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Math Help - Theoretical Coordinate Geometry

  1. #1
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    Theoretical Coordinate Geometry

    Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

    The line y=2kx cuts the curve y=x^2-4x+1 at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

    Now I tried putting 2kx=x^2-4x+1
    and this worked down to x=2+k \pm \sqrt[2]{k^2+4k+3}
    But I feel that would be finding the co-ordinates of a and b.
    Again, any help would be appreciated.
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  2. #2
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    Opinion

    Without knowing what a and b are, how can you determine what the midpoint is?
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    That's what I was wondering. Maybe there's some relationship between the midpoint and the axis of the curve...
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  4. #4
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    Quote Originally Posted by FreeT View Post
    Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

    The line y=2kx cuts the curve y=x^2-4x+1 at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

    Now I tried putting 2kx=x^2-4x+1
    and this worked down to x=2+k \pm \sqrt[2]{k^2+4k+3}
    But I feel that would be finding the co-ordinates of a and b.
    Again, any help would be appreciated.
    Hi FreeT,

    It would seem that you are on the right track.

    Once you have found the coordinates of a and b in terms of k,
    you can then determine the midpoint of ab.

    a(x_1, y_1) and b(x_2, y_2)

    Midpoint of ab = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
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  5. #5
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    Solve the equation x^2-4x+1=2kx for x in terms of k.
    That will give the x-coordinates of a~\&~b in terms of k.
    Use the equation y=2kx to find the y-coordinates of a~\&~b in terms of k.
    The midpoints are just the averages.
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    Quote Originally Posted by FreeT View Post
    Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

    The line y=2kx cuts the curve y=x^2-4x+1 at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

    Now I tried putting 2kx=x^2-4x+1
    and this worked down to x=2+k \pm \sqrt[2]{k^2+4k+3}
    But I feel that would be finding the co-ordinates of a and b.
    Again, any help would be appreciated.
    The tricky point is avoiding actually solving the equation. Here is one way.

    By substitution,
    2kx = x^2 - 4x +1
    so
    x^2 - (4+2k)x + 1 = 0

    If a and b are the roots of this equation, then their sum must be the negative of the coefficient of x, i.e.
    a+b = 4+2k.

    From here you can find \frac{a+b}{2} and the associated value of y.
    Last edited by awkward; May 23rd 2010 at 04:35 AM. Reason: corrrection
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