Which points on the curve $5x^2+4xy+2y^2-6=0$ are closest to the origin. 
Which points on the curve $5x^2+4xy+2y^2-6=0$ are closest to the origin. 

I have solved countless of problems like this but this one is just giving me such a hard time. I'm supposed to solve this with Lagrange's method. So I want to minimize $f(x,y)=x^2+y^2$ due to the constraint $g(x,y)=5x^2+4xy+2y^2-6=0$. 
Ok easy: Find $x,y$ so the following equations are satisfied:
$2x+\lambda(10x+4y)=0$
$2y+\lambda(4y+4x)=0$
$5x^2+4xy+2y^2-6=0$
Right? But however i do, i get very complicated equations with root terms to solve, getting me nowhere. I would love to see how you would solve this. Thanks. 

Let $x^2+y^2=k$.
Thus, we need to find all points $(x,y)$ such that $$5x^2+4xy+2y^2=6$$ for which $k$ gets a minimal value, which says the equation
$$x^2+y^2=k\cdot\frac{5x^2+4xy+2y^2}{6}$$ has solutions or
$$(5k-6)x^2+4kxy+(2k-6)y^2=0,$$
which for $k\neq\frac{6}{5}$ gives
$$4k^2-(5k-6)(2k-6)\geq0$$ or
$$k^2-7k+6\leq0,$$ which gives
$$1\leq k\leq6.$$
Since $1<\frac{6}{5}$, we see that $1$ is a minimal value of $k$ because the equality occurs for
$$-x^2+4xy-4y^2=0,$$ which gives $x=2y$ and we got two following points:
$\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right)$ and $\left(-\frac{2}{\sqrt5},-\frac{1}{\sqrt5}\right)$
