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 $b$.
 A: I think that the bests approach is to see that the general equation of an ellipse you have written above is the result of a standard ellipse that has suffered a rotation about the axes. Can you find a coordinate transformation such as
\begin{align}
x' &= x \cos \theta - y \sin \theta \\
y' &= x\cos \theta + y \sin \theta
\end{align}
That would render the general equation in the form
$$\frac{(x'-h)^{2}}{a^{2}}+ \frac{(y'-k)^{2}}{b^{2}}=m, \qquad (a, b, h, k, m) \in \mathbb{R}$$
From which you could readily find the area?
A: Hint
Let
$$ax^2+bxy+cy^2+dx+ey+f=0$$
Note
$$\tan 2\theta=\frac{b}{a-c}$$
or
$$\cot \theta=\frac{a-c}{b}+\sqrt{1+\left(\frac{a-c}{b}\right)^2}$$
we have
$$\cot \theta=\frac{1+\sqrt{5}}{2}$$
find $\sin\theta$ and $\cos\theta$ and apply
\begin{cases}
x=X\cos\theta-Y\sin\theta\\
y=X\sin\theta+Y\cos\theta
\end{cases}
A: 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:
$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$
$\lambda_1 = 7 + \sqrt{45}, \lambda_2 = 7 - \sqrt{45}\\
x' = (x-2) (\cos \theta) + (y+4.5)(\sin \theta)\\
y' = (x-2)(\sin \theta) - (y+4.5)(\cos \theta)\\
\theta = \frac 12 \tan^{-1} 2$ 
A: For the curve $ax^2+2hxy+by^2+2gx+2fy+c = 0$, the center is given by 
\begin{align*}
ax+hy+g &= 0\\
hx+by+f &= 0
\end{align*}
and the equation of the ellipse with the axes through the center parallel to the axes is 
\begin{align*}
ax^2+2hxy+by^2 + c' = 0
\end{align*}
where $c' = gx_c + fy_c + c$, where $(x_c, y_c)$ is the center.
A: The answer by Doug M solves your problem. I just want to expand on it to the general ellipse. 
Assume $$ax^2+2bxy+cy^2+dx+ey+f=0$$ is an ellipse ($ac-b^2>0, a>0$). It has center $x=\frac{eb-cd}{2(ac-b^2)}, y=\frac{bd-ae}{2(ac-b^2)}$, which means that when the equation transforms to $$ax'^2+2bx'y'+cy'^2+f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}=0,$$ that is $f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}$ is negative (if not the conic is just a point or doesn't have real points). We need to multiply this by the $g>0$ so that $g(f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)})=-1$. Then $agx'^2+2bgx'y'+cgy'^2-1=0$ and the area is $\frac{\pi}{g\sqrt{ac-b^2}}$ by Doug M's method.
For your equation $g=2$ and $ac-b^2=1$ so $A=\frac{\pi}{2}$.
