Lagrange Multiplier: Distance to the Origin

Question

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?

Answer 1

Hint: the distance to the origo is given by $$d=\sqrt{x^2+y^2}=\sqrt{2-xy}$$

Answer 2

Without using calculus,

$$8=(2x+y)^2+3y^2$$

WLOG $\sqrt3y=2\sqrt2\sin t, 2x+y=2\sqrt2\cos t\iff x=?$

So, we need to minimize $$\left(\dfrac{2\sqrt2\sin t}{\sqrt3}\right)^2+\left(\dfrac{2\sqrt2\cos t-\dfrac{2\sqrt2\sin t}{\sqrt3}}{2}\right)^2$$

$$=\dfrac{8\sin^2t}3+\dfrac{8(\sqrt3\cos t-\sin t)^2}{12}$$

$$=\dfrac{8\sin^2t+3\cos^2t+\sin^2t-2\sqrt3\sin t\cos t}3$$

$$=\dfrac{9(1-\cos2t)+3(1+\cos2t)-2\sqrt3\sin2t}6$$

$$=\dfrac{12-2\sqrt3(\sqrt3\cos2t+\sin2t)}6$$

using $\cos2t=1-2\sin^2t=2\cos^2t-1$

and use $$-\sqrt{a^2+b^2}\le a\cos2t+b\sin2t\le\sqrt{a^2+b^2}$$

Answer 3

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 $$\lambda =-\frac 23 \qquad x=-\sqrt{\frac{2}{3}}\qquad y=-\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$ $$\lambda =-\frac 23 \qquad x=\sqrt{\frac{2}{3}}\qquad y=\sqrt{\frac{2}{3}}\implies x^2+y^2=\frac 43$$ $$\lambda =-2 \qquad x=-\sqrt{2} \qquad y=\sqrt{{2}} \implies x^2+y^2= 4$$ $$\lambda =-2 \qquad x=\sqrt{2} \qquad y=-\sqrt{{2}} \implies x^2+y^2= 4$$