<s>[INST] Lagrange Multiplier: Distance to the Origin Find the points on the curve $x^2+xy+y^2=2$ that are closest to the origin.
Is there a way to use Lagrange multipliers to answer this question? 
 [/INST] Using Lagrange multipliers, consider the function
$$f=x^2+y^2+\lambda  \left(x^2+x y+y^2-2\right)$$ Compute the partial derivatives and set them equal to $0$ to get the equations
$$\frac{\partial f}{\partial x}=2x+\lambda  (2 x+y)=0$$
$$\frac{\partial f}{\partial y}=2y+\lambda  ( x+2y)=0$$
$$\frac{\partial f}{\partial \lambda }=x^2+x y+y^2-2=0$$ and get, as possible solutions
$$\lambda =-\frac 23 \qquad x=-\sqrt{\frac{2}{3}}\qquad y=-\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$
$$\lambda =-\frac 23 \qquad x=\sqrt{\frac{2}{3}}\qquad y=\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$
$$\lambda =-2 \qquad x=-\sqrt{2} \qquad y=\sqrt{{2}} \implies x^2+y^2= 4$$
$$\lambda =-2 \qquad x=\sqrt{2} \qquad y=-\sqrt{{2}} \implies x^2+y^2= 4$$
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