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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Finding the gradient - Parametric equations - SOLVED

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)..
(edited 11 years ago)
Reply 1
Avatar for scherzi
scherzi
4
Once you have your expression, factorise (difference of two squares)
Reply 2
Avatar for james22
james22
16
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)
Reply 3
Avatar for Hasufel
Hasufel
16
you know the gradient is: [4p-4q]/[2p^2-2q^2] = [4(p-q)]/2(p+q)(p-q)]

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