Parametrizing the curve $2 x^2 + x^4 + 2 y^2 = 1$

Question

Do you know how one can find a parametrization of the following curve? $$2 x^2 + x^4 + 2 y^2 = 1$$

Thanks a lot in advance!

Answer 1

$$x^4+ 2x^2= (x^2)^2+ 2(x^2)= (x^2)^2+ 2(x^2)+ 1- 1= (x^2+ 1)^2- 1 \tag{1}$$
We can write $$2x^2+ x^4+ 2y^2= (x^2+ 1)^2- 1+ 2y^2= 1 \tag{2}$$ so $$(x^2+ 1)+ 2y^2= 2 \tag{3}$$

Let $x^2+ 1= \cos\theta$, so that $x= \sqrt{\cos\theta-1}$ and $y= \frac{1}{\sqrt{2}}\sin\theta$. Of course, that only works for $x$ positive. For $x$ negative, use $x= -\sqrt{\cos\theta- 1}$.

(Parcly Taxel, I do NOT see a "rather trivial" parameterization! Please enlighten us.)

Answer 2

We choose the parameter $t$ such that $$2x^2+x^4=\cos^2 t,\qquad 2y^2=\sin^2 t\ .$$ Solving the second equation gives $y(t)={1\over\sqrt{2}}\sin t$. Solving the first equation gives $$x^2=-1+\sqrt{1+\cos^2 t}={\cos^2 t\over\sqrt{1+\cos^2 t}+1}\ ,$$ so that $$x(t)={\cos t\over\sqrt{\sqrt{1+\cos^2 t}+1}}\ .$$ The full parametric representation then is $$\gamma:\quad t\mapsto\left\{\eqalign{x(t)&={\cos t\over\sqrt{\sqrt{1+\cos^2 t}+1}}\cr y(t)&={1\over\sqrt{2}}\sin t\cr}\right.\qquad(0\leq t\leq2\pi)\ .$$

Answer 3

Tweaked from the ellipse

$$ 2 x^2+ x^4= \cos ^2 \phi;\quad 2 y^2= \sin ^2\phi ;$$ to solve a quadratic equation:

$$(x,y)=\pm(\sqrt{\sqrt{1+\cos^2\phi}-1},\frac{\sin \phi}{2}), (\phi,-\pi,\pi) $$

Axes intercepts $$\pm\sqrt{(\sqrt{2}-1)},\frac{\pm1}{\sqrt2}$$

Angle $\phi$ corresponds to the eccentric anomaly of the standard ellipse.