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Thread: parametric locus

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    parametric locus

    hi i have 2 problems for anyone who has time

    1. Tangents to the parabola x^2=4ay drawn from points P(2ap,ap^2) and Q(2aq,aq^2) intersect at right angles at point R. Find the locus of
    a) point R
    b) The mid point of PQ

    2.Find the equation of the locus of point R that is the intersection of the normals at P(2p,p^2) and Q(2q,q^2) in the parabola x=4Y,given that PQ=-4.

    the '^2' only applies to p of the y-coord ap^2.
    the '^2' only applies to q of the y-coord aq^2.

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    Sequential Timske's Avatar
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    Re: parametric locus

    Q1.
    Find the point of intersection of tangents at P and Q to find the coordinates of R.
    P(2ap, ap^2) and Q(2aq, aq^2)

    x^2 = 4ay
    y = x^2/4a
    y' = x/2a
    Sub x values in x/2a to to find gradient
    2ap/2a = p , 2aq/2a = q
    Gradient of tangent at P = p
    Gradient of tangent a Q = q

    using point gradient formula , y-y1 = m(x - x1)
    Tp; y - ap^2 = p(x - 2ap)
    Tp = px - ap^2
    therefore, Tq = qx - aq^2

    solve simulataneously
    y = px - ap^2 ...1
    y = qx - aq^2 ...2
    1 - 2
    - px + qx + ap^2 - aq^ 2 = 0
    a(p^2-q^2) = px - qx
    [a(p-q)(p+q)/(p-q)] = x
    x = a(p+q)
    SUB INTO 1
    y - pa(p+q) + ap^2 = 0
    y - ap^2 - apq + ap^2 = 0
    y = apq
    R is (a(p+q), apq)
    Locus, y = -4a



    2. Midpoint of PQ use midpoint formula [(x1 + x2)/2 , (y1 + y2)/2]
    P(2ap,ap^2) and Q(2aq,aq^2)

    [(2ap + 2aq)/2 , (ap^2+aq^2)/2]
    M - [ap+aq , (ap^2 + aq^2)/2]
    B ECO@USYD 2013

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    Executive Member Nooblet94's Avatar
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    Re: parametric locus

    Quote Originally Posted by Timske View Post
    Q1.
    Find the point of intersection of tangents at P and Q to find the coordinates of R.
    P(2ap, ap^2) and Q(2aq, aq^2)

    x^2 = 4ay
    y = x^2/4a
    y' = x/2a
    Sub x values in x/2a to to find gradient
    2ap/2a = p , 2aq/2a = q
    Gradient of tangent at P = p
    Gradient of tangent a Q = q

    using point gradient formula , y-y1 = m(x - x1)
    Tp; y - ap^2 = p(x - 2ap)
    Tp = px - ap^2
    therefore, Tq = qx - aq^2

    solve simulataneously
    y = px - ap^2 ...1
    y = qx - aq^2 ...2
    1 - 2
    - px + qx + ap^2 - aq^ 2 = 0
    a(p^2-q^2) = px - qx
    [a(p-q)(p+q)/(p-q)] = x
    x = a(p+q)
    SUB INTO 1
    y - pa(p+q) + ap^2 = 0
    y - ap^2 - apq + ap^2 = 0
    y = apq
    R is (a(p+q), apq)
    Locus, y = -4a



    2. Midpoint of PQ use midpoint formula [(x1 + x2)/2 , (y1 + y2)/2]
    P(2ap,ap^2) and Q(2aq,aq^2)

    [(2ap + 2aq)/2 , (ap^2+aq^2)/2]
    M - [ap+aq , (ap^2 + aq^2)/2]
    Locus of the first one should be y=-a, shouldn't it?

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    Sequential Timske's Avatar
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    Re: parametric locus

    Uh yeahy bad because pq=-1
    B ECO@USYD 2013

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    Re: parametric locus

    hey how did you establish pq=-1?

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    Re: parametric locus

    because the tangents are perpendicular, then the gradients of the tangents must be equal to -1
    since the gradients of the tangents to the curve at P and Q are p and q respectively, therefore pq=-1
    2012 HSC: 4U Maths, English Advanced, Economics, Chemistry

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    Re: parametric locus

    ooooooo lol .
    thanks

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