What points on the the curve $\displaystyle x^2+2xy+2y^2= 100$ are closest to the origin? I think that $\displaystyle F(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^2+2xy+2y^2-100)$

Then $\displaystyle F_{x}=2x+2\lambda x+2\lambda y$

$\displaystyle F_{y}=2y+2\lambda x+4\lambda y$

$\displaystyle F_{z}=0$

$\displaystyle F_{\lambda}=x^2+2xy+2y^2-100$

If I set those equal to zero I get

$\displaystyle 2x=-2\lambda x-2\lambda y$

$\displaystyle 2y=-2\lambda x-4\lambda y$

$\displaystyle x^2+2xy+2y^2=100$

I'm having trouble solving that system. I don't know if I've got the partials wrong, or am just doing the algebra wrong. Any helpful thoughts?

Thanks