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Math Help - geometry of parabolas

  1. #1
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    geometry of parabolas

    The points P(2ap,ap^2) and Q(2aq, aq^2) vary on the parabola x^2=4ay. The chord PQ subtends a right angle at the vertex. The tangents at P and Q meet at T, while the normals at P and Q meet at N
    Find the cartesian equation of the locus of N


    So in this qs I have already proved that pq = -4 and that T has coordinates (a(p+q),apq) and that N has coordinates (-apq(p+q),a(p^2+pq+q^2+2))
    i just need to use the coordinates of N and find the locus, also using T and pq but i keep getting it wrong, if someone could help me, i would REALLY appreciate it, the answer is x^2 = 16a(y-6a)

    thankyou !! =D
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  2. #2
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    Quote Originally Posted by flyinhigh123 View Post
    The points P(2ap,ap^2) and Q(2aq, aq^2) vary on the parabola x^2=4ay. The chord PQ subtends a right angle at the vertex. The tangents at P and Q meet at T, while the normals at P and Q meet at N
    Find the cartesian equation of the locus of N


    So in this qs I have already proved that pq = -4 and that T has coordinates (a(p+q),apq) and that N has coordinates (-apq(p+q),a(p^2+pq+q^2+2))
    i just need to use the coordinates of N and find the locus, also using T and pq but i keep getting it wrong, if someone could help me, i would REALLY appreciate it, the answer is x^2 = 16a(y-6a)

    thankyou !! =D
    Given

    pq=-4 (1)

    x=-apq(p+q) (2)

    y=a(p^2+pq+q^2+2) (3)

    we must eliminate p and q to obtain an equation connecting x and y.

    Subst from (1) into (2):

    x = 4a(p+q)

    \Rightarrow x^2=16a^2(p+q)^2

    =16a^2(p^2+2pq+q^2)

    =16a^2(p^2 +pq +q^2 +2+pq-2)

    =16a\times \Big(a(p^2+pq+q^2+2) + a(pq-2)\Big)

    =16a(y -6a), from (1) and (3)

    Grandad
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