## hyperbolic or "hyper - bollock"

Hi folks,

As we know the equation of a horizontal hyperbola is

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ ....................(1)

A rectangular hyperbola is a special case in which the semi major axis a and semi minor axis b are the same. This results in the asymptotes being orthogonal at $y = x$ and $y = -x$. This hyperbola has equation:

$xy = c^2$ ........................(2)

where c is the distance from the centre to the focus.

So I wanted to see how we get from the first equation to the second.

The first equation reduces to $x^2 - y^2 = a^2$ ......................... (3)

And all we have to do is rotate the x-y axes through 45 degrees to move the horizontal hyperbola such that the asymptotes become the new horizontal and vertical axes (x', y').
This is a standard transformation i.e. $\underline{x'} = T \underline{x}$

where $T = \left( \begin{array}{cc} \cos t & -\sin t \\ \sin t & \cos t \end{array} \right)$

In this case t = 45 degrees. So $\underline{x} = T^{-1} \underline{x'}$

So $T^{-1}x' = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) \left( \begin{array}{c} x' \\ y' \end{array} \right)$

Therefore $x = \frac{x' + y' }{\sqrt{2}}$ and $y = \frac{y' - x'}{\sqrt{2}}$

so, substitute these values into (3)

$(\frac{(x' + y')}{\sqrt{2}})^2 - (\frac{(y' - x')}{\sqrt{2}})^2 = a^2$

$(x' + y')^2 - (y' - x')^2 = 2a^2$

$4x'.y' = 2a^2$

$x'.y' = \frac{a^2}{2}$

$x'.y' = c^2$ And this seems to do the trick.

BUT... this gives $\frac{a^2}{2} = c^2$ or $a^2 = 2c^2$

whereas for a horizontal hyperbola $c^2 = a^2 + b^2$ and when a = b then $c^2 = 2a^2$ the "a"s and "c"s are the wrong way around. I have spent hours on this, so I really hope I am not doing anything really stupid. Can anyone spot the mistake?