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

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SUMMARY

The discussion focuses on finding the coordinates on the parabola defined by the equation x² = y that are closest to the point P(3,0). The method involves minimizing the distance D using the equation D = (x-3)² + y². Through derivation and solving the resulting equations, the closest point on the curve is determined to be (1,1). The calculations include substituting y into the distance formula, differentiating, and solving a quadratic equation.

PREREQUISITES
  • Understanding of parabolic equations, specifically x² = y.
  • Knowledge of distance minimization techniques in calculus.
  • Familiarity with differentiation and solving quadratic equations.
  • Basic algebra skills for manipulating equations.
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  • Study optimization techniques in calculus, particularly for functions of multiple variables.
  • Learn about the geometric interpretation of derivatives in relation to curves.
  • Explore distance formulas in coordinate geometry.
  • Investigate applications of quadratic equations in real-world scenarios.
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Students and professionals in mathematics, particularly those studying calculus and geometry, as well as anyone interested in optimization problems involving curves and distances.

leprofece
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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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