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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?

enter image description here

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    $\begingroup$ Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate $\endgroup$ – user150203 Jan 13 at 12:32
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    $\begingroup$ Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there. $\endgroup$ – DonAntonio Jan 13 at 12:34
  • $\begingroup$ @DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+\frac{2}{3})^2+5(y+\frac{4}{5})^2=-4xy+\frac{73}{15}$ $\endgroup$ – Hussain-Alqatari Jan 13 at 13:12
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Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where $$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$ To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = \lambda y + z$ where $\lambda$ is any constant. This gives $$(A\lambda^2 + 2H\lambda + B)y^2 + 2(A\lambda + H)yz + Az^2 +2(G\lambda + F)y + 2Gz + C = 0.$$ Get rid of $yz$ by choosing $\lambda = -H/A.$ Then $$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$ that is $$(AB - H^2)\left(y - {{(AF - GH)}\over{AB - H^2}}\right)^2 + (Az + G)^2 - {{(AF - GH)^2}\over{AB - H^2}} - G^2 + AC = 0.$$ Moving the centre of the ellipse to the origin, this can be put as $$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$ where $$K = {1\over{(AB - H^2)}}\left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)\right],$$ which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are $a = \sqrt{K/(AB - H^2)}$ and $b = \sqrt{K}/A,$ and the required area is $\pi ab = \pi L/(AB - H^2)^{3/2}.$

In particular, the ellipse in the question has area $86\pi/26^{3/2}.$

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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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The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=\pi ab$.

If you insist on using an integral, you can of course solve for $y$ to find two solutions $$ y_\text{up/down}={1\over5}\left( -2x-4\pm\sqrt{-26x^2-24x+3} \right) $$ and then compute the area as $$ \int_{x_\text{left}}^{x_\text{right}}(y_{up}-y_{down})\,dx, $$ where $x_\text{left/right}$ are the values of $x$ giving a vanishing square root.

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