<s>[INST] Rotate $xy=1$ by $\frac{\pi}{4}$ in a negative (clockwise) direction. I was studying hyporbolae for the first time and noticed that $y=\frac{1}{x}$ is a rotated hyperbola. I had seen equations like $y=\frac{1}{x}$ before but never noticed they where hyperbolae.
Anyway using geometry and the general form of a hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, I realised that $xy=1$ is $\frac{x^2}{2}-\frac{y^2}{2}=1$ which has been rotated clockwise by $\frac{\pi}{4}$.
I tried to rotate $xy=1$ using the rotation matrix: $ \left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right)$
$$\left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right) \left( \begin{array}{cc} x \\ \frac{1}{x}\end{array} \right)=\left( \begin{array}{cc} x'\\ y' \end{array} \right)$$
$$\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=x', -\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=y'$$
I used two methods. The first worked and the second did not. I'd like to know where I'm going wrong with the second approach.
In the first approach I squared both sides of the above equations and then used simultaneous equations to remove the $x$. Then dividing across by $2$ gives the desired result.
Just out of curiosity I tried another method. I first multiplied across both equations by $\sqrt{2}x$ and rearranged to quadratic form: $$x^2-\sqrt{2}x'x+1=0, x^2-\sqrt{2}y'x-1=0$$
I equated the solutions of each to get: $$\frac{\sqrt{2}x' \pm\sqrt{2x'^2-4}}{2}=\frac{-\sqrt{2}y' \pm\sqrt{2y'^2+4}}{2}$$
The $\pm$s are ugly but I figured I could get rid of them through a series of squaring. I tried a few different ways starting with roots on opposite sides, roots on the same side etc. but I can't quite get it through. Should this work or am I going down a dead-end road ?
Ok I got it: $$\frac{\sqrt{2}x \pm\sqrt{2x^2-4}}{2}=\frac{-\sqrt{2}y \pm\sqrt{2y^2+4}}{2}$$ simplifies to  $$\frac{x^2}{2}-\frac{y^2}{2}=1$$
 [/INST] You can't get a 4th degree equation by an affine transformation of coordinates.
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