Find the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$ I've been asked to find, if it exists, the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$
To start, I've tried to find the points of the ellipsoid where the distance is minimum and maximum using the method of Lagrange multipliers.
First, I considered the sphere centered on  the point $P$ as if it were a level surface and I construct:
$f(x,y,z)=(x-4)^2+(y-2)^2+z^2-a^2$
And I did the same with the ellipsoid, getting the following function:
$g(x,y,z)=x^2+y^2+2z^2-6$
Then, I tried to get the points where $\nabla$$f$ and $\nabla$$g$ are paralell:
$\nabla$$f=\lambda$$\nabla$$g$ $\rightarrow$ $(2x-8,2y-4,2z)=\lambda(2x,2y,4z)$
So, solving the system I got that  $\lambda=1/2$, so in consquence $x=8$ and $y=4$
But when I replaced the values of $x,y$ obtained in $g=0$ to get the coordinate $z$ I get a complex number. I suspect there is something on my reasoning which is incorrect, can someone help me?

Since by C-S $$2x+y\leq\sqrt{(x^2+y^2)(2^2+1^2)}=\sqrt{5(x^2+y^2)},$$
we obtain
$$(x-4)^2+(y-2)^2+z^2=x^2+y^2+z^2-4(2x+y)+20\geq $$
$$\geq x^2+y^2+z^2-4\sqrt{5(x^2+y^2)}+20=6-2z^2+z^2-4\sqrt{5(6-2z^2)}+20=$$
$$=26-z^2-4\sqrt{5(6-2z^2)}\geq26-4\sqrt{30}.$$
It's obvious that the equality occurs, which gives the minimal value:
$$\sqrt{26-4\sqrt{30}}.$$
Since  $-2x-y\leq\sqrt{5(x^2+y^2)}$, by the same way we can get a maximal value: $$\sqrt{26+4\sqrt{30}}.$$
Done!
