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!
 A: 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.
A: Per the hint, this can be set up as a Lagrange multiplier problem. As usual, we can take the square of the distance as the objective function and the equation of the ellipsoid as the constraint. The distance from a point $p=(x,y,z)$ to the plane $x+y+z=6$ is, per the well-known formula, $${\left|x+y+z-6\right| \over \sqrt3}.$$ So, you would find the gradient of $\frac13(x+y+z-6)^2-\lambda(x^2+y^2+4z^2-4)$, set it to zero and solve the resulting system of equations.  
However, there’s a fairly straightforward way to solve this without any (direct) use of calculus. The distance to a plane is measured along a perpendicular to the plane. If you imagine sliding the reference plane $x+y+z=6$ along its normal toward the ellipsoid, it should become clear that the minimum distance to the ellipsoid is equal to the distance between the reference plane and a plane parallel to it that’s tangent to the ellipsoid. A plane $Px+Qy+Rz+S=0$ is tangent to the ellipsoid iff it satisfies the dual conic equation $$P^2+Q^2+\frac14R^2-\frac14S^2=0.\tag{*}$$ All planes parallel to the given plane have equations of the form $x+y+z=d$. Substituting this into (*) and solving for $d$ yields $\pm3$. The constant term in this form of equation of a plane is the signed distance of the plane from the origin, scaled by the length of the normal. The given ellipsoid is centered on the origin, so the nearer tangent plane is the one with a constant term that has the same sign as the constant term of the reference plane’s equation. The distance between these two planes, and hence also the minimum distance between the ellipsoid and the reference plane, then, is just the difference of the constant terms divided by the length of the common normal, i.e., $${\left|6-3\right| \over \sqrt3} = \sqrt3.$$
A: Let $(r, s, t)$ be a point on the ellipsoid $x^2+y^2+4z^2=4$. Then, the signed distance from the point to the plane is
$$d=\dfrac{r+s+t-6}{\sqrt{1^2 + 1^2 + 1^2}}.$$
Now we just need to minimize $d$ subject to $r^2+s^2+4t^2 - 4 =0$. Lagrange Multipliers should work here.
A: I found similar question in my textbook .hope you can follow on same lines

