Find the points that are closest and farthest from $(0,0)$ on the curve $3x^2-2xy+2y^2=5$ Find the points that are closest and farthest from $(0,0)$ on the curve $3x^2-2xy+2y^2=5$
My attempt:
So, I'm looking to find global extrema of the function $f=x^2+y^2$ (since square root is a motonous function it has extrema at same points as this function).
The set $S = \{ (x,y)\in \mathbb{R} : 3x^2 -2xy + 2y^2 -5 = 0\}$ is not compact, therefore I cannot guarantee that $f$ will have a minimum or a maximum on $S$.
Anyways by applying the usual theorem, gradient of $f$ is colinear with the gradient of $g$. (gradient of g is $0$ only if $x=y=0$ and that point is not in $S$ therefore not relevant so the set containing only gradient of $g$ is linearly independent).Let the coefficient be noted by $\lambda$.
I'm not sure how to go about this from now though. I have 3 equations that each contain $x,y,xy,\lambda x,\lambda y$:
(if my calculations were right)
$$\lambda (-3x+y) + x=0$$
$$\lambda (x-2y)+y=0$$
$$3x^2 -2xy + 2y^2 - 5 = 0$$ 
I'm stuck here and I'm not sure what to do.Hints would help! Thanks in advance!
 A: Note that the set $S$ is compact. It's an ellipsis.
What you've done is fine. Consider the first two equations:$$\left\{\begin{array}{l}(-3\lambda+1)x+\lambda y=0\\\lambda x+(-2\lambda+1)y=0.\end{array}\right.$$If the determinant of the matrix of this system is not $0$, then the system has one and only one solution, which is $(0,0)$. But $g(0,0)\neq0$.
So, what matters here is the case in which $\lambda$ is such that$$\begin{vmatrix}-3\lambda&\lambda\\\lambda&-2\lambda+1\end{vmatrix}\neq0.$$It's easy to see that this takes place if and only if$$\lambda=\frac{5\pm\sqrt5}{10}.$$So, see what happens for these values of $\lambda$? You will get the set of points$$\left\{\left(-1,-\frac{1+\sqrt5}2\right),\left(-1,-\frac{1-\sqrt5}2\right),\left(1,\frac{1-\sqrt5}2\right),\left(1,\frac{1+\sqrt5}2\right)\right\}.$$Now, compute $f$ at each of them.
A: Following your method,note that from here
$$y=\lambda (2y-x) \quad x=\lambda (3x-y) \implies \frac{y}{x}=\frac{2y-x}{3x-y}\implies \frac{y}{x}=\frac{2\frac{y}{x}-1}{3-\frac{y}{x}}$$
Let $$t=\frac{y}{x} \implies t=\frac{2t-1}{3-t}\implies 3t-t^2=2t-1\implies t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2}$$
Then
$$\frac{y}{x}=\frac{1\pm\sqrt{5}}{2}$$
from the given equation you can find the points
$$\left(-1,-\frac{1+\sqrt5}2\right),\left(-1,-\frac{1-\sqrt5}2\right),\left(1,\frac{1-\sqrt5}2\right),\left(1,\frac{1+\sqrt5}2\right)$$
A: Let $x^2+y^2=k$.
Thus, we need to find a minimal and a maximal value of $k$ for which the equation 
$$k(3x^2-2xy+2y^2)=5(x^2+y^2)$$ has solutions $(x,y)$ and to find these solutions for extreme values of $k$.
Hence, $$(3k-5)x^2-2kxy+(2k-5)y^2=0,$$ which gives
$$k^2-(3k-5)(2k-5)\geq0$$ or
$$\frac{5-2\sqrt5}{2}\leq k\leq\frac{5+2\sqrt5}{2},$$
which gives extreme values of $k$ and since for these values $$\frac{x}{y}=\frac{k}{3k-5},$$ we obtain possibility to find needed points.
