Posted by Son on Wednesday, December 7, 2011 at 11:21pm.
using 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)
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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)