Among the points of the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$, determine the ones closest to the origin in $\mathbb{R}^3$.

I'm trying to solve this example using the Lagrange multipliers method, minimizing the squared distance function $f(x,y,z) = x^2+y^2+z^2$ restricted to $g(x,y,z) = \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}-1=0$. Making $\nabla f = \lambda \nabla g$ we get the following system:

$$x = \dfrac{\lambda}{a^2}x$$ $$y = \dfrac{\lambda}{b^2}y$$ $$z = \dfrac{\lambda}{c^2}z$$

My difficulty is in solving this system, I've already tried to substitute in the restriction $g$, but I don't get an expression that allows finding $\lambda$. Are there any tricks that can be done here?

  • $\begingroup$ here $\lambda$ is detecting the six vertices for such an ellipsoid $\endgroup$
    – janmarqz
    Oct 12 '21 at 15:11
  • 1
    $\begingroup$ You will have to test points on the boundary $x = 0, y = 0, z = c$ etc. Also, the question does not state which of $a, b, c$ is smallest. $\endgroup$
    – Math Lover
    Oct 12 '21 at 15:12

You have several solutions. In the first equation, you can have $\lambda=a^2$. If $a\ne b\ne c$ then you have $y=z=0$. From the equation of the ellipse you have $x=\pm a$. So the solution is $(\pm a,0,0)$. Similarly you get the other solutions $(0,\pm b, 0)$ and $(0,0,\pm c)$. The corresponding distances are $a$, $b$, and $c$. One is a minimum, one is a maximum, one is just a critical point.


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