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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Updated on August 01, 2022

Comments

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