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Reply 1
Bump!
Narik
Can someone please please please please please help me with this question - part b.
Thanks! :biggrin:


Essentially, you've got your equation to the normal now from part a) you're looking for where it cuts the curve y^2 = 4ax again, what can you do to find out where they intersect?
Reply 3
Clarity Incognito
Essentially, you've got your equation to the normal now from part a) you're looking for where it cuts the curve y^2 = 4ax again, what can you do to find out where they intersect?


no idea. can you tell me withut riddles please. :wink: :p:
i've tried substituting it into both the normal equation and the y^2 = 4ax one, but no luck. :mad: :frown:
Narik
no idea. can you tell me withut riddles please. :wink: :p:
i've tried substituting it into both the normal equation and the y^2 = 4ax one, but no luck. :mad: :frown:


You're totally on the right lines, there's two substitutions you can do, one which is much longer i.e. subbing in y^2, because you'd have to expand and simplify, why not try subbing in x = y^2/4a?

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