Linear transformation: Two dimensional rotation matrix

Question:
There is an ellipse:
\[
x^2 + \frac{y^2}{4} = 1
\]
If it is on a new coordinate that is rotated by 45$^o$, find the equation of the ellipse with the new coordinate, ($X, Y$).

Answer:
When the coordinate axes are rotated, the old coordinate ($x, y$) will be expressed by the new coordinate ($X, Y$).
\[
\left( \begin{array}{c} x \\ y \end{array} \right)
 =
\left( \begin{array}{cc} \cos\theta & -\sin\theta \\
                                      \sin\theta & \cos\theta
\end{array} \right)
\left( \begin{array}{c} X \\ Y  \end{array} \right)
\]
Therefore, we have
\[
\left\{
\begin{array}{l}
x = X\cos\theta -Y\sin\theta = X\cos 45^o -Y\sin 45^o =\frac{X-Y}{\sqrt{2}} \\
y = X\sin\theta + Y\cos\theta = X\sin 45^o + Y\cos 45^o = \frac{X+Y}{\sqrt{2}}
\end{array}
\right.
\]
Then, they are plugged into the original equation of ellipse.
\[
\frac{1}{2}(X-Y)^2 + \frac{1}{8}(X+Y)^2 = 1
\]
Or you can also express it as
\[
\frac{1}{8}(5X^2 -6XY +5Y^2) = 1
\]

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