A continuous class of curves (solutions to a polynomial equation) degenerating at some mysterious value! Find that value.

Question

Depending on $a$ -- I am mostly interested in $ -1 \leq a \leq +1 $ -- the solutions of $$ y^{2} -\left(1+x\right)x^{2} = a $$ as a subset of $\mathbb{R}^2$,

1) Looks like a human figure at about $ a= 0.05$, then as $a$ decreases, the neck becomes thinner and thinner, until

2) at $a=0$ (precise) it collapses to a point. So the figure is an $\alpha$ sign shape with the self-intersection at $(0,0)$.

3) Immediately as $a$ becomes negative, the shape is a disjoint union of a "loop" on the left hand side of $y$-axis, and a "parabola"-like piece on the right hand side of $y$-axis. For $a$ negative but close to zero together they look like a robot/alien with head magnetically but not physically joined to the body :)) However,

4) Somewhere around $a=-.1481$, we lose the head! The loop degenerates to a point and vanishes immediately past that $a$ values.

Question: What is the exact value of $a$ where this occurs?

Conjecture: From zooming in and in and in, I can see that $a=48148148148...$ is a strong candidate for the answer. But why?!

I have used https://www.desmos.com/calculator/jgziy7rmdf for all of this. Thanks!

Answer 1

Ok! Found it. This occurs when the pre-image of $$ f:\mathbb{R}^2 \to \mathbb{R} $$ given by $$ f(x,y)=y^2 - (1+x)x^2 $$ is NOT a manifold.

Since $f$ is $C^\infty$, this is exactly when $a$ is not a regular value of $f$. In other words, somewhere along $f^{-1}(a)$, the rank of the derivative of $f$ is less than $1$, i.e. derivative is zero.

Computing $Df(x,y)$, it vanishes only at $(0,0)$ and $(-2/3,0)$ which are on the pre-images of $0$ and $4/27$, resp. $$ 4/27=0.14814814814... $$

Answer 2

COMMENT.-My viewpoint is there is no a "mysterious value". We have the cubic curve $$y^2=x^3+x+a$$ whose discriminant is $\Delta=-16(4+27a^2)$. It follows that when $a\ne0$ the equation in RHS has three distict roots so the cubic is elliptic (without oval when $a\gt0$ and with oval when $a\lt0$). In the case of $a=0$ we have the very known nodal cubic (with singularity, of genus $0$, so non elliptic of genus $1$). I guess this nodal cubic is your solution because it is in fact "degenerate" respect of all the other elliptic curves.