Finding the long and short axes of an ellipses after applying an anisotropic scaling

Question

Suppose I have an ellipse with long axis $l_1$, short axis $s_1$ and say it has an orientation of $\frac{\pi}{2}$ (unnecessary but the graphic I remembered after I created the graphic attached that the usual convention is to have the long axis aligned to the x-axis).

Now suppose I apply an anisotropic scaling (scaling along two orthogonal axes); specifcally I scale by some factor $k_1$ in the direction of some vector $\vec{v}_1$ and by some other factor $k_2$ in the direction of a vector $\vec{v}_2$ that is orthogonal to $\vec{v}_1$. For non-trivial values such scaling maps an ellipse to an ellipse.

Given the directions of $\vec{v}_1$ and $\vec{v}_2$, the scaling factors $k_1$ and $k_2$ and the original ellipse parameters $l_1,s_1$ and $\theta_1$ (for completeness - its orientation relative to the x-axis / angle between long axis and x-axis), what are the new long and short axes - $l_1'$ and $s_1'$ respectively?

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Initial attempt:

Assuming it to be centered at the origin, I applied the transformations to the ellipse parameterization and then sought out to transform that parameterization to some form $(l_1' \cos(t)\cos(\theta_1') - s_1' \sin(t)\cos(\theta_1'),l_1' \cos(t)\sin(\theta_1') + s_1' \sin(t)\sin(\theta_1'))$ where $\theta_1'$ is the angle $l_1'$ makes with the x-axis. This turns out to be nearly impossible or at least way above what I'm seeking, since there are so many variables involved.

I also thought of finding the max of the ellipse parameterization after applying the transformations, however also not exactly straight forward since it depends on the scaling factors and directions - but could be numerically achievable I suppose when one provides specific scaling factors and directions. However, I'm not seeking a numerical solution.

So is there a straightforward analytic way to find the long and short axis of the transformed ellipse?

Thanks in advance!

Answer 1

The equation of the starting ellipse (assuming it is centered at the origin) can be written as

$ r^T Q r = 1 $

where $ r =[x, y]^T $

If the semi-axes are $a$ and $b$ with $a \gt b$, and assuming the orientation of the major axis is parallel to the y-axis then

$Q = \begin{bmatrix} \dfrac{1}{b^2} && 0 \\ 0 && \dfrac{1}{a^2} \end{bmatrix} $

Now you want to apply scaling to the points $r$

along a certain given pair of orthogonal unit vectors $v_1$ and $v_2$ such that the component of $r$ along $v_1$ get stretched by $k_1$ and the component of $r$ along $v_2$ gets stretched by $k_2$

The components of $r$ onto $v_1$ and $v_2$ are easily computed by defining

$R = [v_1, v_2] $

Then the components (the coordinates) of $r$ with respect to $v_1 $ and $v_2$ are given by

$r_1 = R^T r $

Next we want to stretch the two entries of $r_1$ by $k_1$ and $k_2$ respectively, so define the diagonal matrix $D$, as follows

$D = \begin{bmatrix} k_1 && 0 \\ 0 && k_2 \end{bmatrix}$

then it would follow that the stretched coordinates are given by $r_2$ where:

$ r_2 = D r_1 = D R^T r $

The image of $r$ is not $r_2$ but $ R r_2 $

i.e.

$r' = R D R^T r = A r $

Having found the image $r'$ of a point $r$ on the ellipse, we can write the equation of the new ellipse.

We have $ r = A^{-1} r' = R D^{-1} R^T r' $

substitute this into the equation of the original ellipse, to obtain,

$ r'^T R D^{-1} R^T Q R D^{-1} R^T r' = 1 $

Define $Q' = R D^{-1} R^T Q R D^{-1} R^T$

then the equation of the new ellipse is

$ r'^T Q' r' = 1 $

To obtain the semi-major and semi-minor axes lengths, we have to perform diagonalization of $Q'$ into

$Q' = R' D' R'^T $

Then the diagonal matrix $D'$ is of the form

$D' = \begin{bmatrix} \dfrac{1}{a'^2} && 0 \\ 0 && \dfrac{1}{b'^2} \end{bmatrix} $

Now $a'$ and $b'$ are the sought lengths of the semi-major and semi-minor axes. The directions of these axis are stored respectively in the matrix $R'$

Answer 2

I can suggest two different strategies.

1. Choose at will five points on the original ellipse and apply to them the anisotropic scaling. The transformed points define a conic which is then the transformed of the ellipse. From them you can get the equation of the new ellipse and then its axes.

2. From the center of the original ellipse trace a line parallel to $\vec v_1$, intersecting the ellipse at diameter $AB$. Construct then the conjugate diameter $CD$ (i.e. tangents at $C$ and $D$ are parallel to $\vec v_1$). The transformed segments $A'B'$ and $C'D'$ are conjugate diameters of the transformed ellipse, and you can then find its axes as explained here.