They arise from partial derivatives of the Lagrange multiplier function. Here below is the original problem:
Goal function: $$f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} $$ with two constraints: $$ \begin{cases} x^2+y^2+z^2-1=0 \\ Ax+By+Cz=0 \end{cases} $$ provided that $a>b>c>0$ and $A^2+B^2+C^2=1$, determine all the possible extremums.
My Lagrange function is
$$L(x,y,z,\lambda,\mu):=f(x,y,z)+\lambda (x^2+y^2+z^2-1)+\mu (Ax+By+Cz)$$
From which naturally arises
$$
\begin{array}{c}
\partial L/\partial x =&2x/a^2+2\lambda x+A\mu=0 \\
\partial L/\partial y =&2y/b^2+2\lambda y+B\mu=0 \\
\partial L/\partial z =&2z/c^2+2\lambda z+C\mu=0 \\
\partial L/\partial \lambda =&x^2+y^2+z^2-1=0 \\
\partial L/\partial \mu =&Ax+By+Cz=0
\end{array}
$$
(Here I want to tag them $(1)$ to $(5)$ from top to bottom, but don't know how to write the Jax code. Is it possible to insert two or more tags into one array made up of several equations and can you help me with this?)
I made some failed attempts to solve them. Every time I failed because I simply didn't know how to take off the coefficients leading $x,y,z$ terms. They might be zero, which really messed things up! For example, in $(1)$ I got $2x(\frac{1}{a^2}+\lambda)+A\mu=0$ but I don't know whether $(\frac{1}{a^2}+\lambda)$ is zero or not. And for $(1)$ to $(3)$, well, I got three uncertain cases. To make things worse, I also could not determine whether $A,B,C$ or $\mu$ is zero or not.... Too many possibilities! which really freaked me out..
So, except the case-by-case discussion about the $A,B,C$ and $\lambda,\mu$, is there a possible way (or trick) to find all the possible solutions for $(x,y,z)$ without much pain? I really need some enlightening. Any useful hint or help that sheds a light on this problem will be appreciated, thanks in advance!
I don't think there is a very simple method to get the answer. But you can simplify the cases:
case 1 : $C=0\cap A*B \neq 0 \implies y=-\dfrac{Ax}{B} $ , the rest should be easy.
case2: $C\neq 0 \implies z= -\dfrac{Ax+By}{C} \implies (1+\dfrac{A^2}{C^2})x^2+(1+\dfrac{B^2}{C^2})y^2+2\dfrac{AB}{C^2}xy=1 \\f=\dfrac{1}{c^2}-(\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}), p^2=\dfrac{(ac)^2}{a^2-c^2},q^2=\dfrac{(bc)^2}{b^2-c^2},q>p,x^2\le 1, y^2 \le 1, x^2+y^2 \le 1$
discuss $AB$ cases (main 3 cases), then it should be easy to find extremums.
case3: $C=B=0 \implies A=1,x=0, f=\dfrac{z^2}{c^2}+\dfrac{y^2}{b^2}\cap y^2+z^2=1$, it should be easy.
$C=A=0$ is same.