Maximum radius of intersection of a plane and an ellipsoid

Question

The question is to use Lagrange multipliers to show that the maximum value of $r = (x^2 + y^2 + x^2)^{1/2}$ subject to the following conditions $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \quad\quad\quad lx + my + nz = 0$$ satisfies $$\frac{l^2a^2}{a^2-r^2} + \frac{m^2b^2}{b^2-r^2} + \frac{n^2c^2}{c^2-r^2} = 0$$ where $a$, $b$, $c$, $l$, $m$, $n$ are arbitrary constants.

I have defined the Lagrangian function $L(x,y,z,\lambda,\mu)$ with two multipliers $\lambda$ and $\mu$, one for each of the constraints. After setting $\nabla L = 0$, I get the following system of equations: $$\frac{x}{r} + \frac{2x\lambda}{a^2} + \mu l = 0$$ $$\frac{y}{r} + \frac{2y\lambda}{b^2} + \mu m = 0$$ $$\frac{z}{r} + \frac{2z\lambda}{c^2} + \mu n = 0$$

All of my attempts to eliminate $\lambda$ and $\mu$ and solve this system of equations have ended in either an algebraic mess or inconsistencies. I have a suspicion that there is an elegant solution there somewhere... can anyone help me find it?

Answer 1

Multiply your first equation with $x$, second with $y$, third with $z$ and add them together. You will get $$(x^2+y^2+z^2)\frac{1}{r}+2\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=r+2\lambda=0$$ Now use $2\lambda=-r$ to plug into your equations, find $x,y,z$ and plug it into the equation of the plane: $$x\left(\frac{1}{r}-\frac{r}{a^2}\right)+\mu l=0$$ $$x=-\frac{\mu l a^2 r}{a^2-r^2}$$ and similar expressions for $y$ and $z$. When you add plug them into the equation of the plane you get $$-\mu r \left(\frac{l^2a^2}{a^2-r^2}+\frac{m^2b^2}{b^2-r^2}+\frac{n^2c^2}{c^2-r^2}\right)=0$$