Ellipse rotation by pi/4

Question

I have an ellipse given by: $(x-y)^2/2m^2+(x+y)^2/2n^2=1$ and it need to be rotated by $\pi/4$, what would the new eq be? thanks.

Answer 1

Hint: If you have a point $(x,y)$, rotation (about the origin) by angle $\theta$ results in a new point which is calculated by $$ \left\{ \begin{array}{ccc} x' &=& x \cos{\theta} - y\sin{\theta}\\ y' &=& x\sin{\theta} + y\cos{\theta} \end{array} \right. $$ So you just need to plug these in the equation and you're done.

Answer 2

Observe that

$$x'=\frac{x-y}{\sqrt2},\\ y'=\frac{x+y}{\sqrt2}$$ forms a new cordinate system rotated by $45°$ degrees.

Hence

$$\frac{x'^2}{m^2}+\frac{y'^2}{n^2}=1.$$