Hyperbolas and Quadrants on Rotation

Question

Let's assume we have a standard hyperbola. On rotating the hyperbola $45^{\circ}$ clockwise, the new hyperbola should lie in the $2$nd and $4$th quadrant. However, the equation of a parabola rotated $45^{\circ}$ clockwise is $x y=\frac{a^2}{2}$.
This equation shows that LHS must be positive since RHS is positive, therefore $x$ and $y$ must have the same sign, and the hyperbola graph is actually in the $1$st and $3$rd quadrant. Why does this happen?

Answer 1

Note that for clockwise rotation by $45^{\circ}$, \begin{align*} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} \cos (-45^{\circ}) & -\sin (-45^{\circ}) \\ \sin (-45^{\circ}) & \cos (-45^{\circ}) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} \cos 45^{\circ} & \sin 45^{\circ} \\ -\sin 45^{\circ} & \cos 45^{\circ} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} \frac{x+y}{\sqrt{2}} \\ \frac{y-x}{\sqrt{2}} \end{pmatrix} \\ x' y' &= \frac{y^2-x^2}{2} \\ x' y' &= -\frac{a^{2}}{2} \end{align*}

which is negative!!

So the transformed hyperbola lies on the Quadrant II and IV.