Solving for a Constant: I.M Gelfand Trig Exercise 8 Explanation & Hint

  • Thread starter Thread starter cyberhat
  • Start date Start date
  • Tags Tags
    Trig
cyberhat
Messages
2
Reaction score
0
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, don't know 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 want to forfeit thinking opportunities either.


P.S I don't know why, but I keep thinking this equation represents a rectangle of sorts.
 
Physics news on Phys.org
cyberhat said:
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.
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.
cyberhat said:
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, don't know 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 want to forfeit thinking opportunities either.


P.S I don't know why, but I keep thinking this equation represents a rectangle of sorts.
 
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 [itex]X= (x_X, y_X)[/itex], [itex]A= (x_A, y_A)[/itex], and [itex]B= (x_B, yb_)[/tex]. Then the equation is <br /> [tex](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[/tex]<br /> <br /> Multiply those out and cancel as much as you can.[/itex]
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
2
Views
2K
Replies
7
Views
9K