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Maths and statistics discussion, revision, exam and homework help.

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Guys can anyone help with Jan 06 q.5(ii)? I got dy/dx as 1/t from the previous q, but when I do y-2p = 1/t (x-p^2) and tidy it up, I don't get what they are looking for...Any ideas?

Cheers
1/t = 1/p
(Original post by Next Level)
http://blogs.thegrangeschool.net/mat...oklet-2012.pdf

Guys can anyone help with Jan 06 q.5(ii)? I got dy/dx as 1/t from the previous q, but when I do y-2p = 1/t (x-p^2) and tidy it up, I don't get what they are looking for...Any ideas?

Cheers
Did you change that t in the gradient to a p when you put it in the eq of a line?
(Original post by BabyMaths)
1/t = 1/p
How does that work?
(Original post by TheJ0ker)
Did you change that t in the gradient to a p when you put it in the eq of a line?
No I don't get how that works? Can you explain please?
(Original post by Next Level)
How does that work?
x=t^2

y=2t

The point is (p^2,2p) so clearly t=p.
(Original post by BabyMaths)
x=t^2

y=2t

The point is (p^2,2p) so clearly t=p.
Ohh gotcha, cheers!

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Last updated: June 20, 2012
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