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January 20, 2017
Posted by **Son** on Wednesday, December 7, 2011 at 11:21pm.

- math -
**Steve**, Thursday, December 8, 2011 at 11:54amusing distance from (2,1/2) to y=x^2,

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

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

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

= x^4 - 4x + 17/4

d = sqrt(x^4 - 4x + 17/4)

d' = 2(x^3 - 1)/sqrt(x^4 - 4x + 17/4)

d' = 0 when x = 1

so, (2,1/2) is closest to (1,1)

______________________

using normal line, we want the line from (2,1/2) normal to the curve. The distance from P to the curve will be minimum along the line normal to the curve.

At (x,y) the slope of x^2 = 2x

so, the normal has slope -1/2x

So, the equation of the normal line from (2,1/2) is

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

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

2x^3 - x = -x + 2

2x^3 = 2

x = 1

So, the normal line from (2,1/2) is the line to (1,1)