Find the locus of points M the difference of the squares of whose distances from two given points A and B is equal to a given value c.

Question

Find the locus of points M the difference of the squares of whose distances from two given points A and B is equal to a given value c. For what values of c does the problem have a solution?

I am trying to understand how to solve this. Here is my attempt at a solution:

Choose a system of coordinates on the plane such that the origin is located at point A and the positive part of the x-axis lies along AB. We take the length of AB as the unit of length. Then the point A will have coordinates (0,0), and the point B will have the coordinates (1,0). The coordinates of the point M we denote by (x,y).

The condition $d (A,M)^2 - d (B,M)^2 = c$ is written in coordinates as follows: $$(\sqrt{x^2+y^2})^2 - (\sqrt{(x-1)^2 +y^2})^2 = c$$ which can be written as: $$ x^2+y^2 - [(x-1)^2 +y^2]= c$$ $$ x^2+y^2 -(x^2 -2x +1 +y^2) =c$$

$$2x -1 = c$$ Is this correct so far?

Answer 1

What you have done is "correct so far". But note that the problem asked for the geometrical description of a certain set $M$ (may be a parabola, or whatever), while you have ended with an equation whose interpretation is dangling.

In the formulation of the problem it was left open whether you want (i) $\ |PA|^2-|PB|^2=c$ or else (ii) $\ \bigl||PA|^2-|PB|^2\bigr|=c$. In the second case the set $M$ is empty when $c<0$, and disconnected when $c>0$.

In order to bring the inherent symmetries of the problem to the fore I'd suggest that you assume $|AB|=d>0$ and choose $A=\bigl({-{d\over2}},0\bigr)$, $B= \bigl({d\over2},0\bigr)$. Furthermore it is unnecessary to introduce square roots here; they are apt to cause second questions about signs, introducing spurious roots, etc.

In the following I shall treat the interpretation (i) $\ |PA|^2-|PB|^2=c$ for given $c\in{\mathbb R}$. This means that we are looking at the set $M$ of points $P=(x,y)$ satisfying $$|PA|^2-|PB|^2=\left(\bigl(x+{d\over2}\bigl)^2+y^2\right)-\left(\bigl(x-{d\over2}\bigr)^2+y^2\right)=c\ .$$ This condition simplifies to $$2d x=c\ .$$ It follows that $M$ is the vertical line $x={c\over 2d}$in the $(x,y)$-plane. Adopting the interpretation (ii) we would get two vertical lines when $c>0$.

Answer 2

What you have done is alright, but can improve in mathematical rigor. Your result already shows that vertical line/lines can be solution to the given problem.

I would state the problem slightly changed as follows keeping in mind difference or sum of distances is still a distance in its physical dimension:

Find the locus of points M the difference of the squares of whose distances from two given points A and B distant $d$ apart is equal to a given value $c^2 $. For what values of c does the problem have a solution?

Let the given points be $(-d,0)$ and $(0,0)$ . Coordinates of the point M are (x,y).

$$({x^2+y^2}) - {((x-d)^2 +y^2}) = c^2 $$ which simplifies to: $$ x = \frac {c^2+d^2}{ 2 \, d} $$

This is a $single$ line parallel to y-axis for each given value of $c$.

It should be a single line, the fact can be verified for $ c=0 ,x = d/2 $ the perpendicular bisector of line joining given points is a single line.

EDIT1:

That is, if the difference of squares is > 0 the line obtained is symmetric about $ x=0 $ to the line if the difference is < 0.