Let $f(x,y)$ be a complex degree $d$ polynomial that has this particular 
form.


$$ f = \frac{f\_{02}}{2} y^2 + \frac{f\_{21}}{2} x^2 y + 
\frac{f\_{12}}{2} x y^2 + \frac{f\_{03}}{6} y^3 + \frac{f\_{40}}{24} x^4+ \ldots $$


This polynomial $f$ can be thought of as an element of $\mathbb{C}^{M\_d}$, where
$M\_d = \frac{d^2+3d-10}{2}$. 
Note that aside from vanishing at the origin, the following derivatives
at the origin also vanish
$$ f\_{10}, f\_{01}, f\_{20}, f\_{11}, f\_{30}=0.$$


Let us now define a subset of 
$$ A\_4^1 \subset \mathbb{C}^M\_d \times \mathbb{C}^2$$ 
given by 


$$ A\_{4}^1:= ( (f,x,y) \in \mathbb{C}^{M\_d} \times \mathbb{C}^2 : 
f(x,y)=0, ~~f\_{x} =0, ~~ f\_{y} =0, 
~~ (x,y) \neq (0,0), ~~ f\_{02} \neq 0, $$


$$ f\_{40} f\_{02} - 3 f\_{21}^2 =0. )$$


I have a question regarding the closure of the space 
$\overline{A\_{4}^1}$. Suppose 
the curve $x(t) = L\_1 t$ and $ y(t) = t^2 $, $t\neq 0$, lies 
in the space $A\_4^1$ for all $t\neq 0 $.
Further suppose that $f\_{02}(t) = L\_2 t^r$, for 
some $r > 0$. Assume that $L\_1$ and $L\_2$ are 
fixed non zero complex numbers (they don't depend on $t$).


What happens to the derivatives $f\_{ij}$ in the limit 
as $t$ tends to zero? We basically want to see what 
happens in the closure when you approach it 
via the path $ x = L\_1 t$, $y = t^2$ and $f\_{02} = L\_2 t^r$.


It is easy to see that $f\_{21}$ will tend to zero, 
using the equation $f\_{y}=0$. Further, using that 
$f\_{21}$ will tend to zero and using $f\_{x} =0$
we get that $f\_{40}$ will go to zero. I expect 
another condition to come up, using the fact that 
$$ f\_{40} f\_{02} - 3 f\_{21}^2 =0.$$


In fact, I expect (but can't prove) that in the limit 


$$ \frac{-f\_{31}^2}{24} + \frac{f\_{50} f\_{12}}{40} =0.$$


In any case even if that last claim is wrong, I still expect
another condition to come up. The remaining coefficients can not
be arbitrary is what I think. May be we get different conditions
depending on what $r$ is? 


This may seem like a random question, but let me explain 
intuitively what I am asking. Look at the form of the function $f$ 
that I have taken. This curve has an $A\_3$ singularity (a tacnode)
at the origin. What this is means is that at the origin, the 
first derivatives vanish, the Hessian has a Kernel ( which we have 
fixed to be $(1,0)$) and the third derivative along the kernel
of the Hessian is zero. 
The condition 
$$ f\_{40} f\_{02} - 3 f\_{21}^2 =0$$
is the condition for an $A\_4$ singularity.
Hence, the space $A\_4^1$ is the space of curves
with an $A\_4$ singularity at the origin 
and one node at a point distinct form the origin. I wish to 
know how much more singular the curves becomes if the two 
points come together in the particular way I said
i.e $ x = L\_1 t$, $y = t^2$ and $f\_{02} = L\_2 t^r$. 
The conditions $f\_{02}=0$, $f\_{21} =0$ and $f\_{40} =0$
 imply that the curve is at least as singular as 
a $D\_6$-node. I expect it to be as singular 
as a $D\_7$ node which is the condition 


$$ \frac{-f\_{31}^2}{24} + \frac{f\_{50} f\_{12}}{40} =0.$$



Closure of singular points
If I understand correctly, you ask what can be the results of the collision of two singular points, of $A\_4$ and $A\_1$ types. In general there does not seem to exist an ultimate effective method to treat such questions. Only in some simple cases, for example in this case.


A somewhat simpler question is: given a point of some singularity type, to which other types can it split by deformation? Of course, it is enough to classify only the "primitive splittings", i.e. those that cannot be factorized through others.


In your particular case, if one restricts to ADE types, you ask: Which types deform to $A\_4+A\_1$? For ADE's you can use the classical criterion (Grothendieck/Brieskorn/Lyashko) that says: a type S (one of ADE's) deforms to a bunch of types $(S\_1,..,S\_k)$ iff the disjoint union of Dynkin diagrams of $(S\_1,..,S\_k)$ is obtained from that of $S$ by erasing some vertices.


Therefore you get immediately, that $D\_7$, $E\_6$ and $A\_6$ deform to $A\_4+A\_1$ while $D\_6$ does not deform.


Unfortunately for higher singularities no such simple general "iff" criterion is known. In your particular case, however the deformation of any other singularity to $A\_4+A\_1$ factorizes through these "prime" splittings.


You can see some additional results and references in [my paper](http://lanl.arxiv.org/abs/0708.1228).


