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    Why is the answer to part b) of question 7 on the June 2014 FP1 R paper (https://7cba9babeb0db0ff9468853e0b2d...20Edexcel.pdf) y-px = -2ap-ap^3 ?

    I can see why it is -2ap as opposed to 2ap for the normal @ P, but I am confused as to why it is now -px not + px. The ap^2 term is the same in both P and P'.
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    (Original post by theDERS)
    Why is the answer to part b) of question 7 on the June 2014 FP1 R paper (https://7cba9babeb0db0ff9468853e0b2d...20Edexcel.pdf) y-px = -2ap-ap^3 ?

    I can see why it is -2ap as opposed to 2ap for the normal @ P, but I am confused as to why it is now -px not + px. The ap^2 term is the same in both P and P'.
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    (Original post by theDERS)
    Why is the answer to part b) of question 7 on the June 2014 FP1 R paper (https://7cba9babeb0db0ff9468853e0b2d...20Edexcel.pdf) y-px = -2ap-ap^3 ?

    I can see why it is -2ap as opposed to 2ap for the normal @ P, but I am confused as to why it is now -px not + px. The ap^2 term is the same in both P and P'.
    From C2 you'll be used to the gradient depending only on the x-coordinate.

    But for this curve you have two different gradients at x = ap^2. One when y = 2ap and another when y = -2ap. Try drawing a sketch and that should hopefully become clear.

    Implicit differentiation is useful here to avoid this confusion. If you haven't learnt that yet (C4) then use y=\pm\sqrt{2ax} when you differentiate.

    y=\sqrt{2ax} is the part above the x-axis and y=-\sqrt{2ax} is the part below.
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    (Original post by notnek)
    From C2 you'll be used to the gradient depending only on the x-coordinate.

    But for this curve you have two different gradients at x = ap^2. One when y = 2ap and another when y = -2ap. Try drawing a sketch and that should hopefully become clear.

    Implicit differentiation is useful here to avoid this confusion. If you haven't learnt that yet (C4) then use y=\pm\sqrt{2ax} when you differentiate.

    y=\sqrt{2ax} is the part above the x-axis and y=-\sqrt{2ax} is the part below.
    Thanks for the response, I have gone over the question with this in mind and understand it now.
 
 
 
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