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.
 A: If you are positive that it is a parabola, the solution is simple enough:

Give a parametrization of a parabola with vertex at $(\frac14,\frac14)$ and symmetry axis $y=x$ that passes through $(1,0)$

The answer is then
$$
(x,y)=\left(\frac14,\frac14\right)+t(1,-1)+f(t)(1,1)
$$
and one can derive that $f(t)=t^2$ for it to pass through $(1,0)$ at $t=\frac12$. Hence
$$
x=\frac14+t^2+t\\
y=\frac14+t^2-t
$$
Still, I have not considered why you know it to be a parabola.
A: 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 $y$:
$x=(1+2t+t^2)/4=(1+t)^2/4$
$y=(1-2t+t^2)/4=(1-t)^2/4$.
