How can we simplify such ellipse to the form of an ellipse $$\frac{{(x - h)^2}}{{a^2}} + \frac{{(y - k)^2}}{{b^2}} = 1$$ As far as I know and could simplify was when $B$ in such second-degree equations is zero. Because then even if we have the general second-degree equation $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ We have $B = 0$ then we can complete the square and still obtain the desired equation of the ellipse in the form of $$\frac{{(x - h)^2}}{{a^2}} + \frac{{(y - k)^2}}{{b^2}} = 1$$
I am not sure how to go about doing this! As an example I have been trying to simplify $3x^2 + 2xy +3y^2 = 8$ But to no avail. Thanks for any advice!
Hint.
Supposing that $3 x^2 + 2 x y + 3 y^2 - 8=0$ represents an ellipse, making the change of variables
$$ \left(\matrix{x\\ y}\right) = \left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\left(\matrix{X\\ Y}\right) $$
we get
$$ \sin (2 \theta ) (X^2-Y^2)+2 X Y \cos (2 \theta )+3 X^2+3 Y^2-8=0 $$
now choosing $\theta = \frac{\pi}{4}$ to eliminate the product $XY$ we arrive at
$$ 4 X^2 + 2 Y^2 - 8 = 0 $$