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.
 A: Hint For a linear transformation $T : \Bbb R^2 \to \Bbb R^2$ given by
$$\pmatrix{u\\v} = \pmatrix{a&b\\c&d} \pmatrix{x\\y} ,$$
the areas of a (nice) region $E \subset \Bbb R^2$ (e.g., our ellipse) and its image $T(E) \subset \Bbb R^2$ are related by
$$\textrm{area}(T(E)) = |{\det T}| \,\textrm{area}(E) .$$
On the other hand, we know that there is a linear transformation $T$ that maps our ellipse to the unit ball, and so for such a transformation rearranging the previous equation gives
$$\textrm{area}(E) = \frac{\textrm{area}(T(E))}{|{\det T}|} = \frac{\pi}{|{\det T}|} .$$
So, we need only write $|{\det T}|$ in terms of $A, B, C$, and in particular show that $$|{\det T}| = |a d - b c| = \frac{1}{2} \sqrt{4 A C - B^2} .$$

Additional hint Substituting using the transformation formula for $T$ gives that in $xy$-coordinates the unit circle is $$(a^2 + c^2) x^2 + 2(ab + cd) x y + (b^2 + d^2) y^2 .$$ Thus, the desired transformation satisfies $$A = a^2 + c^2, \quad B = 2 (a b + c d), \quad C = b^2 + d^2.$$

A: 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}
