rotation of hyperbola

Question

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 .

Answer 1

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

Answer 2

$xy = c$

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 $