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Locus of points with distance sum =8

  1. Mar 23, 2005 #1
    Locus of points such that the sum of whose distances from A(0,-2) and B(0,2) is 8

    I plotted the points its a straight line on the y axis. I sub both points into the distance formula (seperate distance formulas) and then made the eqn = 8 I squared both sides getting rid of the square root and then simplified and got
    [tex] 2x^2 +2y^2+8=64 [/tex] I think I did this question wrong plz help me :cry:
  2. jcsd
  3. Mar 23, 2005 #2
    their talking about the sum of the distances from A and B. So you need to come up with an equation for the the distance from A and the distance from B seperatly, and then add them...

    that sum should equal 8

    to see the motivation behind this problem look up the definition of an ellpise, they even give a method to solve this problem here.
    Last edited: Mar 23, 2005
  4. Mar 23, 2005 #3
    Here they are separately

    Point A = square root[(x^2)+(y+2)^2]

    Point B= square root [(x^2)+(y-2)^2]

    I did make two different equations but when you say for the distance from A and from B I dont get FROM WHAT?

    I added the above two equations and set it =8 and then solved but I think I did something wrong.

    shouldnt I be finding an equation for the locus why am I verifying that point A's distance + point B's =8?
  5. Mar 24, 2005 #4


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    The distance from A and from B TO any point that satisfies the conditions you were given! i.e. any point (call it C, for example) in the plane such that if you add how far C is from A, to how far C is from B, you get 8. Stated another way: the sum of the distances to point C from point A and from point B is 8. There is more than one such point 'C'...indeed they trace out a curve, and this set of points on the curve that satisfies that condition is called the locus.

    Edit: the animation on that link MathStudent gave you illustrates it very well.
  6. Mar 24, 2005 #5
    Yes I see what you mean the animation is good as well but it is hard to find the equation for the locus did you see how many steps their are? Isnt there an easier way?
  7. Mar 24, 2005 #6
    Sure. They just simplified the equation to a nice form. The equation

    [tex]\sqrt{x^2 + (y+2)^2} + \sqrt{x^2 + (y-2)^2} = 8[/tex]

    defines the locus just as well as the standard ellipse equation.
    Last edited: Mar 24, 2005
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