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Finding the gradient - Parametric equations - SOLVED

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    Need some help guys

    Question

    The parametric equations of a curve are x=2t^2 and y=4t. Two points on the curve are P(2p^2,4p) and Q(2q^2,4q)

    Show that the gradient of the chord joining the points P and Q is 2/(p+q)

    So i tried by finding the gradient by using y step over x step and i did not get anything like 2/(p+q)


    This is a 2 mark question, so it must be something simple:/

    BTW this is part B of the question

    part a was .. show that the gradient of the normal to the curve at P is -P (I got that right)..
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    Once you have your expression, factorise (difference of two squares)
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    Your method is correct, what did you do. Post your working so we can find the mistake. Remember that x^2-y^2=(x+y)(x-y)
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    you know the gradient is: [4p-4q]/[2p^2-2q^2] = [4(p-q)]/2(p+q)(p-q)]

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