Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola. 
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also for $\sqrt {x}-\sqrt {y}=1 $)   
 A: Rearranging, we get $\sqrt{y} = 1 - \sqrt{x}$, which becomes $y = 1 - 2\sqrt{x} + x$ when we square both sides. Rearranging again and squaring both sides, we get $(y-x-1)^2 = y^2 + x^2 + 1 - 2xy - 2y + 2x = 4x$.
Generally, when there is an equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if $B^2 - 4AC = 0$, then the curve is one of the following:


*

*A parabola

*2 Parallel Lines (certainly this is not the case)

*1 Line (certainly this is not the case)

*No curve (certainly this is not the case)


Therefore, $\sqrt{x} + \sqrt{y} = 1$ is a part of a parabola.
(Credits partly go to WyzAnt)
A: From your equation, you get:
$$y = (1-\sqrt{x})^2= 1-2\sqrt(x)+x$$
$$(x - y + 1)^2 = 4x$$
Now transform the variables:
$$t = x+ y,\ \ s = x-y$$
$$(s+1)^2 = 2(s+t)$$
Or:
$$t =\frac{s^2+1}{2}$$
Which is a parabola. This is only part of a parabola, since when $x$ or $y$ are smaller than zero, or larger than $1$, the original equation is undefined, since if either $x$ or $y$ were greater than one, the square root would have to return a negative value. This condition means your parabola is cropped by the square $[0,1]\times[0,1]$
A: Squaring, we get $x + y + 2\sqrt{xy} = 1$; moving the $2\sqrt{xy}$ term to one side, and squaring again, we obtain $$
x^2 -2xy + y^2 -x -y + 1 = 0$$
The value of the determinant $$\begin{vmatrix}A & B \\ B & C\\\end{vmatrix}
$$ will tell us which type of curve is represented by a quadratic relation $$Ax^2 + 2Bxy + Cy^2 + Dx + Ey+F = 0;$$ we have a parabola if and only if the determinant is 0.  (Details) Here $A=1, B=-1, C=1$, so the determinant is $\begin{vmatrix}1 & -1 \\ -1 & 1\\\end{vmatrix} = 0$, and this is indeed a parabola, truncated because it omits the points where $x<0$ or $y<0$.
A: Square both sides:
$$x + 2 \sqrt{xy} + y = 1.$$
Isolate the radical and square:
$$xy = \left(\frac{1-x-y}{2}\right)^2.$$
Rotate the axes by $\pi/4$ to new axes ${X,Y}$:
$$\frac{Y^2-X^2}{2} = \left(\frac{1 - \sqrt{2}Y}{2}\right)^2.$$
Simplfying gives:
$$X^2 = 2\sqrt{2}Y - 1,$$
which is the equation for a parabola.
