Area of the ellipse $Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$ 
Prove that the area of the ellipse $$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0,$$ where $AC-B^2>0$, is equal to $$S=\frac{- \pi \Delta}{(AC-B^2)^{3/2}},$$ where $$\Delta =\begin{vmatrix}A&B&D\\B&C&E\\D&E&F\end{vmatrix}.$$

I could prove that area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$ because I know its vertices, center, etc., but in this question I do not know the vertices of the ellipse, the major axis equation, the minor axis equation, the center, etc. So I could not solve it. Can someone assist me in this question?
 A: Assume WLOG that $A>0$ (otherwise multiply the equation by $-1$). We have
$$
\left[\matrix{x\\ y}\right]^T\left[\matrix{A & B\\ B & C}\right]\left[\matrix{x\\ y}\right]+2\left[\matrix{D\\ E}\right]^T\left[\matrix{x\\ y}\right]+F=0.
$$
After completing the square we get
$$
\left(\left[\matrix{x\\ y}\right]+\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]\right)^T\left[\matrix{A & B\\ B & C}\right]\underbrace{\left(\left[\matrix{x\\ y}\right]+\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]\right)}_{=z}=\\
=\underbrace{\left[\matrix{D\\ E}\right]^T\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]-F}_{=\alpha^2}.
$$
If we factorize the matrix of the quadratic form (which is positive definite)
$$
M=\left[\matrix{A & B\\ B & C}\right]^{1/2}
$$
the ellipse equation becomes
$$
\left\|\frac{1}{\alpha}Mz\right\|=1
$$
which means that the image of the ellipse under the linear mapping $y=\frac{1}{\alpha}Mz$ is the unit circle with area $\pi$. Area deformation under linear/affine maps is the determinant, i.e. the area of the ellipse $S$ satisfies
$$
\pi=\det(\frac{1}{\alpha}M)S=\frac{1}{\alpha^2}\det(M)S\qquad\Rightarrow\qquad
S=\frac{\pi\alpha^2}{\det(M)}=\frac{\pi\alpha^2}{\sqrt{AC-B^2}}.
$$
Finally, to calculate $\alpha^2$ we use the Schur determinant formula
$$
\Delta=\det\left[\matrix{A & B & D\\ B & C & E\\D & E & F}\right]=
\det\left[\matrix{A & B\\ B & C}\right]\cdot (-\alpha^2)=-(AC-B^2)\alpha^2
$$
which gives
$$
S=\frac{-\pi\Delta}{(AC-B^2)^{3/2}}.
$$
A: Read up about three invariants to gain a broader understanding:
$$ I_1= A + B $$
$$ I_2 = (A C - B^2) $$
$$ I_3 = \Delta $$
Several other geometrical properties are also linked to these invariants.
