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

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To find the nearest point on the curve defined by 5x² - 6xy + 5y² = 4 to the point P = (0,0), participants suggest using derivatives or rotation of axes to simplify the equation. Some recommend employing the quadratic formula to solve for one variable, while others note that the cyclic symmetry of the variables implies extrema occur at y = ±x. There is a discussion about the complexity of solving the equations, with one participant expressing concern over the lengthy calculations involved. Overall, the conversation emphasizes various mathematical strategies to tackle the problem effectively. Understanding these methods is crucial for exam preparation.
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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?
 
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Thread 'Problem with calculating projections of curl using rotation of contour'

Thread 'Problem with calculating projections of curl using rotation of contour'

Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...
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