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The standard form of a parabola in cartesian coordinates is,
c(t) = t s1 + at2 s2
where a > 0 and t R. The standard form of the same parabola in polar coordinates centered at the focus p is,

x(q) = p + L

1+cos(q)
seiq
where s is a unit vector in the i-plane of the parabola. i.e. s2=1 and si = 0. Find i,L,s,p,q in terms of a,s1,s2,t. Hint: Find the location of the vertex x(q0) where the parabola has maximum curvature and equate [^v](q0) = s2.

  • L = 1

    2a
    s = -s2, p = 1

    4a
    s2, cot(q) = (4at)-1 - at

  • L = 1

    2a
    s = -s2, p = 1

    4a
    s2, cot(q) = (4at)-1 - at

  • L = 1

    2a
    s = -s2, p = 1

    4a
    s2, cot(q) = (4at)-1 - at

  • L = 1

    2a
    s = -s2, p = 1

    4a
    s2, cot(q) = (4at)-1 - at

  • L = 1

    2a
    s = -s2, p = 1

    4a
    s2, cot(q) = (4at)-1 - at
  • *****
    calculator


    Carlos Rodriguez <carlos@math.albany.edu>
    This problem was contributed by a student. It is offered as it is with no warranty of any kind
    Last modified:Mon Mar 07 2011 18:08:31 GMT-0500 (Eastern Standard Time)