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Coordinate Geometry

  1. Aug 3, 2007 #1
    Question Statement
    The straight line y=mx -2 intersects the curve y^2=4x at the two points P(x1,y1) and Q(x2,y2). Show that
    1) m>-1/2, m not equal to 0.
    2) x1 + x2 = 4(m+1)/m^2
    3) y1 + y2 = 4/m

    If the point O is the origin and the point T is a point such that OPTQ is a parallelogram, show that, when m changes, the equation of the locus of T is y^2 + 4y = 4x

    My attempt:

    The first part of the question is pretty straightforward, but the second part of the question (regarding the parallelogram) is a bit confusing.

    My solution so far:
    Let T be (x,y)
    Gradient of OP = gradient of TQ

    y1/x1 = y-y2 / x-x2
    xy1 - x2y1 = yx1 - xy2
    yx1 = xy1 + xy2 - x2y1......(1)

    Similarly, Gradient of PT = gradient of OQ

    yx2 = xy2 -xy2 + x2y1......(2)

    (1) + (2)
    (x1 + x2)y = (y1 + y2)x
    4y(m+1)/ m^2 = 4x/m
    y(m+1)/m = x

    But I can't get rid of the m afterward :frown:
  2. jcsd
  3. Aug 3, 2007 #2
    If your parellogram is in the form OPTQ this means simply T is the sum of the vectors P and Q
    From this we conclude T=(x1+x2,y1+y2)
    and from the first part you know the values
    just try the equation given to you whether T is on that or not
    Last edited: Aug 4, 2007
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