Find area of $x^2+axy+y^2=1$, $|a|\leq1$ I was wondering how to find the area of
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
I have solved
$$\rho^{2}(\theta)=\frac{1}{1+\frac{a}{2}\sin 2\theta}$$
However, integrating this equation using trigonometric substitution is very cumbersome.

The equation of the ellipse can be written compactly using the position vector $r=[x,y]^T$ as
$ r^T Q r = 1 $
where $Q = \begin{bmatrix} 1 && \frac{1}{2} a \\ \frac{1}{2} a && 1 \end{bmatrix} $
To find the area you need the product of the semi-minor and semi-major axes.
If you diagonalize $Q$ and write it as $ Q = R D R^T $ , then the original equation becomes
$ r^T R D R^T r = 1 $
The eigenvalues of $D$ are the square of the reciprocals of the semi-major and semi-minor axes.  Therefore its determinant
$|D| = \dfrac{1}{a^2 b^2} $
Since $R$ is a rotation matrix $ | Q | = | R D R^T | = | D | $
Therefore, $ a b = \dfrac{1}{\sqrt{|D|}} = \dfrac{1}{\sqrt{|Q|}} $
From $Q$, we get $|Q| = 1 - \dfrac{a^2}{4} $
Therefore, the area is given by
$\text{Area} = \pi a b = \dfrac{\pi}{ \sqrt{ 1 - \dfrac{a^2}{4} }} = \dfrac{2 \pi}{\sqrt{4 - a^2}} $
