Parametrization of Parabola After seeing recent curve I wonder if a parametrization of red curve
of equation $ \sqrt{x}+ \sqrt{y}=1 $ can be found for extended domain/range. Parametrization  $ ( x= \cos^4 t, y=\sin^4 t \; )$ is bounded $\pm1$ for $(x,y).$

This interesting curve is a parabola, intersection of a cone touching three coordinate planes and another plane $z=1$.
3D equation of this cone with vertex at origin and touching the three orthogonal planes can be factored: ( actually I back calculated)
$$ x^2+y^2+z^2-2 xy-2 yz-2 zx=0 $$
$$ (x^2+y^2+z^2-2 xy +2 xz-2 ay)- 4 a x =0 $$
$$ (y-x-z)^2 - 4 ax =0 $$
$$y=x+z -2 \sqrt{zx} = ( \sqrt{z} -\sqrt{x})^2 $$
$$ \sqrt{y}=   \sqrt{z} -\sqrt{x} $$
So, combination of signs there are 8 cones with their 24 parabola intersections that can be packed around the origin touching the 3 orthogonal planes along contact lines at $45^\circ$ to the axes.
$$ \pm \sqrt{x} \pm \sqrt{y} \pm \sqrt{z} =0 $$
Taking for the present case
$$z=1 \rightarrow  \sqrt{x} +\sqrt{y} =1\;$$
Intersection of cones with planes parallel to generators result in parabolic arc intersections. The cones touch the coordinate planes. Hence all the parameter lines on surface are parabolas.

One approach is to convert the equation to the standard bivariate polynomial form for a conic section.  Start with the given equation $\sqrt{x}+\sqrt{y}=1$ and square both sides:
$x+2\sqrt{xy}+y=1$
$2\sqrt{xy}=1-(x+y)$, square again:
$4xy=1-2(x+y)+(x+y)^2$
Using the quarter-square multiplication formula $4xy=(x+y)^2-(x-y)^2$ we get
$1-2(x+y)+(x-y)^2=0$
Note that the variable terms are a linear term involving one combination of $x$ and $y$ and a squared linear term involving an independent linear combination of $x$ and $y$.  This combination guarantees a parabola.
The derived equation lends itself to the identification
$x-y=t$, whereupon
$x+y=(1+t^2)/2$
By taking appropriate linear combinations we solve for $x$ and $y$:
$x=(1+2t+t^2)/4=(1+t)^2/4$
$y=(1-2t+t^2)/4=(1-t)^2/4$.
