Hard algebra problem Given   
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$   
Now, it is necessary to find 
$$\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=?$$  
Is this possible and how? a,b,c are given constants.   
I think, ? is probably a complicated function $F=F(a,b,c)$
 A: The expression 
$$\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}$$ isn't constant on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$ Notice that I get different values on the points $(x,y,z) = (a,0,0)$ and $(x,y,z)=(0,b,0)$. 
A: Without loss of generality we can assume $a\ge b\ge c>0$.
Then
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}\le\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}.$$
Noting that
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}
=\frac1{a^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)$$
$$\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}=\frac1{c^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right),$$
we have
$$\frac1{a^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{c^2},$$
where equality holds, for example, for $(x,y,z)=(a,0,0)$ and $(x,y,z)=(0,0,c)$. (There are other sets of values $(x,y,z)$ for which the equality holds).
More generally (i.e. if we do not assume $a\ge b\ge c>0$),
$$\frac1{\max(|a|,|b|,|c|)^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{\min(|a|,|b|,|c|)^2}.$$
Edit. We can further show that, when $a\ge b\ge c>0$, $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ can take any value $k\in \left[\frac1{a^2}, \frac1{c^2}\right]$. To show this we can consider $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ (where $(x,y,z)=(at,\ 0,\ c(1-t))$) as a continuous function of $t\in [0,1]$ and use the intermediate value theorem.
