<s>[INST] 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.
 [/INST] 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
$$ A=\frac4{\sqrt{4-a^2}}\int_0^{\pi/2}\cos^2 td= \frac{2 \pi}{\sqrt{4-a^2}}\tag{1} .$$
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