How to go from this equation to the equation of an hyperbola I've seen that $x*y=1$ graphs an hyperbola, but I am struggling to get that equation to the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$. How can I do this? Ultimately, what I want is to be able to graph it by hand by looking at the equation.
 A: The formula $x^2/a^2 - y^2/b^2 = 1$ is for hyperbolas whose axes are aligned with the coordinate axes. $xy=1$ does not have this feature, so it doesn't fit that equation form. 
See this page for more details. 
A: Consider the following fact: we say that an algebraic curve $f(x,y)=0$ is a blow-up of the algebraic curve $g(x,y)=0$ if $g(0,0)=0$ and $f(x,y)-g(x,y)$ is constant. Hence the hyperbola whose equation is $xy=1$ is a blow-up of the double line $xy=0$ (union of the line $x=0$ and the line $y=0$) while the hyperbola having equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a blow-up of the double line $y=\pm\frac{b}{a}$. The double lines we got are just the asymptotes of our hyperbolas: in the first case, they are parallel to the coordinate axis, in the second one, they are not.
Anyway, by having the asymptotes and a point on we can graph any hyperbola.
A: As the nature of a Curve is invariant under Rotation of axes
Using Rotation of axes,  $ x=h\cos\theta-k\sin\theta,y=h\sin\theta+k\cos\theta$ on the Rectangular Hyperbola $$x^2-y^2=a^2,$$
$$(h^2-k^2)(\cos2\theta)-hk\sin2\theta=a^2$$
Set $\sin2\theta=\pm1$
