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If I've got 2 points in the complex plane, A and B, of respective affixes a and b, how could I find the set of all points M, of affix z, such that the difference between the distances AM and BM equals a given constant d, ie such that |a-z|-|b-z|=d ?

It feels like there should be a general solution for this. If d=0, then it would just be the perpedicular bisector of the segment [AB]. But once I give d a different value, whether I work it out in imaginary and real parts, or in modulus and complex argument, I get really clunky equations. There's got to be an elegant solution out there, right ?

What's more, if you've got such a point M, then the point symetric to M according to the segment [AB] also works out, so there seems to be a general symmetry to this problem.

Any help on this would be appreciated, thanks.

The Pointer
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Eli
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    The set of points whose difference of distances to two fixed points $A$ and $B$ is a branch of hyperbola with foci $A$ and $B$. – Jean Marie Dec 24 '21 at 22:56

2 Answers2

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As 'Jean Marie' said, the set of points whose difference of distances from $F_1$ and $F_2$ is a fixed number, constitutes a hyperbola in the complex plane. In what follows it is assumed the $x$ coordinate is the real part of $z$, and $y$ is the imaginary part. To find the equation of this hyperbola, in the most direct way is as follows:

  1. The equation of the hyperbola in the rotated coordinate frame is

$ \dfrac{u^2}{a^2} - \dfrac{v^2}{b^2} = 1$

where the rotated coordinate is centered at the center of the hyperbola, and its $u$ axis extends along the vector pointing from $F_1$ to $F_2$.

  1. Define the vector $\Delta F = F_2 - F_1 $, then $| \Delta F | = 2 c $ where $c$ is the distance between each of the two foci ($F_1$ , $F_2$) and the center of the hyperbola which is at $C = \frac{1}{2} (F_1 + F_2) $.

  2. The relation between $c , a, b $ is as follows $c = \sqrt{a^2 + b^2} $

  3. Also, we have a following relation that involves $d$, namely,

$ d = 2 a $

  1. Thus, from $(4.)$, we can find $a$, then from $(3.), (2.)$ we can find $b$.

  2. Now the equation of the hyperbola is almost ready. The rotation matrix that expresses the hyperbola with the equation given in $(1.)$ in the $x, y$ coordinates, we need to relate $(x, y)$ to $(u, v)$

$ \begin{bmatrix} x \\y \end{bmatrix} = C + R \begin{bmatrix} u \\ v \end{bmatrix} $

  1. The rotation matrix R is given by

$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

where $\theta$ is the angle that $\Delta F$ makes with the positive $x$ axis.

  1. Finally, the equation of the hyperbola is

$ (r - C)^T Q (r - C) = 1 $

where $Q = R D R^T$ with $D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && - \dfrac{1}{b^2} \end{bmatrix} $ and $r = [x, y]^T $

  1. To find all the points on this hyperbola, define $w = R^T (r - C) $, then

$ r = C + R w $ and $w^T D w = 1$, from which it follows that $\dfrac{w_1^2}{a^2} - \dfrac{w_2^2}{b^2} = 1 $

  1. Hence, we can take $w_1 = a \sec \theta$ and $w_2 = b \tan \theta $

  2. From $(9.)$ we have a parameterization for vector $w = [w_1, w_2]^T $ which when we plug into $ r = C + R w $ we get $r$.

Hosam Hajeer
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Let $2\lambda=|a-b|$, $2c=a+b$, $\theta=\arg (b-a)$ and $2\lambda \cos v=d$ then $$z=c\pm \lambda\cosh [e^{i\theta}(u+vi)]$$

Also see another posts of mine here, here and here.

Ng Chung Tak
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