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# Finding the gradient - Parametric equations - SOLVED

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1. 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)..
2. Once you have your expression, factorise (difference of two squares)
3. 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)
4. you know the gradient is: [4p-4q]/[2p^2-2q^2] = [4(p-q)]/2(p+q)(p-q)]

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Updated: June 3, 2012
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