Parameterizing a quadric curve

Question

While studying my course in Algebraic Curves, I found myself struggling with parameterizing the following curve:

$F=X_0^2X_1^2-X_0^2X_2^2+X_1^2X_2^2$

where $V(F) \in \mathbb {P}_\mathbb C^{2}$

I know I can parameterize it using Puiseux series, but as I know that the curve is irreducible and parameterizable, as for Eisenstein criteria on the affine dishomogeneus polynomial and evaluating as a polynomial with respect variable $X$ for the prime $p=X^2$*, and it has 3 singular points, so it is parameterizable (for an irreductible curve of degree d, the maximum number of points is $(d-1)(d-2)/2$ getting the equality iff the curve is parameterizable), I want to know if there would be another way to parameterize the curve without using Puiseux series.

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Edit: the dishomogeneus polynomial is $f=X^2-Y^2+X^2Y^2$ so you can apply Eisenstein as $X^2$ isn't divisor of $X^2-1$ and $X^4$ doesn't divide $X^2$, so Eisenstein criteria does work, and we can guarantee that f is irreductible, so F does.

Answer 1

I would de-homogenize in a slightly different way, although this does not matter (but i like the symmetry), to get the equation $$ -X^2-Y^2+X^2Y^2=0\ , $$ i.e. after adding one on both sides: $$ (1-X^2)(1-Y^2)=1\ . $$ From here we take $X$ as parameter, compute the series for $1/(1-X^2)=1+\dots$ involving only even powers of $X$, then take the square of the series to extract $iY$.

Answer 2

Using Pythagorean triples, I suggest $$ X_0 = u\,v,\;\; X_1 = t\,u,\;\; X_2 = v\,t $$ where $$ t = 2\,a\,b,\;\; u = a^2-b^2,\;\; v = a^2+b^2 $$ because $$ F/(X_0X_1X_2)^2 = 1/X_2^2 - 1/X_1^2 + 1/X_0^2 $$ and $\, v^2 = t^2+u^2.\,$ Now you can pick any parametrization of $\,a\,$ and $\,b\,$ that you like.