Question

Find the transformed equation of the hyperbola xy = 4 when rotated 45 degrees.

Answer 1

The given hyperbola is 
xy = 4

The transformation matrix when the hyperbola is rotated by an angle of θ is
$T=\begin{bmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta\end{bmatrix}$

Note that for θ = 45°, cosθ = sinθ = 1/√2.
Therefore in the transformed coordinate system,
$\begin{bmatrix} x' \\ y'\end{bmatrix} = \frac{1}{ \sqrt{2} } \begin{bmatrix} 1&1\\-1&1\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix}$

That is,
$x'= \frac{1}{ \sqrt{2} } (x + y) \\ y'= \frac{1}{ \sqrt{2} } (-x+y) \\ x'y'= \frac{1}{2} (y^{2}-x^{2}) =4 \\ y^{2}-x^{2}=8 \\y=\pm \sqrt{x^{2}+8}$

The graphs of the original and the rotated hyperbola are shown in the graph below.