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 $b$.

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}
