? 發問於 科學及數學數學 · 1 十年 前

Tangent and normal exercise

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  • 1 十年 前
    最佳解答

    part A

    When t=-1

    x= 4/(2+-1)^2 = 4

    y= 12/(2+-1) = 12

    therefore,

    P is (4,12)

    When t=0

    x= 4/(2+0)^2 = 1

    y= 12/(2+0) = 6

    therefore,

    Q is (1,6)

    Equation of PQ is

    (y-6) / (x-1) = (12-6) / (4-1)

    (y-6) / (x-1) = 2

    (y-6) = 2x-2

    y-2x-4=0

    part B

    dy/dx = (dy/dt)/(dx/dt)

    dy/dt= -12/(2+t)^2

    dx/dt= -8/(2+t)^3

    dy/dx = 3(2+t) / 2

    part C

    since the normal is perpendicular to PQ,

    slope of tangent = slope of PQ

    3(2+t) / 2 = (12-6) / (4-1)

    3(2+t) / 2 = 2

    3(2+t) = 4

    6+3t = 4

    t=-2/3

    coordinate of the point is

    (9/4 , 9)

    slope of normal is -1/2

    equation of normal is

    -1/2 = (y-9)/(x-9/4)

    -x+9/4=2y-18

    -4x+9=8y-72

    0 = 8y+4x-81

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