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    Surely there must be a mistake here in the mark scheme? [Part B]

    I keep getting  y+px=ap^{3}-2ap
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    (Original post by edothero)
    Surely there must be a mistake here in the mark scheme? [Part B]

    I keep getting  y+px=ap^{3}-2ap
    'Fraid not - markscheme is fine.

    Point P has parameter p, and P' has parameter -p.

    So, just replace p with -p throughout the equation for the normal at P, to get the equation for the normal at P'.

    Note: Question says "write down", implying no calculcation is necessary.
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    (Original post by ghostwalker)
    'Fraid not - markscheme is fine.

    Point P has parameter p, and P' has parameter -p.

    So, just replace p with -p throughout the equation for the normal at P, to get the equation for the normal at P'.

    Note: Question says "write down", implying no calculcation is necessary.
    This may sound stupid, but I'm trying to understand fully just in case a similar question pops up in the exam. How did you gather that P' has parameter -p and that P has parameter p? I don't think my brain is working properly today haha
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    (Original post by edothero)
    This may sound stupid, but I'm trying to understand fully just in case a similar question pops up in the exam. How did you gather that P' has parameter -p and that P has parameter p? I don't think my brain is working properly today haha
    By looking at the coordinates for P, P', and knowing the standard parameterisation for the parabola.
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    (Original post by edothero)
    Surely there must be a mistake here in the mark scheme? [Part B]

    I keep getting  y+px=ap^{3}-2ap
    Ah I was stuck on this question before, basically your gradient function should be in terms of y, so the gradient changes as well as the y co-ordinate

    y^2 = 4ax
    d(y^2=4ax)/dx = 2y(dy/dx)=4a so dy/dx = 4a/2y
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    For part b your finding the equation of the normal for p' not just p. Thats why the answer in the mark scheme is correct. I think you may have confused p and p'

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