# 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, 2022

• 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 years
ok 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 years
Well you should be!
• user127778 over 9 years
how... $\sqrt{(8/3)^2 + (8/3)^2 + (2/3)^2} \neq \sqrt3$
• TonyK over 9 years
That's the maximum distance.
• user127778 over 9 years
You're killing me man, what's the minimum distance...?
• TonyK over 9 years
I'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 years
This 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.