<s>[INST] Prove that area of ellipse $Ax^2+Bxy+Cy^2=1$ is equal to $\frac{2\pi}{\sqrt{4AC-B^2}}$ Question:
If $B^2-4AC \lt 0$, the equation
$$Ax^2+Bxy+Cy^2=1$$
represents an ellipse. Prove that the area of the ellipse is
$\frac{2\pi}{\sqrt{4AC-B^2}}$.
I know that, if $2a$ and $2b$ are the major and minor axis of the ellipse respectively, its area is $\pi ab$.
I honestly have no idea how to go about this. Any help is appreciated.
 [/INST] In polar coordinates, the ellipse is given by
$$r^2 (\theta)= \frac1{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}$$
Denote $\Delta = \sqrt{(A-C)^2+B^2}$ and integrate its area as follows
\begin{align}
A=\frac12 \int_0^{2\pi} r^2(\theta)\> d\theta
&= \frac12 \int_0^{2\pi} \frac{1}{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}\> d\theta\\
&= \int_0^{2\pi} \frac{1}{(A+C)+(A-C)\cos 2\theta +B\sin2\theta}\> d\theta\\
 &= \int_0^{2\pi} \frac{1}{(A+C)+\Delta\cos 2\theta }\>d\theta\\
&= 4\int_0^{\pi/2} \frac{d(\tan\theta)}{(A+C+\Delta)+(A+C-\Delta)\tan^2\theta }\\
&=\frac{2\pi}{\sqrt{(A+C)^2 - \Delta^2}}
=\frac{2\pi}{\sqrt{4AC-B^2}}\\
\end{align}
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