Parametrization of Parabola

Question

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.

Answer 1

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 $y$:

$x=(1+2t+t^2)/4=(1+t)^2/4$

$y=(1-2t+t^2)/4=(1-t)^2/4$.

Answer 2

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.