<s>[INST] Area of an ellipse An ellipse has equation : 
$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$$
Can you provide an optimum method to find it's area?
 [/INST] When rotating conics in implicit form
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}
$$
around the origin there are 5 invariants:
$$
\begin{array}{rl}
I_1&=A+C\\
I_2&=(A-C)^2+B^2\\
I_3&=D^2+E^2\\
I_4&=(A-C)(D^2-E^2)+2DEB\\
I_5&=F\tag{2}
\end{array}
$$
Assuming that we have rotated to eliminate $B$, we have
$$
Ax^2+Cy^2+Dx+Ey+F=0\tag{3}
$$
Translating the center to the origin gives
$$
Ax^2+Cy^2+F-\frac{D^2}{4A}-\frac{E^2}{4C}=0\tag{4}
$$
which is the same as
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{5}
$$
if we set
$$
a=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}}\quad\text{and}\quad
b=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}}\tag{6}
$$
Using $\text{Area}=\pi ab$ and $(4)$ and rewriting in terms of the invariants to remove the rotation, we get
$$
\begin{align}
\text{Area}
&=\pi\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{\sqrt{AC}}\\
&=\pi\frac{\color{#C00000}{2CD^2+2AE^2}-\color{#00A000}{8ACF}}{\color{#0000FF}{8(AC)^{3/2}}}\\
&=\pi\frac{\color{#C00000}{I_1I_3-I_4}-\color{#00A000}{2(I_1^2-I_2)I_5}}{\color{#0000FF}{(I_1^2-I_2)^{3/2}}}\tag{7}
\end{align}
$$
Therefore, $(2)$ and $(7)$ give the area in terms of the coefficients in $(1)$.
Invariants Under Rotation and Translation
$I_1$ and $I_2$ are invariant under rotation and translation, but there is one more: the constant coefficient when the center is translated to the origin. Writing $F-\dfrac{D^2}{4A}-\dfrac{E^2}{4C}$ in terms of the rotational invariants and expanding yields
$$
I_6=F-\frac{AE^2-BDE+CD^2}{4AC-B^2}\tag{8}
$$
Thus, the conic satisfying $(1)$, when rotated and translated becomes
$$
(I_1-\sqrt{I_2})x^2+(I_1+\sqrt{I_2})y^2+2I_6=0\tag{9}
$$
</s>