Finding the curve coordinates of the point nearest to P in the curve

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Discussion Overview

The discussion revolves around finding the coordinates of the point on the curve defined by the equation 5x² - 6xy + 5y² = 4 that is nearest to the point P = (0,0). Participants explore various methods for solving this problem, including derivatives and algebraic manipulation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests using derivatives to solve for x or y in the context of minimizing the distance to point P.
  • Another participant proposes a rotation of axes to eliminate the xy term in the equation, although they acknowledge that this method may not be familiar to all students.
  • A different approach is mentioned, involving the quadratic formula to solve for one variable, or leveraging the cyclic symmetry of the variables to find extrema at y = ±x.
  • One participant expresses uncertainty about the lengthy algebraic manipulation they attempted, questioning if it is the correct approach.

Areas of Agreement / Disagreement

Participants present multiple competing views and methods for solving the problem, with no consensus on a single approach or solution. The discussion remains unresolved regarding the best method to find the nearest point on the curve.

Contextual Notes

Some participants express concern about the applicability of certain methods, such as rotation of axes, given that students may not have studied it. There is also uncertainty regarding the complexity of the algebraic manipulations proposed.

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Find the curve coordinates of the point nearest to P in the curve
5x2 -6xy +5y2 = 4
P = (0,0)
oK x2 + y2 =D2

But how can i solve for x or y ?
Maybe by expliciting derivative
 
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Hint: I would try a rotation of axes to eliminate the $xy$ term. :D
 
MarkFL said:
Hint: I would try a rotation of axes to eliminate the $xy$ term. :D
Ouh (Shake) I have a lot of time without studying that
another method?'
because it appears in an exam and students haven't studied rotation
 
Well, you could try the quadratic formula to solve the constraint for one of the variables, or for much faster results you could use the fact that the two variables are cyclically symmetric and so the extrema will occur for $$y=\pm x$$. :D
 
MarkFL said:
Well, you could try the quadratic formula to solve the constraint for one of the variables, or for much faster results you could use the fact that the two variables are cyclically symmetric and so the extrema will occur for $$y=\pm x$$. :D

Well I think is 5x2 -6xy +5y2
= 4
Solving x2 = 4-5y2+6xy
introducing in 1
4-5y2+6y(4-5y2+6xy)1/2

but it is very long maybe it isnot so

is that the way?
 
Last edited:

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