# Determine points on an ellipsoid that are closest to the origin using Lagrange multipliers

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?

• here $\lambda$ is detecting the six vertices for such an ellipsoid Oct 12 '21 at 15:11
• 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. 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.