How to rotate an hyperbola by $45^\circ$ so that I have an equation of the form $xy=c$

Question

I am trying to show that If I rotate an hyperbola of the form $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$ by $45^\circ$ I get an equation of the form $x'y'=c$.

Using the following rotation coordinates:

$x=x' \cos 45^\circ -y' \sin 45^\circ$

$y=x' \sin 45^\circ + y' \cos 45^\circ $

I have that

$$ \begin{align} \cfrac{(x' \cos 45^\circ -y' \sin 45^\circ)^2}{a^2} - \cfrac{(x' \sin 45^\circ + y' \cos 45^\circ)^2}{b^2}&=1 \\ b^2(\cfrac{1}{2})(x'-y')^2-a^2(\cfrac{1}{2})(x'+y)^2&=a^2b^2 \\ (x')^{2}b^2+y^{2}b^{2}-2b^2(x'\cdot y') -a^2(x')^2-a^{2}(y')^2 -2a^{2}(x' \cdot y') &=2a^2b^2 \\ (x')^2(b^2-a^2)+(y'^2)(b^2-a^2) &=2(a^2b^2-x'y') \end{align}$$

But this is not of the form $x'y'=c$ as I wanted.

Can you guys help ?

Answer 1

The hyperbola $xy=c$ has asymptotes that are orthogonal (they meet at a right angle.) (The asymptotes are the lines $x=0,y=0$.)

The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ does not have orthogonal asymptotes. (The asymptotes are the lines $ay=bx$ and $ay=-bx$, which are not at right angles in general.)

So you can't rotate one to get the other, unless $a^2=b^2$. This is because rotations preserve relative angles between lines.

Note that if $a^2=b^2$, your rotation gives you $0=2(a^2b^2-x'y')$ or $x'y'=a^2b^2$, which is the form you wanted.

Answer 2

If $ a \ne b$ then it is not possible because rotated axes do not exactly fall along x- and y- axes.We can switch between them by rotation only for the rectangular ( aka equi-angular) hyperbola, whose asymptotes cut at $ 90^0$at origin.