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. 
 A: In general, no. Even with calculus, there is no general procedure for proving that an arbitrary implicit function $f(x,y)$'s zero level set lies only in the first quadrant.
If your curve lies within a bounded region of space, you can cut up that region into small patches, and try to bound the function on each patch, to prove that each patch that lies outside the first quadrant cannot contain a portion of the zero level $f(x,y)=0$. This may require using calculus, and very small patches.
One simpler strategy is to detect if the curve crosses either the $x$ or $y$ axis (by plugging in $y=0$ and finding all roots of the resulting univariate function of $x$, which in itself is not always an easy problem). If there are no crossing points, you still need to somehow determine that there are no connected components of your curve that lie entirely outside of the first quadrant (such as is the case for $(x+4)^2+(y+4)^2-1=0,$ for example).
A: (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 
$$\begin{cases}
x\geq 0\\
y\geq 0\end{cases}
\qquad 
\text{i.e. positive quadrant only}$$
