Is the affine curve $y^2=x^4+y^4$ in $\mathbb A^2$ singular?

Question

Is the affine curve $y^2=x^4+y^4$ in $\mathbb A^2$ singular or nonsingular? Find the singularities and show the types of the singularities if the curve is singular.

Let $f(x,y)=x^4+y^4-y^2$.

$f(x,y)=0$ at $(0,0), (0,1), (0,-1)$.

$\frac{\delta f}{\delta x}(x,y)=4x^3=0$ at $(0, y)$.

$\frac{\delta f}{\delta y}(x,y)=4y^3-2y=0$ at $(x, 0), (x, \pm \frac{1}{\sqrt{2}})$.

The only consistent point in these three equations is $(0,0)$. So does that mean there is one singularity $P_1=(0,0)$? And if so, what 'type' is this singularity?

Answer 1

You can use fooplot and parametric equations to see the graph.

Graph of $x^4+y^4=y^2$ at http://fooplot.com/plot/4bk34e5n3r

EDIT: The fooplot graph is not working, so here is an animated GIF

Answer 2

You are right about the location of the one singularity on the curve.

One way to classify the singularity is to draw the set over $\Bbb R$.

If that is not your style, note that the equation may be rewrittten into $x^4 = y^2 - y^4$. Around the origin $y^2$ dominates $y^4$, so the curve looks like $x^4 = y^2$, or $x^2 = |y|$.