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?
 A: 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
$$
(I_1-\sqrt{I_2})x^2+(I_1+\sqrt{I_2})y^2+2I_6=0\tag{9}
$$
A: The rotation of axes can be avoided if all that is needed is the area of the ellipse.
What we do need is to translate and normalize the ellipse to be centered at the origin:
$$Ax^2 +  Bxy + Cy^2 = 1$$
In this form the ellipse has area $\frac{2\pi}{\sqrt{4AC-B^2}}$.  The ellipse equation can be reduced to the above form by substituting $x+h$ for $x$ and $y+k$ for $y$ and choosing $h,k$ that make the linear order terms vanish (then divide through to get constant 1 on the right side).
Details:  Assuming the original equation is that of an ellipse (and not some other general conic), we replace $x$ by $x+h$ and $y$ by $y+k$, giving this:
$$Ax^2 + Bxy + Cy^2 + (2Ah+Bk+D)x + (Bh+2Ck+E)y = -(Ah^2+Bhk+Ck^2+Dh+Ek+F)$$
Solve the simple linear system for $h,k$:
$$2Ah + Bk = -D$$
$$Bh + 2Ck = -E$$
and then divide both sides by the right hand shown above to get the desired form.
A: As  the area of a closed curve is an invariant property under the Rotation of axes,using this or this, we can remove $xy$ term to  convert the given equation into $$\frac{(X-a^\prime)^2}{(A^\prime)^2}+\frac{(Y-b^\prime)^2}{(B^\prime)^2}=1$$
Consequently, the area will be $\pi A^\prime B^\prime$
A: Using the canonical form, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, the area of an ellipse is $ab\pi$.
Hence, one way of doing it is to obtain $a,b$ from $A,\ldots,F$. This can be done by expanding the canonical form (with origin $x_0,y_0$) into your form.
$$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$$
Expand the above expression and map your polynomial coefficients to $A,\ldots,F$. 
