calculus

find the minimum distance from the point (4,1) to the parabola y^2=4x.

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  1. The distance z from (4,1) to (x,y) is

    z = √((x-4)^2 + (y-1)^2)
    = √((y^2/4 - 4)^2 + (y-1)^2)
    = (1/4)√(y^4-16y^2-32y+272)

    dz/dy = (1/2)(y^3-8y-8)/√(y^4-16y^2-32y+272)

    dz/dy=0 when y^3-8y-8=0
    y = -2 or y = 1±√5
    The minimum z is at y=1+√5
    x = (3+√5)/2
    z = (25-5√5)/2

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  2. or:
    let the point of minimum distance from A(4,1) be P(a,b)
    slope AP = (b-1)/(a-4)
    2y dy/dx = 4
    dy/dx = 2/y
    so at A the tangent has a slope of 2/b

    but AP is a normal to the tangent at A , so the slopes of the tangent and the slope of AP must be negative reciprocals of each other, i.e.
    (a-4)/(1-b) = 2/b
    a-4 = 2(1-b)/b , but b^2 = 4a ---> a = b^2/4
    b^2/4 - 4 = (2 - 2b)/b
    b^3 - 16b = 8 - 8b
    b^3 - 8b - 8 = 0

    b = -2, or b = 1 ± √5
    reaching the same stage as Steve's method

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