if A represents the area of the ellipse 3(x^2)+4xy+3(y^2)=1 then the value of (3sqrt(5)/(pi))*A
The ellipse can be written:
$\displaystyle E\equiv (x,y)\begin{pmatrix}{3}&{2}\\{2}&{3}\end{pmatrix}\ begin{pmatrix}{x}\\{y}\end{pmatrix}=1
$
By the Spectral Theorem, there is an orthogonal transformation (i.e. preserving distances) $\displaystyle (x,y)^t=P(x',y')^t$ such that the ellipse's equation is:
$\displaystyle \lambda_1x'^2+\lambda_2y'^2=1$
where $\displaystyle \lambda_1,\lambda_2$ are the corresponding eigenvalues. In this case $\displaystyle \lambda_1=5,\lambda_2=1$. So,
$\displaystyle E\equiv \dfrac{x'^2}{(1/\;\sqrt[]{5})^2}+\dfrac{y'^2}{1^2}=1$
Then,
$\displaystyle A=\pi ab=\pi/\;\sqrt[]{5}$.
Regards.
The equations of any rotation are:
$\displaystyle G_{\alpha } \equiv\begin{Bmatrix} x=x'\cos \alpha-y'\sin \alpha\\y=x'\sin \alpha+y'\cos \alpha\end{matrix}$
Substitute in the ellipse's original equation and make $\displaystyle \textrm{coef}(x'\cdot y')=0$, you will obtain $\displaystyle \alpha=\pi/4$ and you can fiind the canonical equation:
$\displaystyle
E\equiv \dfrac{x'^2}{(1/\;\sqrt[]{5})^2}+\dfrac{y'^2}{1^2}=1
$
Regards.