How does $axy+bx+cy+d=0$ represent a rectangular hyperbola?

Question

I'm told a curve of this form $$axy+bx+cy+d=0$$ is a rectangular hyperbola.

But I don't seem to understand where are it's $x^2$ and $y^2$ terms since $x^2-y^2=a$ is also a rectangular hyperbola.

Sorry for the boring question, but did I learn something wrong here?

Answer 1

1. Yes, $x^2-y^2=a$ with $a\neq 0$ is equation of a rectangular hyperbola.
   Its asymptotes are $y=\pm x$ and axes of symmetry $x=0$ and $y=0.$

2. I assume you know that $y={1 \over x}$ also defines a rectangular hyperbola obtained by a rotation of the first one, where $a=\pm 1.$ The asymptotes are now $x=0$ and $y=0,$ the axes are $y=\pm x.$

The equation $axy+bx+cy+d=0$ (here $a$ is not that from the above part) defines a hyperbola of the second type: its axes are parallel to $y=\pm x,$ thus they are orthogonal.
They pass through the center of the hyperbola $\left(-\frac{c}a,-\frac{b}a\right).$

For more information see this wikipedia article.

Answer 2

$$x^2 -y^2 =a^2$$

when rotated by $45^{\circ}$, becomes

$$ xy = a^2/2 $$

Our curve $$ xy + \dfrac{b}{a}x + \dfrac{c}{a}y \color{blue}{+ \dfrac{bc}{a^2}} = \dfrac{-d}{a} \color{blue}{+\dfrac{bc}{a^2}}$$

is

$$ \Big(x+\dfrac{c}{a}\Big)\Big(y+\dfrac{b}{a}\Big)=\dfrac{bc-ad}{a^2}$$

in the form $$ XY = K$$

So it is also a rectangular hyperbola.

Answer 3

You need to rotate the coordinates by $45°$ to see the correspondence, which is achieved by the transformation

$$x=u+v,\\y=u-v.$$

If we plug this in $x^2-y^2$, we obtain

$$(u+v)^2-(u-v)^2=4uv,$$ which gets closer to the desired form. (Conversely, $xy=u^2-v^2$.)