Qing asked in 科學及數學數學 · 7 years ago

maths m2 mock question

suppose that p(a,2(a^2)-2) is a moving point on the curve C:y=2(x^2)-2

a. find the coordinates of P when |vector OP| is minimum.

b. show OP is the normal to the curve C at P.

c. find the shortest distance from the point Q(5,-1) to the curve C.

不會C PART,求解。

2 Answers

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  • 7 years ago
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    圖片參考:http://imgcld.yimg.com/8/n/HA00273367/o/2013022021...

    (a) OP^2 = a^2 + (2a^2 - 2)^2

    = 4a^4 - 7a^2 + 4

    dOP/da = 16a^3 - 14a = 0 => a = 0 or a = √14/4 or -√14/4

    The coordinates of P are (√14/4,-1/4)

    (b) dy/dx = 4x

    At P, dy/dx = √14 and the slope of OP is -1/√14

    So, the slope of tangent at P * slope of OP is -1. This means that OP is the normal to the curve C at P

    (c) Let |MQ| is the shortest distance from the point Q(5,-1) to the curve C

    Let M(p,q)

    Slope of tangent at M = 4p

    Slope of MQ = (q + 1)/(p - 5)

    As 4p * (q + 1)/(p - 5) = -1

    4p(q + 1) = 5 - p

    4pq = 5 - 5p

    But q = 2p^2 - 2

    Thus 4p(2p^2 - 2) = 5 - 5p

    8p^3 = 5 + 3p

    8p^3 - 3p - 5 = 0

    (p - 1)(8p^2 - 8p + 5) = 0

    p = 1 is the only real root

    So, M(1,0) and MQ = √17

  • 7 years ago

    c) Let D = distance (PQ)

    D = √ { (a-5)^2 + [2(a^2) -2 - (-1)]^2}

    = (4a^4 - 3a^2 -10a +26)^(1/2)

    dD/da = (1/2)* (4a^4 - 3a^2 -10a +26)^(-1/2) * (16a^3 - 6a - 10)

    = (8a^3 - 3a - 5) * (a^4 - 3a^2 -10a +26)^(-1/2)

    when dD/da = 0,

    (8a^3 - 3a - 5) = 0

    (a-1)(8a^2 + 8a +5) = 0

    a = 1 is the only real root

    so the required distance

    = D(1)

    = (4*1^4 - 3*1^2 -10*1 +26)^(1/2)

    = √17 #

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