Calculus 3: Relative Minimums and Maximums

I'm given this curve 3x² + 3xy + 3y² =1 that's located on the xy plane and I have to find the the point that's closes to the origin. My question is, do I substitute a variable into the distance formula (see example 3: http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx) then find the partial derivatives?

Question

Calculus 3: Relative Minimums and Maximums

I'm given this curve 3x² + 3xy + 3y² =1 that's located on the xy plane and I have to find the the point that's closes to the origin. My question is, do I substitute a variable into the distance formula (see example 3: http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx) then find the partial derivatives?

anonymous · Answer

Here is thhe fixed link to the example: http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

anonymous · Answer

Also do not use Lagrange multiplier

anonymous · Answer

I tried setting it as z = 3x² + 3xy + 3y² -1 and plugging z into D²=x² + y ² + z² which gives me a long complicated equation. Is this the correct way in finding the point closes to the origin?

amoodarya · Answer

you must rotate it 45 degree 
by rotation matrix
then you see an ellipsoid