<s>[INST] How to determine the area of a rotated ellipse? The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=\pm\sqrt{a^2-x^2}$, nor in terms of $y$; like $x=\pm\sqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=r\cos(\theta)$, and $y=r\sin(\theta)$, and then integrating, but finding the limits of integration [with respect to $\theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?

 [/INST] Write the equation as:
$$\begin{pmatrix}x & y & 1\end{pmatrix}
\left(\begin{array}{ccc|c} \\&Q& &P \\ \\ \hline &P^T& &R \end{array}\right)
\begin{pmatrix}x \\ y \\ 1\end{pmatrix}
=\begin{pmatrix}x & y & 1\end{pmatrix}
\left(\begin{array}{cc|c}6 & 2 & 4 \\ 2 & 5 & 4 \\ \hline 4&4&1 \end{array}\right)
\begin{pmatrix}x \\ y \\ 1\end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$\nabla(6x^2+4xy+5y^2+8x+8y+1)=0 \Rightarrow \begin{cases}12x+4y+8=0\\4x+10y+8=0\end{cases} \Rightarrow x=-\frac 6{13}, y=-\frac 8{13}$$
So the translation vector $t=-\frac 1{13}\begin{pmatrix}6\\8\end{pmatrix}$.
Divide the equation by $-(P\cdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-\frac{1}{P\cdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $\frac 1{a^2}$ and $\frac 1{b^2}$.
So:
$$\det Q' = \frac{1}{a^2b^2}$$
The area of an ellipse is $\pi a b$ so that:
$$\text{Area} = \pi a b = \frac{\pi}{\sqrt{\det Q'}} = \frac{\pi\cdot |P\cdot t+R|}{\sqrt{|\det Q|}}
=\frac{\pi\cdot |-\frac 1{13}(4\cdot 6 + 4\cdot 8)+1|}{\sqrt{|6\cdot 5-2\cdot 2|}}
=\frac{43\pi}{13\sqrt{26}}$$
The general formula is:
$$\text{Area}=\frac{\pi|P^TQ^{-1}P-R|}{\sqrt{|\det Q|}}$$
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