MHB Finding the point on a parabola closest to a given point

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The discussion focuses on finding the coordinates on the parabola defined by x² = y that are closest to the point P(3,0). The approach involves setting up the distance formula and deriving it to find critical points. The calculations lead to the equation 4y² - 13y + 9 = 0, which is solved to find y = 1 and subsequently x = 1. The closest point on the parabola to P is determined to be (1,1). The solution process demonstrates the application of calculus in optimizing distances to curves.
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Find the curve coordinates nearest to P
a)x2 = y P(3,0)

ok to the boox exercise model I must do so

(x-3)2+ (y)2 = D

then x2-6x +9+y2 = D

introducing y into x
y-6sqrt(y) +9 + (y)2 = D

derivating
1-3/(sqrt(y) +0+2y

mcm = sqrt(y) -3 +2y = 0
solving
sqrt(y) = -2y+3
Squaring
y = 4y2-12y +9
4y2-13y +9 = 0
x = 1
y = 1
 
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leprofece said:
Find the curve coordinates nearest to P
a)x2 = y P(3,0)

ok to the boox exercise model I must do so

(x-3)2+ (y)2 = D

then x2-6x +9+y2 = D

introducing y into x
y-6sqrt(y) +9 + (y)2 = D

derivating
1-3/(sqrt(y) +0+2y

mcm = sqrt(y) -3 +2y = 0
solving
sqrt(y) = -2y+3
Squaring
y = 4y2-12y +9
4y2-13y +9 = 0
x = 1
y = 1

Looks like you already figured this one out. :)
 
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