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    Hi, I'm currently studying FP3 and am having trouble understanding Ex2F, the finding loci subsection, so any help would be much appreciated. In particular, could anyone explain to me how to solve Q1b and 1c:

    'The tangent at P (ap^2, 2ap) and the tangent at Q (aq^2, 2aq) to the parabola y=4ax meet at R.

    a) FInd the coordinates of R -> (apq, ap + aq) using FP1 methods

    The chord PQ passes through the focus (a, 0) of the parabola.
    b) Show that the locus of R is the line x=-a.

    Given instead that the chord PQ has gradient of 2,
    c) Find the locus of R.'

    https://9ccd9b5f157a67423e5533d7dfd0...XFSYzg/CH2.pdf

    In the worked solutions ^ (p36), the major part that's puzzling me is the fact that it jumps from 2a + 2apq = 0 to the locus being x=-a. What's the working out in between those 2 equations, and where has x come from; the x-coordinate of R? But R doesn't lie on PQ so why is it being used? Same thing with part c, how is the locus of R y = a(p + q)?

    Also, in my working out for b) I got the chord PQ to be y(p + q) = 2x-2a by inserting (a, 0) rather than the co-ordinates of P as in the worked solutions. As pq = -1 that means my equation and the one in the solutions are equivalent, but I can't get to x=-a using this either.

    And finally and more in generally can anyone explain to me what the link is between the chord and the locus? The chord connects P & Q, so how come R is involved and how come we can use it to find the locus of R? I just don't understand what's the logic behind the methods to find loci in general, especially when not using parametric equations.
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    (Original post by karim_elo)
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    (Original post by karim_elo)
    Hi, I'm currently studying FP3 and am having trouble understanding Ex2F, the finding loci subsection, so any help would be much appreciated. In particular, could anyone explain to me how to solve Q1b and 1c:

    'The tangent at P (ap^2, 2ap) and the tangent at Q (aq^2, 2aq) to the parabola y=4ax meet at R.

    a) FInd the coordinates of R -> (apq, ap + aq) using FP1 methods

    The chord PQ passes through the focus (a, 0) of the parabola.
    b) Show that the locus of R is the line x=-a.

    Given instead that the chord PQ has gradient of 2,
    c) Find the locus of R.'

    https://9ccd9b5f157a67423e5533d7dfd0...XFSYzg/CH2.pdf

    In the worked solutions ^ (p36), the major part that's puzzling me is the fact that it jumps from 2a + 2apq = 0 to the locus being x=-a. What's the working out in between those 2 equations, and where has x come from; the x-coordinate of R? But R doesn't lie on PQ so why is it being used? Same thing with part c, how is the locus of R y = a(p + q)?

    Also, in my working out for b) I got the chord PQ to be y(p + q) = 2x-2a by inserting (a, 0) rather than the co-ordinates of P as in the worked solutions. As pq = -1 that means my equation and the one in the solutions are equivalent, but I can't get to x=-a using this either.

    And finally and more in generally can anyone explain to me what the link is between the chord and the locus? The chord connects P & Q, so how come R is involved and how come we can use it to find the locus of R? I just don't understand what's the logic behind the methods to find loci in general, especially when not using parametric equations.

    Where is the question?
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    (Original post by RDKGames)
    Link doesn't work.
    Sorry it was working for me when I posted it but went down...

    http://www.physicsandmathstutor.com/...hapter%202.pdf

    Here it is. Question and worked solution on page 36. Thanks again.
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    (Original post by karim_elo)
    Hi, I'm currently studying FP3 and am having trouble understanding Ex2F, the finding loci subsection, so any help would be much appreciated. In particular, could anyone explain to me how to solve Q1b and 1c:

    'The tangent at P (ap^2, 2ap) and the tangent at Q (aq^2, 2aq) to the parabola y=4ax meet at R.

    a) FInd the coordinates of R -> (apq, ap + aq) using FP1 methods

    The chord PQ passes through the focus (a, 0) of the parabola.
    b) Show that the locus of R is the line x=-a.

    Given instead that the chord PQ has gradient of 2,
    c) Find the locus of R.'

    https://9ccd9b5f157a67423e5533d7dfd0...XFSYzg/CH2.pdf

    In the worked solutions ^ (p36), the major part that's puzzling me is the fact that it jumps from 2a + 2apq = 0 to the locus being x=-a. What's the working out in between those 2 equations, and where has x come from; the x-coordinate of R? But R doesn't lie on PQ so why is it being used? Same thing with part c, how is the locus of R y = a(p + q)?

    Also, in my working out for b) I got the chord PQ to be y(p + q) = 2x-2a by inserting (a, 0) rather than the co-ordinates of P as in the worked solutions. As pq = -1 that means my equation and the one in the solutions are equivalent, but I can't get to x=-a using this either.

    And finally and more in generally can anyone explain to me what the link is between the chord and the locus? The chord connects P & Q, so how come R is involved and how come we can use it to find the locus of R? I just don't understand what's the logic behind the methods to find loci in general, especially when not using parametric equations.
    Coordinates of R(apq, a(p+q))

    In b, pq=-1. Sub that into coordinates of R, R(-a,a(p+q))

    For all a, the X coordinate of R lies on X=-a.

    In c, p+q=1, replace the y coordinate of R with p+q=1, to give R(apq, a).

    You see that for all a, the y coordinate of R lies on y=a.


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