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

[leprofece]

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

 

[MarkFL]

Hint: I would try a rotation of axes to eliminate the $xy$ term. :D

 

[leprofece]

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

 

[MarkFL]

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

 

[leprofece]

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?