rotation of hyperbola How can we rotate the rectangular hyperbola
xy=c   ( c is any constant) 
Into a form of standard hyperbola that is
(x/a)$^2$ - (y/b)$^2$ = 1
By rotating the hyperbola . 
 A: I take $c^2$ instead $c$ for convenience like:
plugging in transformation relations for rotation  by $45^0$
$$ x =  (x_1 - y_1)/\sqrt2 \, ;  y = (x_1 + y_1)/\sqrt2 $$
into the equation of the rectangular hyperbola $ x\, y = c^2 $ and
you get it into standard  hyperbola form with new coordinate labels: 
$$ (x_1/\sqrt2 c)^2  - (y_1/\sqrt2 c)^2  = 1. $$
Axes Rotation
A: $xy = c$
let $x = u-v\\ y = u+v$
$xy = u^2 - v^2 = c$
Now the transformation that I have just done has a little bit of spacial compression to it.  
If you do a traformation along the lines of $x = au-bv, y = bu+av$ then there will be compression on the order of  $\sqrt {a^2 + b^2}$  So, it is not a bad idea to choose $a,b$ such that $a^2 + b^2 = 1$
or $x = \cos \phi u - \sin\phi v\\ y = \sin\phi u + \cos\phi v$
and by trig identity that $cos^2 \phi + sin^2 \phi = 1$
$x = \frac {\sqrt 2}{2} u-\frac {\sqrt 2}{2}v\\ y = \frac {\sqrt 2}{2}u+\frac {\sqrt 2}{2}v\\
xy = \frac {u^2}2 + \frac {v^2}2 = c\\
 \frac {u^2}{2c} + \frac {v^2}{2c} = 1 $
A: $$
\begin{bmatrix}X\\Y\end{bmatrix}=\mathcal{Rot}_{\text{cw}}(45)\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\cos45&-\sin45\\\sin45&\cos45\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\
=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\
\implies X=\frac{x-y}{\sqrt{2}}\quad;\quad Y=\frac{x+y}{\sqrt{2}}
$$
$$
xy=c\implies \frac{x-y}{\sqrt{2}}.\frac{x+y}{\sqrt{2}}=\frac{x^2-y^2}{\sqrt{2}}=c\\\frac{x^2}{\sqrt{2}c}-\frac{y^2}{\sqrt{2}c}=1
$$
$$
\color{blue}{xy=c \iff \frac{x^2}{\sqrt{2}c}-\frac{y^2}{\sqrt{2}c}=1}
$$
