Generalized Conics with the Sharp Corners

Abstract

This paper analyses the properties of generalized conics that are defined by N focal points with weights. The generated shapes can be convex or concave. By varying the weights, it is possible to obtain up to N corners associated with the focal points. From the shape analysis perspective, the generalized conics extend the capabilities of the ellipse by adding the single extra parameter. In general, they enrich the potential of approaches to describe or represent the shape.

A. Gabdulkhakova—Supported by the Austrian Agency for International Cooperation in Education and Research (OeAD) within the OeAD Sonderstipendien program, financed by the Vienna PhD School of Informatics.

Appendices

Appendix A

Proof of Theorem  1

The pair of focal points, $F_1 \text { and } F_2$, generates the weighted multifocal ellipse, such that the weight, $0<\mu \le 1$, corresponds to $F_2$ (Fig. 3b). Let us denote $\delta (F_1,F_2)=2f$, $\delta (F_1,P)=n$, $\delta (F_2,P)=m$, and $\widehat{F_1F_2P}=\alpha$. By definition the level set contains a group of points mapped to the same distance value. So the distance value at $F_2$ is identical to the one at P. According to the normalized version of Eq. 6, it equals:

$$\begin{aligned} \delta (F_1,F_2) + \mu \delta (F_2,F_2) = \delta (F_1,F_2) = 2f \end{aligned}$$

(12)

As noted from Eq. 12, the distance value corresponding to the level set with the corner equals the length of the line segment $\overline{F_1F_2}$. Let us substitute it in the analogical equation for P:

$$\begin{aligned} \delta (F_1,P) + \mu \delta (F_2,P) = 2f \end{aligned}$$

(13)

$$\begin{aligned} n + \mu m = 2f \end{aligned}$$

(14)

$$\begin{aligned} \implies n = 2f - \mu m \end{aligned}$$

(15)

To derive an alternative estimate of m, consider a triangle $\bigtriangleup F_1PF_2$. According to the law of cosines [8]:

$$\begin{aligned} m^2 + 4f^2 - 4mf\cos \alpha = n^2 \end{aligned}$$

(16)

Substituting the value of n from Eq. 15 in Eq. 16 leads to:

$$\begin{aligned} m^2 + 4f^2 - 4mf\cos \alpha = 4f^2 - 4\mu mf + m^2\mu ^2 \end{aligned}$$

(17)

$$\begin{aligned} \implies m = \frac{4f(\mu -\cos \alpha )}{\mu ^2-1} \end{aligned}$$

(18)

The important assumption about the point P in continuous space states that it is infinitely close to $F_2$. This implies that the length of m converges to zero. Then Eq. 18 is further simplified:

$$\begin{aligned} m = \frac{4f(\mu - \cos \alpha )}{\mu ^2 - 1} = 0 \end{aligned}$$

(19)

$$\begin{aligned} \implies \mu = \cos \alpha \end{aligned}$$

(20)

According to Eq. 20, the angle formed at the corner of the level set depends on the weight of the focal point and not on the distance between the focal points.

Appendix B

Proof of Theorem  2

The pair of focal points, $F_1$ and $F_2$, generates the weighted multifocal hyperbola (Fig. 6b). The level set passing through $F_2$ contains the sharp corner. Let us denote $\delta (F_1,F_2)=2f$, $\delta (F_1,P)=n$, $\delta (F_2,P)=m$, and $\widehat{F_1F_2P}=\beta$. Assuming the normalized version of Eq. 9, the distance value at $F_2$ equals:

$$\begin{aligned} |\delta (F_1,F_2) - \mu \delta (F_2,F_2)| = \delta (F_1,F_2) = 2f \end{aligned}$$

(21)

Similar to the proof for the egg-shape, consider the triangle $\bigtriangleup F_1PF_2$ and derive the following relations:

$$\begin{aligned} n&= 2f + \mu m\end{aligned}$$

(22)

$$\begin{aligned} m&= \frac{-4f(\mu +\cos \beta )}{\mu ^2-1} \end{aligned}$$

(23)

In continuous space, m is infinitely small. In discrete space, it can be assigned to zero, resulting in: $\mu = -\cos \beta$. So the angle at the corner of the hyperbolic shape depends only on the weight of the respective focal point.