I'm working on Matlab and I have the parameters a,b and c of a hyperbola. I'm working on polar coordinates because an Matlab example did so. I'm trying to get a matrix $2 \times N$ where each column is a set of (x,y) points of a hyperbola. The problem is that I have the resulting expressions but I have to demonstrate them.
The expressions given to create a hyperbola centered at (h,k) from a Matlab example are the following:
$x = a·\cosh(\mu)·\cos(\phi) - b·\sinh(\mu)·\sin(\phi) + h$
$y = a·\cosh(\mu)·\sin(\phi) - b·\sinh(\mu)·\cos(\phi) + k$
where $\phi$ is the angle when getting the polar vector between the two foci.
Considering a hyperbola centered in $(0,0)$ with foci in $[-100 0]$ and $[100 0]$, the parameters $a,b$ and $c$ of the hyperbola are variable but I get the polar coordinates of the hyperbola:
$\rho = 200$
$\phi = 0$
$\mu$ is an array to evaluate a part of the hyperbola
$\mu = [-2, -1.99, -1.98, .... 0, ... 1.98, 1.99, 2.00]$
Then the previous equations become as follows and evaluating them for all $\mu$ vector it provides the desired matrix of pairs (x,y) creating an hyperbola:
$x = a·\cosh(\mu)$
$y = b·\sinh(\mu)$
THE PROBLEM is that:
I'm trying to demonstrate how I can get the above mentioned expressions from the standard hyperbola equation:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
where the center is in $(h,k)$ to the expressions above mentioned considering that $(h,k) = (0,0)$. I don't know how to make appear a $\cosh(\mu)$ and a $sinh(\mu)$ in the expressions. I don't understand how to include the $\mu$ angle inside this equation.
Thanks, Alex
$x = a·\cosh(\mu)·\cos(\phi) - b·\sinh(\mu)·\sin(\phi) + h$
$y = a·\cosh(\mu)·\sin(\phi) + b·\sinh(\mu)·\cos(\phi) + k$
If $ r = [x, y]^T$ , and the center $C = (h, k) $, and the rotation matrix $R$ is given by,
$ R = \begin{bmatrix} \cos \phi && - \sin(\phi) \\ \sin(\phi) && \cos(\phi) \end{bmatrix} $
and the vector
$ u = [ \cosh(\mu) , \sinh(\mu) ]^T $
then
$ r - C = R D u $
where $ D = \operatorname{diag} \{ a , b \}$
So that
$ u = D^{-1} R^T (r - C) \tag{1} $
Now since $\cosh^2(\mu) - \sinh^2(\mu) = 1 $, then
$ u^T Q_0 u = 1 \tag{2}$
where $Q_0 = \operatorname{diag} \{ 1 , -1 \} $
Using $(1)$ in $(2)$, gives the equation as
$ (r - C)^T R D^{-1} Q_0 D^{-1} R^T (r - C) = 1 \tag{3} $
But
$ Q' = D^{-1} Q_0 D^{-1} = \operatorname{diag} \{ \dfrac{1}{a^2}, -\dfrac{1}{b^2} \} $
So $(3)$ is the rotated version of the standard equation which is
$ (r - C)^T Q' (r - C) = 1 \tag{4}$