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Coordinate Geometry question help!

  1. Apr 1, 2005 #1
    Hi there, i've been struggling with this question for days :cry: :confused: :grumpy: :yuck: , the first part where they call me to prove that equation, i could do it.....the second one i don't know how to do...could anyone help me? how do i find the locus equation? its so confusing..... ty in advance.

    If the normal at P(ap^2, 2ap) to the parabola y^2=4ax meets the curve again at Q(aq^2,2aq), prove that p^2 + pq + 2 = 0. Prove that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y^2(x + 2a) + 4a^3 = 0.

    ^ means to the power of...
  2. jcsd
  3. Apr 1, 2005 #2
    I know I'm supposed to give hints not complete solutions, but I really didn't know what kind of hint to give...

    The equation of the tangent to the parabola at the point (at^2, 2at) is:
    y = (1/t)x + at (I'll leave the proof for you.)

    Using this equation for points P and Q, we get:
    y = (1/p)x + ap
    y = (1/q)x + aq

    Solving these simultaneously twice (once for x and once for y), we get:
    (1) pq = x/a
    (2) p+q = y/a

    Re-writing (1) as q=x/(ap), then using it in (2) gives us:
    p^2 = (py-x)/a

    Now using p^2+pq+2=0 with the values for p^2 and q we just obtained gives us:
    (py-x)/a + x/a + 2 = 0
    p = (-2a)/y

    Now re-writing (2) as q=(y^2+2a)/ay and using the value of p we just obtained in (1) yields:
    pq = (-2y^2 - 4a^2)/y^2 = x/a

    Upon multiplying it out we get:
    -2ay^2 - 4a^3 = xy^2

    xy^2 + 2ay^2 + 4a^3 = 0
    y^2(x+2a) + 4a^3 = 0, as required.
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