What is the relationship between ellipse and hyperbola?

Question

I am working with lines both in polar ($\rho-\theta$) and Cartesian co-ordinates. Initially, the line is given by $(\rho_0, \theta_0)$ and the covariance in the parameter space is $$R=\left(\begin{array}{cc} A & B\\ B & C \end{array}\right).$$

I am able to plot the error covariance ellipse in the parameter space using Choleski decomposition as shown how to plot a 2D covariance error ellipse?, or using convectors and values (see ellipse equation from eigenvectors and eigenvalues).

To show the expression for the covariance ellipse using the eigen method, assume the eigenvectors to be $d_1=(a_{11}, a_{12})$ and $d_2=(a_{21}, a_{22})$ while the eigenvalues $v_1$ and $v_2$. Then the expression for the covariance ellipse is

$$ K_1\left(\rho-\rho_0\right)^2+K_2\left(\rho-\rho_0\right)\left(\theta-\theta_0\right)+K_3\left(\theta-\theta_0\right)^2=1. $$

Where $$ K_1 = \left(\frac{a_{11}^2}{v_1^2}+\frac{a_{12}^2}{v_2^2}\right),\\ K_2 = 2\left(\frac{a_{11}a_{21}}{v_1^2}+\frac{a_{12}a_{22}}{v_2^2}\right),\\ K_3 = \left(\frac{a_{21}^2}{v_1^2}+\frac{a_{22}^2}{v_2^2}\right). $$

The eigenvalues can be scaled to satisfy a different confidence level using Chi-squared table (I do not remember where I saw this).

My question is how can I transform this confidence interval into a cartesian space around the line. I need a closed form expression either using the eigenvectors and eigenvalues or the covaraince matrix.

Further information

Hoping an example might clarify things better. The shaded region in the following image is formed from lines that lie on the edge of a covariance ellipse.

The initial line is shown in dark. It seems the other lines are tangent to some hyperbola. The red hyperbola, in the figure, is an approximate solution I found using an implementation that I saw somewhere. Accordingly, if a point $P$ lies $m$ distance from the point $(\rho_0\cos(\theta_0), \rho_0\sin(\theta_0))$ on the line, the distance from this point to the hyperbola $d_n$ is given by:

$$d_n= \pm\sqrt{C*m^2+B*m+A}$$

Problem with this approximation:

1. It is not perfectly aligned with the edge of the shaded region.

2. I have no idea where this came from.

I am more comfortable with geometry than algebra, but I will appreciate any suggestion.

Answer 1

Your question is important. Permit me to reformulate it as I have met the issue some years ago.

Consider a family of straight lines with "normal equations":

$$x \cos(\theta) + y \sin(\theta) - p=0$$

for which we have fluctuations $\Delta \theta, \Delta p$ around a nominal value. Observation and/or simulation give rise to hyperbolic locii (plural of "locus"). Just as Gaussian behavior gives rise to incertitude ellipses. How to interpret a hyperbolic locus? How can we get certain characteristics out of it (for example, how its "narrowness" can be quantified in statistical terms ?...).

One good answer is through the duality "pole-polar". Have a look at the graphics below. I made it with Geogebra software that has a "Polar" tool (underlining its importance in geometry). All is with reference to circle centered in $0$ with radius $R=2$, but it could any other circle or even any other conic curve.

An example of the pole-polar association: point M has for its polar line the green line which has two properties: it is orthogonal to $OM$, and if $S$ denotes the intersection point with $OM$, one has $OM.OS=R^2.$ Moreover, as $M$ is external to the circle, there is another means for obtaining its polar line: take the two tangent lines issued from $M$ to the circle ; let U and V be their contact points ; then line $UV$ is the polar line of $M$.

The main thing to be observed is that the polar lines of points situated on the hyperbola $\frak{H}$ are enveloping a certain ellipse $\frak{E}.$ This process can be done in a dual way: had we taken points on ellipse $\frak{E}$, we would have obtained tangents to hyperbola $\frak{H}$. In particular, if we take the polar line of the center $C$ of $\frak{E}$, we obtain the main axis of $\frak{H}$.

With this possible transfer, one can take advantage of reliable methods and techniques developed for gaussian distributions.

Moreover, what has been said for 2D is at once extensible to 3D, because duality is not limited to 2D: instead of a point-line coupling, it is a point-plane coupling with respect to a sphere.

Edit : All this is accessible to simple analytic computations. Let us take an example. Consider the second image below. One can see a hyperbola and an ellipse with resp. parametric equations:

$$\frak{H}\cases{x=\tan(t)\\y=\dfrac{0.2}{\cos(t)}+3} \ \ \ \ \ \ \frak{E}\cases{X=\dfrac{-0.2\sin(t)R^2}{0.2 \cos(t)+3}\\Y=\dfrac{R^2}{0.2 \cos(t)+3}}$$

(where the radius of the circle is $R=2$).

How has been found the second system for $\frak{E}$ knowing the first system (for $\frak{H}$)?

By using transformation:

$$\cases{X=\dfrac{-R^2y'}{xy'-yx'}\\X=\dfrac{R^2x'}{xy'-yx'}}.$$

We refer to this French site for more explanation. I will look for an equivalent English speaking site.