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.
 A: Solving
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
for $y$, one has
$$ y_{1,2}=\frac12(-ax\pm\sqrt{a^2x^2-4(x^2-1}))=\frac12(-ax\pm\sqrt{4-(4-a^2)x^2})$$
where $4-(4-a^2)x^2\ge0$ or $|x|\le\frac{2}{\sqrt{4-a^2}}$. So the are is
\begin{eqnarray}
A&=&\int_{-\frac{2}{\sqrt{4-a^2}}}^{\frac{2}{\sqrt{4-a^2}}}|y_1-y_2|dx\\
&=&\int_{-\frac{2}{\sqrt{4-a^2}}}^{\frac{2}{\sqrt{4-a^2}}}\sqrt{4-(4-a^2)x^2}dx.
\end{eqnarray}
Letting $x=\frac{2}{\sqrt{4-a^2}}\sin t$, then
$$ A=\frac4{\sqrt{4-a^2}}\int_0^{\pi/2}\cos^2 td= \frac{2 \pi}{\sqrt{4-a^2}}\tag{1} .$$
A: It is the equation of an ellipse invariant through transformations
$$(x,y) \to (y,x) \ \ \& \ \ (x,y) \to (-x,-y)$$

Fig. 1: the ellipse for $a=0.5$.
therefore having $y=x$ and $y=-x$ as its axes of symmetry.
The intersection of the curve with its semi-axes are obtained by setting either by setting $y=x$ or $y=-x$ in its equation, giving
$$A=\frac{1}{\sqrt{2+a}}\sqrt{2} \ \ \text{and} \ \ B=\frac{1}{\sqrt{2-a}}\sqrt{2}$$
or the contrary according to the sign of $a$. (multiplication by $\sqrt{2}$ is for accounting the fact that we must take the diagonal of the square with sidelength $x$)
therefore, its area is
$$\pi AB=\frac{2 \pi}{\sqrt{4-a^2}}\tag{1}$$

Remark: if you want to use the polar equation you have found, this formula gives the enclosed area:
$$\int_0^{2 \pi}\frac12 r^2 d\theta=\int_0^{2 \pi}\frac{1}{2(1+\tfrac{a}{2} \sin(2 \theta))} d\theta$$
which is indeed equal to (1) (I have checked it).
A: 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}} $
A: Integrate the area as follows
\begin{align}
Area =&\>\frac12 \int_0^{2\pi} \rho^2(\theta)\> d\theta= \frac12 \int_0^{2\pi} \frac1{1+\frac12 a\sin2\theta}d\theta
= \int_0^{2\pi} \frac{1}{2+a\cos 2\theta }\>d\theta\\=& \>4\int_0^{\pi/2} \frac{d(\tan\theta)}{(2+a)+(2-a)\tan^2\theta }
=\frac{2\pi}{\sqrt{4-a^2}}
\end{align}
