I'm looking for a way in which I can rotate a non rectangular hyperbola; in particular I'd like to get the equation of a non rectangular hyperbola referred to its asymptotes. To do it I need to transform in a proper way the cartesian axes, but online I didn't find anything about that.
Thank you very much.
Consider for example the hyperbola with equation :
$$(y-\tfrac12x)(y-2x)=-\tfrac12\tag{1}$$
Let us consider red axis with equation $y=\tfrac12x$ (resp. blue axis with equation $y=2x$) as the $X$ axis (resp. Y axis).
If we want them to be oriented the way it is given in the figure, consider for example all the lines with equation $y=\tfrac12x+Y \ \ (a) $ : they are parallel to the red line and intersect the (vertical) $x$ axis in point $(0,Y)$.
Remark : had we desired to have a wider step on the $Y$ axis, we would have taken for example $y=\tfrac12x+3Y$.
Let us invert (a) into $Y=-\tfrac12x+y \ \ (a')$.
If we do the same thing for the other axis, we must take care : in this case, we will have to write not $y=2x+X$ but $y=2x-X \ \ (b)$ in order for example that when $X=1$ for example, this line is (of course parallel to the blue asymptote but) under it in order to proceed in the good direction ("eastward"). Inverting (b) will give $X=2x-y \ \ (a').$
This will define the following change of coordinates :
$$\begin{cases}X&=&2x-y& \ (a')\\Y&=&-\tfrac12x+y & \ (b')\end{cases}\tag{2}$$
With these notations (2), relationship (1) become naturally :
$$XY=\tfrac12$$
Edit : How could the change of variables formula (2) could be found directly ? Written under the matrix-vector form :
$$\begin{pmatrix}X\\Y \end{pmatrix}=\underbrace{\begin{pmatrix}2&-1\\-\tfrac12&1\end{pmatrix}}_{C}\begin{pmatrix}x\\y \end{pmatrix}\tag{2'}$$
it is difficult to attach a special significance to the entries of matrix $C$. This is normal because we have expressed the new coordiantes as functions of the old ones. We have to inverse it, because the good matrix expresses the old coordinates as functions of the new ones. In our case, here it is :
$$\underbrace{\begin{pmatrix}x\\y \end{pmatrix}}_{\binom{old}{coord.}}=\underbrace{\begin{pmatrix}0.4&0.4\\-0.2&0.8\end{pmatrix}}_{C^{-1}}\underbrace{\begin{pmatrix}X\\Y \end{pmatrix}}_{\binom{new}{coord.}}\tag{3}$$
Indeed, the meaning of columns of matrix $C^{-1}$ is similar to the meaning they have for a transformation matrix : the first (resp. 2nd) column is constituted by the (old) coordinates of the new directing vectors of the axes (check it !).
Conclusion : with this interpretation, it is very easy to begin by finding relationship (3), then invert this matrix to obtain (2').