I would like to know how to derive the formula for the right-side branch of $Axy + Bx + Cy = D$, where the constant and coefficients are positive integers, and expressed as an equation and in terms of $A$, $B$, $C$ and $D$.
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Try parametrizing – Karn Watcharasupat Jul 19 '18 at 04:33
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What happens if you try solving for $x$? The denominator then is $Ay+B$ so if that's not zero there is a unique $x.$ I haven't thought about "right side branch" -- it seems there would be only one branch anyway – coffeemath Jul 19 '18 at 06:11
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@coffeemath, there are two branches (see here) – NeonNarwhal Jul 19 '18 at 10:30
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The equation for the right side or the left side is the same.
The only difference is the domain.
We need to find the vertical asymptote and select the domain accordingly.
For example if you have $$3 xy+2x-2 y=5.$$
Solving for $y$ we get $$ y=\frac {5-2x}{3x-2}$$
The vertical asymptote is $x=2/3$, therefore the right branch occurs at $$y=\frac {5-2x}{3x-2}, x>2/3. $$
Mohammad Riazi-Kermani
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Take
$$|A|xy+|B|x+|C|y=|D|$$
and let's assume $x$ and $y$ are real.
The solution is
$$\\y=\frac{-|B|x+|D|}{|A|x+|C|}.$$
The graph of the function $f$ with $y=f(x)$ is a hyperbola. It has a pole at $x=-\frac{|C|}{|A|}$. The right-side branch is therefore
$$f\colon (-\frac{|C|}{|A|},\infty)\to\mathbb{C},x\mapsto f(x)=\frac{-|B|x+|D|}{|A|x+|C|}.$$
IV_
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