Find area of ellipse $5x^2 -6xy +5y^2=8$

Under the variable changes $x = \frac{u+v}{\sqrt2}$ and $y = \frac{u-v}{\sqrt2}$, the equation $5x^2 -6xy +5y^2=8$ is of the standard ellipse form

$$\frac{u^2}4+v^2=1$$

with the major and minor axes 2 and 1, respectively. Thus, the area is $2\pi$.

Given $$5x^2-6xy+5y^2=8$$ Curve represent ellipse whose center is at $(0,0)$ and it is symmetrical about $y=x$ and $y=-x$ line

Put $y=x$ in $5x^2-6xy+5y^2=8$, getting $=\pm \sqrt{2}.$

So point of intersection of line and ellipse is $(\sqrt{2},\sqrt{2})$ or $(-\sqrt{2},-\sqrt{2})$

Which is notning but distance length of semi major axis i e $a = 2$

Put $y=-x$ in $5x^2-6xy+5y^2=8$, getting $\displaystyle =\pm \frac{1}{\sqrt{2}}.$

So point of intersection of line and ellipse is $\displaystyle \bigg(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\bigg)$ or $\displaystyle \bigg(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigg)$

Which is notning but distance length of semi minor axis i e $b=1$

So we have eccentricity $$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$$

And area of ellipse is $$\pi ab=2\pi$$

I've never seen this approach before, but after messing around with it I think it's a cool way of thinking about it. If you have the position vector $X=[x,y]$, then your equation can be written like $X^TgX=1$ for the matrix $g=\begin{bmatrix} \frac{5}{8} & -\frac{3}{8} \\-\frac{3}{8}& \frac{5}{8}\end{bmatrix}$, sometimes called the "metric". Then, because $g$ is symmetric we can find an orthonormal rotation matrix $R$ such that $R^TgR$ is diagonal, where the two diagonal components correspond to the inverse squares of the semi-major and semi-minor axes. We can write this as $R^TgR = \begin{bmatrix} \lambda_1 & 0 \\0& \lambda_2\end{bmatrix}$, where the area is $\frac{\pi}{\sqrt{\lambda_1\lambda_2}}=\frac{\pi}{\sqrt{\det(R^TgR)}}=\frac{\pi}{\sqrt{\det(g)}}$ because $\det(R)=1$.

So, you first need to divide the matrix you have by $8$ to make sure the RHS is $1$. Then, you don't even have to compute the eigenvalues, because when you multiply them together you get the determinant anyways! Of course you can compute them if you want, but in general just taking the determinant is easier if you just want the area.