Simple method to determine if a graph exists only in the positive quadrant. 
How can one tell by inspection or by some simple tests, preferably without using calculus, that a graph exists only in the positive quadrant only (or for that matter in any one quadrant only)? 

An example is the graph
$$(bx-ay)^2=2ab\left(bx+ay-\frac {ab}2\right)$$
from the question/answer here.
It is not immediately obvious. Setting RHS$>0$ does not help either. 

(Revised solution)
The equation of the curve (assuming $a, b>0$) is 
$$\begin{align}
(bx-ay)^2&=2ab\left(bx+ay-\frac {ab}2\right)\\
b^2x^2-2abxy+a^2y^2&=2ab(bx+ay)-a^2b^2\end{align}$$
$$\small\begin{array}
{l|l}
\hline
\text{Form quadratic in $x$:}
&\text{Form quadratic in $y$:}\\
\;\;b^2x^2-2ab(b+y)x+a^2(y^2+b^2)-2a^2by=0 
&\;\;a^2y^2-2ab(ax+x)y+b^2(x^2+a^2)-2ab^2x=0\\
\;\;b^2x^2-2ab(b+y)x+a^2(y-b)^2=0
&\;\;a^2y^2-2b(a+x)y+b^2(x-a)^2=0\\
\text{For real $x$}, 
&\text{For real $y$}, \\
\;\;[-2ab(b+y)]^2\geq 4\cdot b^2\cdot a^2(y-b)^2
&\;\;[-2ab(a+x)]^2\geq 4\cdot a^2\cdot b^2(x-a)^2\\
\;\;(y+b)^2-(y-b)^2\geq 0
&\;\;(x+a)^2-(x-a)^2\geq 0\\
\;\;4by\geq 0
&\;\;4ax\geq 0\\
\;\;y\geq 0
&\;\;x\geq 0\\
\hline
\end{array} 
$$
i.e., for real $x,y$, the curve exists for 
$$\begin{cases}
x\geq 0\\
y\geq 0\end{cases}
\qquad 
\text{i.e. positive quadrant only}$$
