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# FP1 June 2014 R Question 7 Watch

1. 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'.
2. (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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3. (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 . One when and another when . 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 when you differentiate.

is the part above the x-axis and is the part below.
4. (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 . One when and another when . 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 when you differentiate.

is the part above the x-axis and 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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