The given hyperbola is
xy = 4
The transformation matrix when the hyperbola is rotated by an angle of θ is
[tex]T=\begin{bmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta\end{bmatrix}[/tex]
Note that for θ = 45°, cosθ = sinθ = 1/√2.
Therefore in the transformed coordinate system,
[tex]\begin{bmatrix} x' \\ y'\end{bmatrix} = \frac{1}{ \sqrt{2} } \begin{bmatrix} 1&1\\-1&1\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix}[/tex]
That is,
[tex]x'= \frac{1}{ \sqrt{2} } (x + y) \\ y'= \frac{1}{ \sqrt{2} } (-x+y) \\
x'y'= \frac{1}{2} (y^{2}-x^{2}) =4 \\ y^{2}-x^{2}=8 \\y=\pm \sqrt{x^{2}+8} [/tex]
The graphs of the original and the rotated hyperbola are shown in the graph below.
