Transform of the Cartesian plane that maps hyperbolic arcs $xy = C$ to line segments I have the finite set of curves: $$y = \frac{C}{x}, \qquad C = 2, 3, \ldots, C_{\max},$$ with $C$ and $x$ positive integers, $2 \le x \le C$ ($x$ varies on a finite domain). 
Is it possible to apply some transformation to the Cartesian coordinate plane, so that all such curves become parallel straight segments?
How would such a transformation be created?
P.S.
One thing I forgot to mention: If possible, I would also like all the lines (or segments, more precisely) to have slope $-1$.
 A: Rearranging gives the equations $$xy = C, \qquad x > 0,$$ defining the branch of the given hyperbolae $y = \frac{C}{x}$ in the (since $C > 0$) find quadrant. We'll give two sets of coordinates well-adapted to the geometry of the hyperbolae (in particular, in which they are represented by straight lines).
Taking the logarithm of both sides of our equations gives for each hyperbola that
$$\log x + \log y = \log C,$$
so in the naive coordinates $(s, t)$ defined by $$s := \log x, \quad t := \log y,$$ these hyperbolae have the simple form
$$s + t = \log C.$$
This equation is affine in $s, t$, and in these coordinates the hyperbolae (or rather, the branches of the hyperbolae in the first quadrant) are lines with slope $-1$ as desired, and hence are all parallel.
Remark One can also define the so-called 
hyperbolic coordinates: Via the slightly more sophisticated coordinate transformation
$$u := \frac{1}{2}\log \frac{x}{y}, \quad v := \sqrt{xy},$$
the hyperbola $xy = C$ corresponds to the vertical line $v = \sqrt{C}$, as for the previous transformation, these lines are all parallel.
A: One way can be to first square the points of the plane:
$$(x,y)\mapsto(x^2-y^2,2xy)$$
This sends the hyperbolas to horizontal lines
$$(x,y)\mapsto(x^2-C^2/x^2,2C)$$
To get the slope that you want we can then rotate the appropriate angle ($-\pi/4$):
$$(x,y)\mapsto\left(\frac{\sqrt{2}}{2}(x+y),\frac{\sqrt{2}}{2}(-x+y)\right)$$
Composing these two transformations we get: 
$$(x,y)\mapsto\left(\frac{\sqrt{2}}{2}(x^2-C^2/x^2+2C),\frac{\sqrt{2}}{2}(2C-(x^2-C^2/x^2))\right)$$
