<s>[INST] 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 $b$.
 [/INST] 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 $A=\frac{\pi}{2}$.
</s>