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  • 2 years ago
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    Function of the cubic curve:

    y = -x^3 + bx^2 + cx + d

    (-2k, 0) is a point:

    0 = -(-2k)^3 + b(-2k)^2 + c(-2k) + d

    0 = 8k^3 + 4bk^2 - 2ck + d

    (k, 0) is also a point:

    0 = -(k)^3 + b(k)^2 + c(k) + d

    0 = -k^3 + bk^2 + ck + d

    Assuming P is the local maximum, take the derivative:

    y' = -3x^2 + 2bx + c

    0 = -3(3k)^2 + 2b(3k) + c

    0 = -27k^2 + 6bk + c

    Three equations, three unknowns (b, c, and d). If we take the first equation and subtract the second:

    0 = 9k^3 + 3bk^2 - 3ck

    0 = 3k^2 + bk - c

    Adding this to the third equation:

    0 = -24k^2 + 7bk

    24k = 7b

    b = 24k/7

    Solving for c:

    0 = 3k^2 + (24k/7)k - c

    0 = 3k^2 + 24/7 k^2 - c

    c = 45/7 k^2

    Solving for d:

    0 = -k^3 + (24/7 k)k^2 + (45/7 k^2)k + d

    0 = -k^3 + 24/7 k^3 + 45/7 k^3 + d

    d = -62/7 k^3

    The function is:

    y = -x^3 + 24/7 k x^2 + 45/7 k^2 x - 62/7 k^3

    If Q is (kn, 0):

    0 = -(kn)^3 + 24/7 k (kn)^2 + 45/7 k^2 (kn) - 62/7 k^3

    0 = 7k^3 n^3 - 24k^3 n^2 - 45k^3 n + 62k^3

    0 = 7 n^3 - 24 n^2 - 45 n + 62

    n = -2, 1, or 31/7

    Since n > 3, n must be 31/7, so Q is at (31/7 k, 0).

    The y-coordinate of P is:

    y = -(3k)^3 + 24/7 k (3k)^2 + 45/7 k^2 (3k) - 62/7 k^3

    y = -27k^3 + 216/7 k^3 + 135/7 k^3 - 62/7 k^3

    y = 262/7 k^3

    So the area of OPQ is:

    A = 1/2 b h

    A = 1/2 (31/7 k) (262/7 k^3)

    A = 4061/49 k^4

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