Find the area of figure enclosed by the curve $5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0$
My approach, the above is an equation of ellipse as $3^2-40\le 0$. The area of a std ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi ab$. I need to transform the above equation so that major axis is either x-axis or y-axis. I am unable to do so
By letting $x=X+2$ and $y=Y-\frac{9}{2}$ the equation is reduced to $$ 5X^2+6XY+2Y^2=(X,Y)\begin{pmatrix} 5 & 3 \\ 3 & 2\end{pmatrix}(X,Y)^T=\frac{1}{2} $$ where the involved (symmetric, positive definite) matrix has determinant $1$.
It follows that the enclosed area equals the area enclosed by $$ \tilde{X}^2+\tilde{Y}^2 = \frac{1}{2}, $$ i.e. $\large{\frac{\pi}{2}}$. This works since the area only depends on the product of the eigenvalues, which is the determinant.