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Determine the locus equation

  1. Jun 4, 2007 #1
    here's the problem:

    in a backyard, there are two trees located at grid points A(-2, 3) and B(4,-6)

    a) a dog walks through the backyard so that it is at all times twice as far from A as it is from B. find the equation of the locus of the dog.

    so PA = 2(PB)
    (x+2)^2 + (y-3)^2 = 4((x-4)^2 + (y+6)^2)

    i asked a teacher and they said it's a ellipse and the equation is
    5x^2(2) - 28x + 2y^2 + 6y + 113
    but that doesnt make sense because the course hasnt dealt with ellipses yet, just lines and circles. does it have to be an ellipse?



    b) the family cat is also walking in the backyard. the line segments between the cat and the two trees are always perpendicular. find the equation for the locus of the cat.

    so this means it path is equidistant from AB?
    so PA = PB
    (x+2)^2 + (y-3)^2 = (x-4)^2 + (y+6)^2

    with my math i end up with y = 2.17 - 0.67x.. but this doesnt look right. there must be something wrong.

    any help is appreciated!

    ~Amy
     
  2. jcsd
  3. Jun 4, 2007 #2

    Dick

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    The teacher is clearly wrong about the equation in a). And is right in saying that it is an ellipse only the sense that a circle is a special case of an ellipse. For b), perpendicular does not mean equidistant. Draw a picture. Since the two lines are perpendicular they form a right triangle with the line between the two trees. Think Pythagoras.
     
  4. Jun 4, 2007 #3
    im drawing a blank. so -a- IS a circle? also, when it comes to -b-, does it have anything to do with part -a-?

    thanks

    `Amy
     
  5. Jun 4, 2007 #4

    Dick

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    Yes, a) is a circle. You'll probably want to figure the center and radius. b) doesn't really have much to do with a). Did you draw that picture?
     
  6. Jun 4, 2007 #5
    i have a drawing of point A and point B

    i dont see how a circle would work just by looking at the drawing.

    are you sure for -b- that its not PA = PB
    (x+2)^2 + (y-3)^2 = (x-4)^2 + (y+6)^2

    how is 'equidistant' different then 'perpendicular'?

    ~Amy
    ~Amy
     
  7. Jun 4, 2007 #6

    Dick

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    Consider the triangle PAB. The angle at point P is a right angle, correct?
     
  8. Jun 4, 2007 #7
    yes it is
    ~Amy
     
  9. Jun 4, 2007 #8

    Dick

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    Please continue. What does our most favoritest theorem about right triangles tell us about the relation between PA, PB and AB?
     
  10. Jun 4, 2007 #9
    *square root sign*( (x2 - x1)^2 + (y2 - y1)^2
     
  11. Jun 4, 2007 #10

    Dick

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    Where did the x's and y's come from? We were talking about PA, PB and AB. The three sides of the right triangle.
     
  12. Jun 4, 2007 #11
    (i got that equation from my book)

    PA = PB?
     
  13. Jun 4, 2007 #12

    Dick

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    Is it true that for EVERY right triangle the two sides are equal? What would Pythagoras say to that?
     
  14. Jun 4, 2007 #13
    no. PA^2 +PB^2 = AB^2?

    ~Amy
     
  15. Jun 4, 2007 #14

    Dick

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    Right!!! That's the equation you want, yes?
     
  16. Jun 4, 2007 #15
    so i should eventually have:

    (x^2 + 4x + 4 + y^2 - 6y + 9) + (x^2 - 8x +1b + y^2 + 12y + 36) = AB
    and then i figure it out from there?

    ~Amy
     
  17. Jun 4, 2007 #16

    Dick

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    Yes, except you want AB^2=6^2+9^2, right?
     
  18. Jun 4, 2007 #17
    (not sure where you got the 6^2 or 9^2 from) :shy:

    i'll try to figure it out tomorrow (i have to get going). thanks for the help!

    ~Amy
     
  19. Jun 4, 2007 #18

    Dick

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    Ok, except the right hand side should be AB^2 (an easily calculated number), right?
     
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