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

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What's left will involve only x_X and y_X. In fact, if you plot the graph of that equation, you'll see that it is a circle with center at the midpoint of the segment from A to B. That is, the set of all points X such that AX^2- BX^2= AB^2 is a circle with center at the midpoint of the segment from A to B.In summary, the question is about finding the set of points for which AX^2- BX^2 is constant, given two points A and B in the plane. The answer, as given by the book, is a circle with its center at the midpoint of the segment from A to B. The process of
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cyberhat
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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.
 
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  • #2
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.
 
  • #3
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
[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]

Multiply those out and cancel as much as you can.
 

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

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