I had the equation $3x^2-2xy+3y^2+4x+4y-4=0$ which I wanted to prove it was a ellipse. To do so, I used orthogonal diagonalization and did a change in coordenates in order to obtain: $$\frac{(x'+\sqrt{2})^2}{(2)^2}+\frac{(y')^2}{(\sqrt{2})^2}=1$$ Which proves the initial equation is an ellipse. I want now to calculate the absolute minimum and maximum of $f(x,y)=xy$ on the boundary of the ellipse. To do so, I parametrized the elipse into $x'=2cos(t)-\sqrt2$, $y'=sin(t)\sqrt(2)$. Now, since I have changed the coordinates do I have to change the coordinates of $f(x,y)=xy$ too? With the change of coordinates it would become $f(x',y')=\frac{x'^2}{2}-\frac{y'^2}{2}$
I need to know this because in order to calculate the absolute minimum and maximum I need to calculate the critical points ($f'(t)=0$).
Please ask anything if I didn't explain myself correctly.
HINT: define the Lagrange function as $$f(x,y,\lambda)=xy+\lambda(3x^2-2xy+3y^2+4x+4y-4)$$ and compute the partail derivatives with respect to $x,y,\lambda$