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Find the area of figure enclosed by the curve $5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0$

My approach, the above is an equation of ellipse as $3^2-40\le 0$. The area of a std ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi ab$. I need to transform the above equation so that major axis is either x-axis or y-axis. I am unable to do so

Samar Imam Zaidi
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    You have a rotated and translated ellipse. Take a look at the answers for https://math.stackexchange.com/questions/280937 – Andrei Jun 28 '20 at 21:22
  • actually, if you have ellipse without any linear terms, just some $a x^2 + bxy + c y^2 = d$ with $a,d, 4ac - b^2$ all positive, the area can be found by formula. That is, the translation is enough. – Will Jagy Jun 28 '20 at 21:22
  • Try this https://math.stackexchange.com/questions/594693/how-to-put-2x2-4xy-6y2-6x-2y-6-in-canonical-form/594707#594707 – Anurag A Jun 28 '20 at 21:23

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By letting $x=X+2$ and $y=Y-\frac{9}{2}$ the equation is reduced to $$ 5X^2+6XY+2Y^2=(X,Y)\begin{pmatrix} 5 & 3 \\ 3 & 2\end{pmatrix}(X,Y)^T=\frac{1}{2} $$ where the involved (symmetric, positive definite) matrix has determinant $1$.
It follows that the enclosed area equals the area enclosed by $$ \tilde{X}^2+\tilde{Y}^2 = \frac{1}{2}, $$ i.e. $\large{\frac{\pi}{2}}$. This works since the area only depends on the product of the eigenvalues, which is the determinant.

Jack D'Aurizio
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