Hyperbolas - Standard Form

Question

This is probably a simple question but if $y = \frac{1}{x}$ is a hyperbola, then how does it comply with the standard form of a hyperbola?

Answer 1

In general, for any angle of rotation $\theta$ from the $x-$axis, we may apply the following change of variable to put a hyperbola in standard form:

$$x = x\cos\theta - y'\sin\theta$$ $$y = x'\sin\theta +y'\cos\theta.$$

Notice our hyperbola $xy = 1$ is rotated by an angle of $\frac{\pi}{4}$ from the $x-$axis.

Hence we let, $$x = \frac{x' - y'}{\sqrt{2}}$$ and $$y = \frac{x' + y'}{\sqrt{2}}.$$

So in standard form, we get

$$xy = \frac{x' - y'}{\sqrt{2}}\frac{x' + y'}{\sqrt{2}} = \frac{x'^2}{2} - \frac{y'^2}{2} = 1.$$

Answer 2

Credits to this guy on Yahoo.

Let $\displaystyle w = \frac{x+y}{\sqrt{2}}$ and $\displaystyle z = \frac{x-y}{\sqrt{2}}$. This means that $\displaystyle x = \frac{w+z}{\sqrt{2}}$ and $\displaystyle y = \frac{w-z}{\sqrt{2}}$.

Plugging this into $\displaystyle y = \frac{1}{x}$, we get $$\frac{w-z}{\sqrt{2}} = \frac{\sqrt{2}}{w+z}$$ or $$w^2 - z^2 = 2$$ which is the desired form.