# Calculus

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Determine the point on the curve y^2=4x that is closest to the point (1,0)

a) (0,0)
b) (1,2)
c) (0.5,sqrt of 2)

• Calculus -

the distance d is found by

d^2 = (1-x)^2 + y^2
= 1-2x+x^2+4x
= (1+x)^2

d = 1+x

this is obviously a minimum when x=0, so (a)

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