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 $)   

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 $y<0$.
