Finding the distance from ellipsoid to plane I'm having problems with finding the distance from the ellipsoid $x^2+y^2+4z^2=4$ to the plane $x+y+z=6$. The question hinted that I'm supposed to find the distance from a point to the plane and minimize it as the point varies on the ellipsoid. But I'm not sure how to approach that. Would really appreciate it if someone could show how this problem can be solved!

I claim that the point on the ellipsoid with the shortest distance to your plane will be such that the vector normal to the ellipsoid at that point will be parallel to the normal to the plane.
The normal at (x, y, z) has the form (2x, 2y, 8z), and the normal to the plane is (1, 1, 1). Therefore, at the closest (and furthest) point on the ellipsoid, there is some $\alpha$ such that $\alpha(1, 1, 1) = (2x, 2y, 8z),\ x = \alpha/2,\ y=\alpha/2,\ z=\alpha/8$, and 
$$4 = x^2+y^2+4z^2 = \frac{\alpha^2}{4} + \frac{\alpha^2}{4} +\frac{\alpha^2}{16} = \frac{9\alpha^2}{16},$$
and $\alpha = \pm 8/3.$
Edit: So, our candidate points on the ellipsoid are $v_1 = (4/3, 4/3, 1/3)$ and $v_2 = (-4/3, -4/3, -1/3).$ We know that the vectors closest to these on the plane will be of the form $w_1 = v_1 + \beta_1 (1, 1, 1)$ and $w_2 = v_1 + \beta_2 (1, 1, 1)$, because the line of shortest length connecting them will be perpendicular to both the plane and the surface. Given that $w_1$ and $w_2$ will be on the plane, we can see that $\beta_1 = 1$ and $\beta_2 = 3$. Therefore, the distance from these points to the plane will be $$\| w_1 - v_1\| = |\beta_1|\|(1, 1, 1)\| = \sqrt{3}$$
and $$\| w_2 - v_2\| = |\beta_2|\|(1, 1, 1)\| = 3\sqrt{3}$$ so the distance is $\sqrt{3}.$
I realise that this doesn't use the hint, but I feel its more direct and straightforward.
