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.

  • $\begingroup$ The equation will be $xy=a$... $\endgroup$ – Jean Marie May 9 at 21:29
  • $\begingroup$ Hello @JeanMarie, xy=a is the equation of a rectangular hyperbola referred to its asymptotes $\endgroup$ – Gennaro Arguzzi May 9 at 21:33
  • $\begingroup$ I know it, but it as well the equation say $XY=a$ when you take as $X$ and $Y$ axes its asymptotes ! $\endgroup$ – Jean Marie May 9 at 21:35
  • $\begingroup$ Ok, now I understood your equation; basically the format of the equation is the same in both cases, orthogonal and non orthogonal cartesian axes (axes=asymptotes). $\endgroup$ – Gennaro Arguzzi May 9 at 21:57

Consider for example the hyperbola with equation :


enter image description here

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 :


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').

  • $\begingroup$ Hello @JeanMarie, thank you so much for your explanation. The only point I'd like to expand is that about the intersection in the point $(0,X)$. Why there is this intersection? It's difficult to see it in the plot. $\endgroup$ – Gennaro Arguzzi May 10 at 8:09
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    $\begingroup$ $(0,X)$ is a bug : it should be $(0,Y)$; I have fixed it. Besides, I'm happy that this explanation has brought something to you. $\endgroup$ – Jean Marie May 10 at 8:17
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    $\begingroup$ I am going to add how we could find directly formulas (2) by a linear algebra argument. $\endgroup$ – Jean Marie May 10 at 8:33
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    $\begingroup$ Edit added..... $\endgroup$ – Jean Marie May 10 at 18:46
  • $\begingroup$ Thank you for your willingness @JeanMarie. I'm studying in detail your edit. $\endgroup$ – Gennaro Arguzzi May 11 at 17:41

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