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} = (1)/( √(2) ) \begin{bmatrix} 1&1\\-1&1\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix}`

That is,

`x'= (1)/( √(2) ) (x + y) \\ y'= (1)/( √(2) ) (-x+y) \\ x'y'= (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.