# 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.

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?

## 2 Answers

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 $c>0$.

• Thank you very much for your time and helpful suggestions. I have thought about how you suggested I approach the problem. I came up with $c=2d^2$, a parabola, as the locus of points M. Commented Nov 22, 2014 at 13:51
• You are on the wrong track. Perform the same calculations as you did in your first approach, but using the disposition suggested in my answer. Then interpret the resulting condition on $P=(x,y)$ in words. Commented Nov 22, 2014 at 15:28
• Ok, using your disposition and performing the same calculations as in my first approach, I get:$$c= (\dfrac {-d}{2} -x)^2+ y^2) - (\dfrac{d}{2} -x)^2+ y^2)$$ $$c=x^2 + dx + \dfrac{d^2}{4} +y^2 - (x^2 - dx + \dfrac{d^2}{4} +y^2)$$ $$c = 2dx$$ and this is where I get stuck. Previously, I substituted $x=d$ to get $c= 2d^2$. However I am unsure how to proceed now with your suggestion to 'interpret the resulting condition on $P= (x,y)$ in words.' I do really want to understand this, so thank you again. Commented Nov 22, 2014 at 17:46

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.