Parametrization of a lemniscate

Question

I wonder how to parametrize lemniscate in the following way:

$\gamma : (-\infty, \infty) \rightarrow \mathbb R^2,$whose image is a lemniscate whose axes of symmetry are $y=x$ and $y+x=0$. The lemniscate is tangent to $x$ axis at $t=0$ and tangent to $y$ axis when $t \to \pm \infty$. When $t>0, \gamma(t)$ lies in the first quadrant and when $t<0, \gamma(t)$ lies in the third quadrant. The lemniscate is described in Foundations of Differentiable Manifolds and Lie Groups, page 25.

Here is a parametrization of such a lemniscate whose domain is $[0,\pi/2]$, but I don't know how to modify this example to meet the conditions above.

Answer 1

Hmm this is an interesting question. The best I was able to do was to achieve this: (the animation goes from t=-5..5, but satisfied the conditions from $[-\infty,\infty]$):
I attempted to first rebrand the function. In the above form it is: $$\gamma:\left\{\begin{array}\\x=\sqrt{\sin(2t)}\cos t \\ y=\sqrt{\sin(2t)}\sin t\end{array}\right.$$ I attempted to rebrand it into a different parameter which has the following parametrization: $$\gamma:\left\{\begin{array}\\ x=\frac{|t|}{t}\sqrt{\sin\left(\pi\left(1-e^{-t^2}\right)\right)}\cos\left(\frac{1}{2}\pi\left(1-e^{-t^2}\right)\right) \\ y=\frac{|t|}{t}\sqrt{\sin\left(\pi\left(1-e^{-t^2}\right)\right)}\sin\left(\frac{1}{2}\pi\left(1-e^{-t^2}\right)\right)\end{array}\right.$$

Answer 2

r[si_] = 2 Tan[si]/Sqrt[1 + (2 Tan[si])^2]; 
th[si_] = 
 Abs[ArcTan[Cos[si], 2 Sin[si]]/2]; ParametricPlot[{r[si] Cos[th[si]],
   r[si] Sin[th[si]]}, {si, 0, 2 Pi}, GridLines -> Automatic, 
 PlotStyle -> {Purple, Thick}, PlotLabel -> "Lemniscate_Bernoulii"]

EDIT1:

An earlier typo was corrected.

$(r,\theta)$ are functions of tangent angle si. Above code in Mathematica gives the full Bernoulli Lemniscate, however not without a jump plot trace at si=$\pi$ along $ x=y$. Difficulty here due to self-intersecting Lemniscate curve at origin.