^3(x^2+y^2))$ for $i=5$, because $-x^2{y}^3-y^5=-y^3(x^2+y^2)$.

All five tangent spaces are two-dimensional: none of the $f_i$ has a linear term by steps 1.2 and 1.3, so each formal partial derivative of [F7] has zero constant term and the $1\times 2$ Jacobian matrix at the origin is the zero matrix $(0,0)$; the kernel of the zero map $k^2\to k$ is $k^2$, so [F8] gives canonical $k$-linear isomorphisms $T_0{X}_i\cong \ker (0:k^2\to k)=k^2$ of dimension $2$, for all five curves at once.

The cone of the first two curves is a doubled line, retained nonreduced: in $R=k[x,y]\mathrm{/}(y^2)$ the class $\bar{y}$ satisfies ${\bar{y}}^2=0$, and $\bar{y}\mathrm{\ne }0$, since if $y\in (y^2)$ then [F13] would give $y=y^{2}h$ for some $h\in k[x,y]$, whereas division by the monic polynomial $y^2\in k[x][y]$ [F20] writes every $g\in k[x,y]$ uniquely as $g=q{y}^2+(a+by)$ with $a,b\in k[x]$, so the classes of $1$ and $\bar{y}$ form a $k[x]$-basis of $R$ and in particular $\bar{y}\mathrm{\ne }0$; hence the cone $\mathrm{Spec}R$ is the line $y=0$, the $x$-axis of ${\mathbb{A}}_k^2$, with a first-order thickening, that is, the doubled line, and no reduction is performed in the cone computation.

The distinct linear factors in each product are pairwise non-proportional; repeated factors record the stated line multiplicities. If $ay=b(\sqrt{3}x-y)$ then comparing coefficients by [F2] gives $b\sqrt{3}=0$ and then $a=0$, so $a=b=0$ because $\sqrt{3}\mathrm{\ne }0$ and $k[x,y]$ is a domain by [F17]; if $a(\sqrt{3}x-y)=b(\sqrt{3}x+y)$ then $\sqrt{3}(a-b)=0$ and $a+b=0$, so $a=b=0$ because $2\mathrm{\ne }0$ and $k$ is a field [F16]; the same coefficient comparisons using $i\mathrm{\ne }0$ and $2i\mathrm{\ne }0$ show that $y$, $x+iy$ and $x-iy$ are pairwise non-proportional, and $x$ and $y$ are non-proportional since $ay=bx$ forces $a=b=0$; hence the cone of $X_1$ and $X_2$ is the line $y=0$ occurring with multiplicity two, the cone of $X_3$ consists of the three distinct lines cut out by $y$, $\sqrt{3}x-y$ and $\sqrt{3}x+y$, each occurring once, the cone of $X_4$ is the product of the two coordinate axes each occurring twice, and the cone of $X_5$ is the line $y=0$ occurring three times together with the two distinct further lines cut out by $x+iy$ and $x-iy$.

Conclusion: the five singular plane curves have the tangent spaces $T_0{X}_i\cong k^2$ by step 3.2 and the multiplicities $2,2,3,4,5$ by step 2.2, with the tangents of the third, fourth and fifth visualized as three distinct lines, the two coordinate axes each doubled, and a triple line with two further lines by steps 2.3 and 4.2; the curves $X_1$ and $X_2$ are different closed subschemes by step 2.4, yet their multiplicity $2$, their tangent cone $\mathrm{Spec}(k[x,y]\mathrm{/}(y^2))$ and their tangent space $k^2$ all agree, so two different equation singularities, distinguished by the source as a tacnode and a ramphoid cusp, share one tangent cone.

Boundary and scope dispositions: every curve and cone is nonempty, the origin being a $k$-rational point of each $X_i$ by step 2.1; the zero cases are the zero Jacobian matrix $(0,0)$ of step 3.2 with kernel all of $k^2$ and the nilpotent $\bar{y}$ with ${\bar{y}}^2=0$ and $\bar{y}\mathrm{\ne }0$ of step 4.1; each curve has one defining equation, one Jacobian row and one cone equation by steps 1.1, 3.1 and 3.2; the degenerate case is the nonreduced cone $k[x,y]\mathrm{/}(y^2)$, retained with its nilpotent and reduced nowhere in steps 3.1, 4.1 and 4.2; the endpoint of the degree filtration is the lowest nonzero homogeneous part, which exists because each $f_i$ is a nonzero polynomial with a finite expansion by steps 1.2 and 1.3 and is attained at degrees $2,2,3,4,5$ by step 2.2, and the extremal lines $y=0$ and the conjugate pair $x\pm iy=0$ are treated in steps 2.3 and 4.2; no choice principle is used, since $\sqrt{3}$ and $i$ are single roots supplied by algebraic closedness in step 2.3, every ideal, cone, Jacobian and factorization is exhibited explicitly, and all cited suppliers are choice-free; no biconditional is asserted or needed, the only two-sided statement used being the characterization of units in step 2.4, whose forward direction is applied to the unit $h$ and never in reverse, so both iff cases are vacuous.