Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.

The center:
find the minimum of $(5x^2 + 6xy + 2y^2 + 7x + 2y)$
$\frac {\partial}{\partial x}(5x^2 + 6xy + 2y^2 + 7x + 2y) = 10x + 6y + 7= 0\\
\frac {\partial}{\partial y}(5x^2 + 6y + 2y^2 + 7x + 2y) = 6x + 4y + 6= 0$
Solve this system of equations
$x = 2, y = -4.5$
$2(x-2)^2 + 6(x-2)(y+4.5)+5(y+4.5)^2 = 0.5$
$4(x-2)^2 + 12(x-2)(y+4.5)+10(y+4.5)^2 = 1$
$\mathbf x^T\begin{bmatrix}
4&6\\
6&10\end{bmatrix}\mathbf x =1$
Now we can find the eigenvalues of that matrix, but rather than solving for them I am just going to call them $\lambda_1,\lambda_2$
$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$
$\lambda_1\lambda_2 = \frac 1{(ab)^2}$
and $\lambda_1\lambda_2$ = determinant of the matrix above = $4$.
$ab = \frac 12$
$A = \frac \pi2$
The transformed equation:
$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$
$\lambda_1 = 7 + \sqrt{45}, \lambda_2 = 7 - \sqrt{45}\\
x' = (x-2) (\cos \theta) + (y+4.5)(\sin \theta)\\
y' = (x-2)(\sin \theta) - (y+4.5)(\cos \theta)\\
\theta = \frac 12 \tan^{-1} 2$ 
