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if the line y=2x+b and the parabola y^2=8(x+2) meet exactly at one point,then what is the value of b

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    Non-Calculus way:

    Let's find their intersection, ...
    (2x+b)^2 = 8(x+2)
    4x^2 + 4bx + b^2 - 8x - 16 = 0
    4x^2 + (4b-8)x + b^2 - 16 = 0

    this is a quadratic with
    A = 4
    B = 4b -8
    C = b^2 - 16

    IF this is to have only one root, then the B^2 - 4AC, or the discriminant, must be zero.
    so
    (4b-8)^2 - 4(4)(b^2 - 16) = 0
    16b^2 - 64b + 64 - 16b^2 + 256 = 0
    64b = 320
    b =5

    Using Calculus:
    for y^2= 8x + 16
    2y(dy/dx) = 8
    dy/dx = 8/2y ----> slope of the tangent

    but we know the slope of the tangent is 2 from y - 2x+b

    then 8/2y = 2
    4y = 8, then y = 2
    sub that into parabola
    4 = 8x + 16
    x = -12/8 =- 3/2

    so the point of contact is (- 3/2,2) which must satisfy
    y = 2x + b
    2 = 2(-3/2) + b
    2 = -3 + b
    b = 5 , just like before.

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