# 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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