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Coordinate geometry

  1. Jul 9, 2005 #1
    Hi all (<--new user) I've spent the last 6 months trying to learn A-level mathematics in a vacuum after 9 years of letting my GCSE knowledge slide into oblivion, this is the first time however, I've hit a question for which how the book arrives at its answer is completely frying my brain. The question (the last of 12 questions) is as follows:


    a point P(a,b) is equidistant from the y axis and from the point (4,0). Find a relationship between a and b.

    The book gives its answer as b^2 = 8a - 16

    My problem is that the answer is correct, but I can't figure out the exact process they followed to reach it. Apart from plotting the points on a graph I am having difficulty trying to use the information in ways that don't end up looking ridiculous.

    Can anyone offer some suggestions?
     
  2. jcsd
  3. Jul 9, 2005 #2

    Hurkyl

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    Definitions are always a good place to start. What does:

    P(a,b) is equidistant from the y axis and from the point (4,0)

    mean?
     
  4. Jul 9, 2005 #3
    sorry :redface: using cartesian coordinates P(a,b) is equistant from the y axis and from the point (4,0)

    gonna be tricky to word this but if we call the point (4,0) Q...then the the point P is just as far away from the y axis as it is away from Q (ie: the line PQ is the same length as a horizontal line from P to the y axis)

    (a,b) are any coordinates

    does this help any? :redface:
     
  5. Jul 9, 2005 #4
    Not really. What is the algebraic definition of 'distance from' a point to a point? Or a point to an axis?

    (Think Pythagoras for the first one, 'common sense' for the second one).

    Write down the two distances as algebraic expressions, and put an "=" sign between them. That's equidistance, isn't it?
     
  6. Jul 9, 2005 #5

    Hurkyl

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    Yes: what you did is the first step to solving the problem.

    You said (I'm paraphrasing)

    "The distance from the point (a, b) to the y-axis is the same as the distance from the point (a, b) to the point (4, 0)"

    Now, as rachmaninoff suggested, this sounds an awful lot like an equation... what do the two halves of that sentence mean algebraically?
     
  7. Jul 9, 2005 #6
    sorry folks :redface: trying to explain the question (it was worded that way) even to myself is probably why I myself am having so much difficulty

    Given that P is at the point(a,b), if a point R lies on the y axis at a the point (0,b)
    PR = PQ (where Q is the point (4,0))
     
  8. Jul 9, 2005 #7
    if RP = PQ then an equation would be:
    root ((a-0)^2 + (b2-b1)^2) = root((4-a)^2 + (0-b)^2

    a^2 = 16 -8a + a^2 + b^2
    0 = 16 - 8a +b^2
    b^2 = 8a -16 :biggrin:
     
    Last edited: Jul 9, 2005
  9. Jul 9, 2005 #8
    cheers for making me think folks :biggrin:
     
    Last edited: Jul 9, 2005
  10. Jul 9, 2005 #9

    Hurkyl

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    Yay, you got it!
     
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