zdebe5l8

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.

So I have this ellipse equation:

I'm asked to get the lengh of the semi-major and semi-minor axis, and it's orientation.

dabCrupedeedaejrg

Beginner2022-04-06Added 10 answers

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 $2\times 2$ matrix, $b$ is a $2\times 1$ 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

$A=\left[\begin{array}{ccc}5& & -6\\ -6& & 10\end{array}\right]$

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 $R=\left[\begin{array}{ccc}\mathrm{cos}\theta & & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & & \mathrm{cos}\theta \end{array}\right]$

3. Calculate the diagonal matrix $D=\left[\begin{array}{ccc}{D}_{11}& & 0\\ 0& & {D}_{22}\end{array}\right]$ where

${D}_{11}={A}_{11}{\mathrm{cos}}^{2}\theta +{A}_{22}{\mathrm{sin}}^{2}\theta +{A}_{12}\mathrm{sin}2\theta$

${D}_{22}={A}_{11}{\mathrm{sin}}^{2}\theta +{A}_{22}{\mathrm{cos}}^{2}\theta -{A}_{12}\mathrm{sin}2\theta$

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

${\mathrm{cos}}^{2}\theta =\frac{1}{2}(1+\mathrm{cos}2\theta )$

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

$R=\frac{1}{\sqrt{13}}\left[\begin{array}{ccc}3& & -2\\ 2& & 3\end{array}\right]$

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

${p}^{T}RD{R}^{T}+c=0$

To put this in the standard form divide by $(-c)=-(-14)=14$, then ${p}^{T}RE{R}^{T}p=1$

where the matrix $E=\left(\frac{D}{14}\right)$ is equal to

$E=D/14=\left[\begin{array}{ccc}\frac{1}{14}& & 0\\ 0& & 1\end{array}\right]$

Now define the vector $q={R}^{T}p$ 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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