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Math Help - Distance d from P

  1. #1
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    Distance d from P

    Let P = (x, y) be a point on the graph of y = x^2 + 2.

    Express the distance d from P to the point (2, 0) as a function of x.
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  2. #2
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    Quote Originally Posted by symmetry View Post
    Let P = (x, y) be a point on the graph of y = x^2 + 2.

    Express the distance d from P to the point (2, 0) as a function of x.
    Distance between two points (x1y1) and (x2,y2):
    d = sqrt[(x2 -x1)^2 +(y2 -y1)^2] ---------------------------------(i)

    or
    d = sqrt[(difference of the x's)^2 +(difference of the y's)^2] ------(ii)
    Meaning, (x1 -x2)^2 is also (x2 -x1)^2; (y2 -y1)^2 is same as (y1 -y2)^2.

    The point P(x,y) on the graph of y = x^2 +2
    is P(x,(x^2 +2)) in terms of x.

    The other point is (2,0)

    So,
    d = sqrt[(x -2)^2 +((x^2 +2) -0)
    d = sqrt[(x-2)^2 +(x^2 +2)]
    d = sqrt[(x^2 -4x +4) +(x^2 +2)]
    d = sqrt[x^2 -4x +4 +x^2 +2]
    d = sqrt[2x^2 -4x +6] --------------------answer.

    --------------
    Edit.

    Oooppps, wrong. I forgot to square the difference of the y's.

    It should have been like this:
    d = sqrt[(x -2)^2 +((x^2 +2) -0)^2]
    d = sqrt[(x-2)^2 +(x^2 +2)^2]
    d = sqrt[(x^2 -4x +4) +(x^4 +4x^2 +4)]
    d = sqrt[x^2 -4x +4 +x^4 +4x^2 +4]
    d = sqrt[x^4 +5x^2 -4x +8] --------------------final answer.
    Last edited by ticbol; January 20th 2007 at 01:14 AM.
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  3. #3
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    ok

    I thank you for every for the insight and math tips.

    Basically, we apply the distance formula.

    Good job!

    Thank!
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