Distance from Ellipsoid to Plane  Lagrange Multiplier
2,895
This is easy if you don't insist on using Lagrange multipliers.
The normal to the ellipse at the point $(x,y,z)$ is $\nabla(x^2+y^2+4z^2) = (2x, 2y, 8z)$. At minimum or maximum distance to the plane, this must be parallel to the normal to the plane, which is $(1,1,1).$ So $x = y = 4z$.
Plug this back into the equation for the ellipse to get $36z^2 = 4$, or $z = \pm \frac13$. So the nearest and farthest points are $\pm(\frac43,\frac43,\frac13)$.
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user127778
Updated on August 01, 2022Comments

user127778 over 1 year
Find the distance from the ellipsoid $x^2 + y^2 + 4z^2 = 4$ to the plane $x + y + z = 6$.
I'm trying to do it using Lagrange multipliers over the distance equation, but then it just gets overwhelming and I have no idea how to go on? Can someone walk me through the computation?

user127778 over 9 yearsok that makes sense, I had gotten the same conclusion, but when I plug those points into the distance equation, I'm not getting $\sqrt3$

TonyK over 9 yearsWell you should be!

user127778 over 9 yearshow... $\sqrt{(8/3)^2 + (8/3)^2 + (2/3)^2} \neq \sqrt3$

TonyK over 9 yearsThat's the maximum distance.

user127778 over 9 yearsYou're killing me man, what's the minimum distance...?

TonyK over 9 yearsI've given you the two points. You have the equation of the plane. So you get two distances, right? One for each point. The minimum distance is (duh) the smaller of these...

Cheerful Parsnip over 9 years

Ted Shifrin almost 3 yearsThis is what Lagrange multipliers gives you immediately, by the way! You just have to understand what's going on and not get buried in letters.