# I.M Gelfand Trig Help

1. Aug 24, 2011

### cyberhat

Hey guys,

I'm currently a freshman at my local community college. I felt the need to solidify my foundation in Trig so I am currently doing a self-study course.

The question is from I.M Gelfand's book on Trigonometry. Chapter 0, page 9, exercise 8.

8) Two points, A and B, are given in the plane. Describe the set of points for which AX^2-BX^2 is constant.

The chapter focuses on right triangles and Pythagorean theorem. Dunno if this will help but I think exercise 7 was given as a hint to solve exercise 8:

Two points, A and B, are given in the plane. Describe the set of points X such that AX^2+BX^2=AB^2.

The book gave the answer: "A circle with its center at the midpoint AB".

I understood exercise 7 after a couple of minutes, but exercise 8 is making me pull hairs!

My attempt at the solution was to manipulate AX^2-BX^2 and get rid of the negative sign.....But even if it could be done, dunno how much help it would give me.

I'm not looking for complete answers -just a small hint. Can't spend 5 hours a day on one problem but don't wanna forfeit thinking opportunities either.

P.S I dunno why, but I keep thinking this equation represents a rectangle of sorts.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 24, 2011

### Staff: Mentor

Are you sure this is how the problem is phrased? A point in the plane has two coordinates, so I don't understand what AX^2 means in this context.

3. Aug 25, 2011

### HallsofIvy

I suspect that "AX" means the distance from A to X so that we are asking for the set of all points, X, such that the distance for A to X, squared, minus the distance from B to X, squared, is equal to the distance from A to B, squared.

Take $X= (x_X, y_X)$, $A= (x_A, y_A)$, and [itex]B= (x_B, yb_)[/tex]. Then the equation is
$$(x_X- x_A)^2+ (y_X- y_A)^2- (x_X- x_B)^2- (y_X- y_B)^2= (x_A- x_B)^2+ (y_A- y_B)^2$$

Multiply those out and cancel as much as you can.