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Parabola Help

  1. Sep 8, 2007 #1
    [SOLVED] Parabola Help

    1. The problem statement, all variables and given/known data

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    2. Relevant equations


    3. The attempt at a solution

    I can complete a) fine, by differentiating and sticking in y value to get m for tangent then -1/m for normal (-p). Then i used (y - y1) = m (x - x1) to provide normal equation.

    I have tried a few ways for b) but i always end up the normal equation - no good. The only method which gives me q in terms of p is if i make the normal equation equal the parabola equation then solve - the answer i get doesn’t look right at all, and it doesn’t leave me in a good position for c)

    is there something i should know about parabola’s to be able to solve b) or is it a route i haven’t seen?

    Any help would be much appreciated :approve:
  2. jcsd
  3. Sep 8, 2007 #2


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    Well the normal passes through Q, so you could sub the points of Q into the normal and q in terms of p
  4. Sep 8, 2007 #3
    How would you arrange 2q + pq^2 = 2p + p^3 ?
  5. Sep 8, 2007 #4


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    Well normally I'd leave it like that...but I am not sure really how q in terms of p is meant as..
  6. Sep 9, 2007 #5
    it means q = 3p or something like that, and you need it looking like that to be able to do (c), im really think im missing something :confused:
  7. Sep 9, 2007 #6


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    [itex]2q + pq^2 = 2p + p^3 [/itex] is a quadratic in q. Rewrite it in the form [itex]pq^2+ 2q- (2p+p^3)= 0[/itex] and solve that equation for q, using the quadratic formula, leaving p in the formula. That's what "in terms of p" means
  8. Sep 9, 2007 #7
    Great, could you confirm q = -2/p - p ?
  9. Sep 9, 2007 #8


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    There are, of course, two solutions. q= -2/p- p is one. The other is obvious and easy to check in the original equation.
  10. Sep 9, 2007 #9


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    Dearly Missed

    Well, simplify:
  11. Sep 12, 2007 #10
    thanks guys
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