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?

hint
The ellipse can be parametrized as
$$x_e =\sqrt {6}\sin (\phi)\cos (\theta) $$
$$y_e=\sqrt{6}\sin (\phi)\sin (\theta ) $$
$$z_e=\sqrt {3}\cos (\phi) $$
the square of the distance from a point of the ellipse to the point $(4,2,0) $ is
$$D^2=(x_e-4)^2+(y_e-2)^2+z_e^2$$
$$=3\sin^2 (\phi)+23-8x_e-4y_e $$
You can find min and max D, 
by solving the system, $$\frac {\partial D^2}{\partial \phi}=\frac {\partial D^2}{\partial \theta}=0$$
