Finding the axis and orientation of an ellipse

Given ${\displaystyle 5x^2+10y^2-12xy=14}$
When this is translated into matrix-vector form, we define the position vector as $p=\left[\begin{array}{c}x\\ y\end{array}\right]$, then we write the equation as ${\displaystyle p^{T}Ap+b^{T}p+c=0}$
Where A is a symmetric ${\displaystyle 2\times 2}$ matrix, $b$ is a ${\displaystyle 2\times 1}$ vector, and c is a scalar.
The first term is ${\displaystyle p^{T}Ap=x^2{A}_{11}+y^2{A}_{22}+2xy{A}_{12}}$
Comparing with the given equation, one finds that
$A=\left[\begin{array}{ccc}5& & -6\\ -6& & 10\end{array}\right]$
The second term is ${\displaystyle b^{T}p=b_{1}x+b_{2}y}$
Comparing with the given equation, we find that b is the zero vector.
Finally, taking $14$ to the left hand side, we deduce that ${\displaystyle c=-14}$
Therefore, the equation in matrix-vector form is
${\displaystyle p^{T}Ap+c=0}$
The next thing you need to do is diagonlize matrix $A$, i.e. find a rotation matrix $R$ and a diagonal matrix $D$ such that ${\displaystyle A=RD{R}^T}$
There is a standard way to do this diagonalization that you should memorize
1. Calculate $\theta =\frac{1}{2}{\tan }^{-1}\frac{2A_{12}}{A_{11}-A_{22}}$
2. Calculate the rotation matrix $R=\left[\begin{array}{ccc}\cos \theta & & -\sin \theta \\ \sin \theta & & \cos \theta \end{array}\right]$
3. Calculate the diagonal matrix $D=\left[\begin{array}{ccc}{D}_{11}& & 0\\ 0& & D_{22}\end{array}\right]$ where
${\displaystyle D_{11}=A_{11}{\cos }^2\theta +A_{22}{\sin }^2\theta +A_{12}\sin 2\theta }$
${\displaystyle D_{22}=A_{11}{\sin }^2\theta +A_{22}{\cos }^2\theta -A_{12}\sin 2\theta }$
Following the above steps, we find that $\theta =\frac{1}{2}{\tan }^{-1}\frac{-12}{-5}$
Therefore ${\displaystyle \theta }$ (and ${\displaystyle \left(2\theta \right)}$ are in the first quadrant. Using the trigonometric identities
${\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta )$
Since $\tan \left(2\theta \right)=\frac{12}{5}$
Hence $\cos \theta =\sqrt{\frac{1}{2}\left(\frac{18}{13}\right)}=\frac{3}{\sqrt{13}}$ and $\sin \theta =\sqrt{\frac{1}{2}\left(\frac{8}{13}\right)}=\frac{2}{\sqrt{13}}$
Thus for the second step, we have the rotation matrix as
$R=\frac{1}{\sqrt{13}}\left[\begin{array}{ccc}3& & -2\\ 2& & 3\end{array}\right]$
For the third step, we have for diagonal matrix
$D_{11}=5\left(\frac{9}{13}\right)+10\left(\frac{4}{13}\right)-6\left(2\right)\left(\frac{6}{13}\right)=1$
$D_{22}=5\left(\frac{4}{13}\right)+10\left(\frac{9}{13}\right)+6\left(2\right)\left(\frac{6}{13}\right)=14$
With all these calculations, we can now write the equation of the ellipse as
${\displaystyle p^{T}RD{R}^T+c=0}$
To put this in the standard form divide by ${\displaystyle (-c)=-(-14)=14}$, then ${\displaystyle p^{T}RE{R}^{T}p=1}$
where the matrix ${\displaystyle E=\left(\frac{D}{14}\right)}$ is equal to
$E=D/14=\left[\begin{array}{ccc}\frac{1}{14}& & 0\\ 0& & 1\end{array}\right]$
Now define the vector ${\displaystyle q=R^{T}p}$ so that ${\displaystyle p=Rq}$, then it follows that ${\displaystyle q^{T}Eq=1}$
Hence $q_1^2{E}_{11}+q_2^2{E}_{22}=\frac{q_1^2}{14}+\frac{q_2^2}{1}=1$
Thus the coordinate vector of q lies on an ellipse (in standard orientation) with semi-major axis ${\displaystyle =\sqrt{14}}$, and semi-minor axis ${\displaystyle =\sqrt{1}=1}$. Our ellipse which is in terms of the vector p is just a rotation of q-ellipse by the angle ${\displaystyle \theta }$ because ${\displaystyle p=Rq}$, and R is a rotation matrix by angle ${\displaystyle \theta }$ (counter clockwise).