Given that $ \phi = lx+my+nz = 0$ and $ \psi = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1 = 0$,
What are the max and min values of $ \kappa = \frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}$?
I wrote the problem using Lagrange multipliers $\lambda$ and $\mu$:
To optimise $\kappa + \lambda \phi + \mu \psi$.
But the equations are extremely complicated and I got stuck calculating.
Hint.
No need for Lagrange multipliers.
Due to it's homogeneous nature, this problem can be transformed into an equivalent one.
Making $y = \lambda x, z=\mu x$ we have
$$ \min(\max)_{\lambda,\mu} \left(\frac{\frac{1}{a^4}+\frac{\lambda^2}{b^4}+\frac{\mu^2}{c^4}}{\frac{1}{a^2}+\frac{\lambda^2}{b^2}+\frac{\mu^2}{c^2}}\right)\ \ \text{s. t.}\ \ l+m\lambda+n\mu=0 $$
which can be considered as a tangency problem.
NOTE
Taking
$$ \kappa = \left(\frac{\frac{1}{a^4}+\frac{\lambda^2}{b^4}+\frac{\mu^2}{c^4}}{\frac{1}{a^2}+\frac{\lambda^2}{b^2}+\frac{\mu^2}{c^2}}\right) $$
after substituting $\mu = -\frac 1n(l+m\lambda)$ we obtain
$$ \lambda^2+\alpha(\kappa,\theta)\lambda+\beta(\kappa,\theta) = 0 $$
where $\theta = \{a,b,c,l,m,n\}$ Now solving for $\lambda$
$$ \lambda = \frac 12\left(-\alpha(\kappa,\theta)\pm\sqrt{\alpha^2(\kappa,\theta)-4\beta(\kappa,\theta)}\right) $$
but at tangency we need
$$ \alpha^2(\kappa,\theta)-4\beta(\kappa,\theta) = 0 $$
which is a second degree polynomial on $\kappa$. Solving for $\kappa$ we obtain the $\kappa^*$ values at tangency hence
$$ \kappa^*_{min}\le \kappa\le \kappa^*_{max} $$