Distance of closest approach of ellipse and line segment

Similarly to the definition of the distance of closest approach (DCA) for two arbitrary ellipses^{1}
we define the DCA for an arbitrary ellipse and a line segment.

This distance is used in some models to calculate the repulsive force between a pedestrian and a wall^{2}.

The quantity we are looking for is \(l\) with:

\[l = \parallel \vec{OP} \parallel − r − d\]\(P\) is the nearest point on \([AB]\) to the ellipse. See also notes of P. Bourke^{3}.

Knowing \(\alpha\) we can easily calculate \(r\). To solve the above mentioned equation

one has to find the quantity \(d\), which would be the necessary
amount to translate \([AB]\) along the direction of \(OP\) such that it becomes tangential to the ellipse.

We have

\[x_{R^\prime} = x_R − d \cdot \cos(\alpha) \;\; \rm{and} \;\; y_{R^\prime} = y_R − d \cdot \sin(\alpha).\]The parametric definition of the line segment \([AB]\) is:

\[x_R = x_A +u\cdot(x_B −x_A)\;\; \rm{and}\;\; y_R = y_A + u\cdot(y_B −y_A); u \in [0, 1].\]\(x_{R^\prime}\) is on the ellipse, which implies

\[\frac{\Big(x_R − d \cdot cos(\alpha)\Big)^2}{a^2} + \frac{\Big(y_R − d \cdot sin(\alpha)\Big)^2}{b^2} = 1, \;\;\;\;\;\;\;\;\;\; [1]\]or

\[p \cdot d^2 +q \cdot d+s=0, \;\;\;\;\;\;\;\;\;\; [2]\]with

\[\begin{equation} \begin{cases} p = cos(\alpha)^2 + sin(\alpha)^2 > 0,\\ q = −2\cdot \Big(\frac{x_R \cdot \cos(\alpha)}{a^2} + \frac{y_R \cdot \sin(\alpha)}{b^2}\Big),\; \rm{and}\\ s = \frac{x_R^2}{a^2} + \frac{y_R^2}{b^2}− 1.\\ \end{cases} \end{equation}\]In case the point \(R\) is known the solution of Eq. (2) is

\[d = \frac{-q - \sqrt{\Delta}}{2p},\;\; \rm{and}\;\; \Delta = q^2 - 4p\cdot s.\]However, in general \(R\) is not known. Developing Eq. (1) with respect to \(u\) yields

\[p_1\cdot u^2 + q_1\cdot u + s_1 = 0,\]with

\[\begin{equation} \begin{cases} p_1 = \frac{x_{BA}^2}{a^2} + \frac{y_{BA}^2}{b^2} \ge 0,\\ q_1 = 2\cdot \Big(\frac{x_A - d\cdot \cos(\alpha)}{a^2}\cdot x_{BA} + \frac{y_A - d \cdot \sin(\alpha)}{b^2}\cdot y_{BA}\Big),\; \rm{and}\\ s_1 = \Big(\frac{x_A - d\cdot \cos(\alpha)}{a}\Big)^2 + \Big(\frac{y_A - d\cdot \cos(\alpha)}{b}\Big)^2− 1,\\ \end{cases} \end{equation}\]with the substitutions \(x_{BA}=x_B-x_A\) and \(y_{BA} = y_B - y_A\).

Since the line \((AB)\) is tangential to the ellipse we get

\[\Delta = q_1^2 - 4\cdot p_1\cdot s_1 = 0,\]which leads to

\[q_1^2 = 4\cdot p_1\cdot s_1. \;\;\;\;\;\;\;\;\;\; [3]\]Supposing that \(P, A\) and \(B\) are not collinear, we solve (3) and get

\[d_{1,2} = \frac{\pm a\cdot b\cdot \sqrt{p_1} - x_{BA} \cdot y_B + y_{BA}\cdot x_A}{y_{BA}\cdot \cos(\alpha) - x_{BA}\cdot \sin(\alpha)}\]and

\[d = \min(|d_1|, |d_2|).\]For \(d\) we calculate \(u\) as

\[u = \frac{-q_1}{2\cdot p_1}.\]If the inequality \(0 \le u \le 1 \) does not hold or \(P, A\) and \(B\) are collinear, then this would mean the point \(R\) is an end point of \([AB]\), i.e. \(A\) or \(B\). In that case we solve (2) twice with \(R:=A\) and \(R:=B\) and get \(d= \min(|d_A|, |d_B|)\).

**References:**