Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I do to plot this function by myself not by WolframAlpha. Thank you in advance.
 A: Since the equation implies $\sqrt{x}+\sqrt{y}\geq 0$, therefore it is equivalent to $\sqrt{x}+\sqrt{y}=\sqrt{5}$. Therefore because $y\geq 0$, $y=(\sqrt{5}-\sqrt{x})^2=x+5-2 \sqrt{5}\cdot \sqrt{x}$. And we know from the condition of $\sqrt{x}, \sqrt{y}\geq 0$ that $0\leq x, y\leq 5$. Therefore, finally, the original statement is the same as $y=x+5-2\sqrt{5} \sqrt{x} (x\in[0, 5])$.
To do a simple graphical analysis (though may not be necessary), We consider the $1/4$ arc of $(x-5)^2+(y-5)^2=25 (0\leq x, y\leq 5)$, which we can rewrite as
\begin{align*}
y&=5-\sqrt{10x-x^2}\\
&=x+5-\sqrt{10x+2x\sqrt{10x-x^2}}\\
&\geq x+5-\sqrt{10x+2x\cdot \sqrt{25}}\\
&=x+5-2\sqrt{5}\sqrt{x}
\end{align*}
, thus indicating that the whole graph lies 'below' the circle.

A: $$
\begin{align}
& (x^{\frac{1}{2}}+y^{\frac{1}{2}})^{2} = 5\\
\implies & x^{\frac{1}{2}}+y^{\frac{1}{2}} = \pm \sqrt{5}\\
\implies & \sqrt{x}+\sqrt{y} = \pm \sqrt{5}
\end{align}
$$
The inequality $0\le x,y\le 5$ holds for real $x$ and $y$, as mentioned in the previous answers.
Method 1: Plot $f(x,y)=|\sqrt{x} + \sqrt{y} - \sqrt{5}|$
We take real numbers $x$ and $y$ from $0$ to $5$ and plot $f(x,y)=|\sqrt{x} + \sqrt{y} - \sqrt{5}| < \epsilon$ for very small $\epsilon$ (here $1e-9$ on a $100\times 100$ grid):

Method 2: Plot individual points
$$
x = 4y\\
\implies y = \dfrac{5}{9}\\
\implies x = \dfrac{20}{9}
$$
Similarly, $x=9y\implies (x,y)=(45/9,5/9)$, $x=16y\implies (x,y)=(80/25,5/25)$, and in general $x=n^{2}y\implies (x,y)=(\dfrac{5n^2}{(n+1)^2},\dfrac{5}{(n+1)^2})$.
Since the equation is symmetric for in $x$ and $y$, we plot both $(x,y)$ (shown in blue) and $(y,x)$ (shown in red).

Note: the point at the center of the curve is both red and blue (overlapping). Different choices of $x(y)$ give different distribution of points.
A: One strategy is to try a different coordinate system. Since the equation is symmetric in $x$ and $y$, then a new coordinate system where the axes are the "diagonals" in the usual coordinate system could be helpful.
So with that in mind, let $u=x+y$ and $v=x-y$. Then the $u$- and $v$-axes are those diagonal axes as in the image below. And
$$\begin{align}\left(\sqrt{x}+\sqrt{y}\right)^2&=5\\
x+2\sqrt{xy}+y&=5\\
u+\sqrt{(u+v)(u-v)}&=5\\
(u+v)(u-v)&=(5-u)^2\\
-v^2&=25-10u\\
u&=\frac{v^2}{10}+\frac{5}{2}
\end{align}$$
So it's a parabola in the $(u,v)$ coordinate system centered on the $u$-axis, shifted "up" by $\frac{5}{2}$, with leading coefficient $\frac{1}{10}$.
Since $(u,v)=(1,0)$ corresponds to $(x,y)=(1/2,1/2)$; and $(u,v)=(0,1)$ corresponds to $(x,y)=(1/2,-1/2)$, you can draw it this way:

This parabla-sketching technique is to start at $(5/2,0)$ on the $u$-axis. Be sure to use distances on the $uv$ system, not the $xy$ system.

*

*Then move one unit in the $v$ direction, and $\frac{1}{10}$ "up".

*Then one unit in the $v$ direction, $\frac{3}{10}$ "up".

*Then one unit in the $v$ direction, $\frac{5}{10}$ "up".

*Then one unit in the $v$ direction, $\frac{7}{10}$ "up".
And so on.

This parabola is to much, since it includes points with $x>5$ or $y>5$. Those were introduced in the algebra above when $(5-u)$ was squared. So the original equation's graph is the picture above, but cropped at $x=5$ and at $y=5$.
