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2022-04-05

Finding the axis and orientation of an ellipse with matrices
So I have this ellipse equation:
$5{x}^{2}+10{y}^{2}-12xy=14$
I'm asked to get the lengh of the semi-major and semi-minor axis, and it's orientation.

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Given $5{x}^{2}+10{y}^{2}-12xy=14$
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 ${p}^{T}Ap+{b}^{T}p+c=0$
Where A is a symmetric $2×2$ matrix, $b$ is a $2×1$ vector, and c is a scalar.
The first term is ${p}^{T}Ap={x}^{2}{A}_{11}+{y}^{2}{A}_{22}+2xy{A}_{12}$
Comparing with the given equation, one finds that

The second term is ${b}^{T}p={b}_{1}x+{b}_{2}y$
Comparing with the given equation, we find that b is the zero vector.
Finally, taking $14$ to the left hand side, we deduce that $c=-14$
Therefore, the equation in matrix-vector form is
${p}^{T}Ap+c=0$
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 $A=RD{R}^{T}$
There is a standard way to do this diagonalization that you should memorize
1. Calculate $\theta =\frac{1}{2}{\mathrm{tan}}^{-1}\frac{2{A}_{12}}{{A}_{11}-{A}_{22}}$
2. Calculate the rotation matrix
3. Calculate the diagonal matrix  where
${D}_{11}={A}_{11}{\mathrm{cos}}^{2}\theta +{A}_{22}{\mathrm{sin}}^{2}\theta +{A}_{12}\mathrm{sin}2\theta$
${D}_{22}={A}_{11}{\mathrm{sin}}^{2}\theta +{A}_{22}{\mathrm{cos}}^{2}\theta -{A}_{12}\mathrm{sin}2\theta$
Following the above steps, we find that $\theta =\frac{1}{2}{\mathrm{tan}}^{-1}\frac{-12}{-5}$
Therefore $\theta$ (and $\left(2\theta \right)$ are in the first quadrant. Using the trigonometric identities
${\mathrm{cos}}^{2}\theta =\frac{1}{2}\left(1+\mathrm{cos}2\theta \right)$
Since $\mathrm{tan}\left(2\theta \right)=\frac{12}{5}$
Hence $\mathrm{cos}\theta =\sqrt{\frac{1}{2}\left(\frac{18}{13}\right)}=\frac{3}{\sqrt{13}}$ and $\mathrm{sin}\theta =\sqrt{\frac{1}{2}\left(\frac{8}{13}\right)}=\frac{2}{\sqrt{13}}$
Thus for the second step, we have the rotation matrix as

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
${p}^{T}RD{R}^{T}+c=0$
To put this in the standard form divide by $\left(-c\right)=-\left(-14\right)=14$, then ${p}^{T}RE{R}^{T}p=1$
where the matrix $E=\left(\frac{D}{14}\right)$ is equal to

Now define the vector $q={R}^{T}p$ so that $p=Rq$, then it follows that ${q}^{T}Eq=1$
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 $=\sqrt{14}$, and semi-minor axis $=\sqrt{1}=1$. Our ellipse which is in terms of the vector p is just a rotation of q-ellipse by the angle $\theta$ because $p=Rq$, and R is a rotation matrix by angle $\theta$ (counter clockwise).

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