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I have recently come across a question and was wondering if anyone could help me simplify it, or at least come up with a better way to solve this. I am currently trying to find the Principal Curvatures of the Elliptic Paraboloid, parametrized as:$$\varphi(u,v) = (au\cos v, bu\sin v, u^2)$$I have found the first and second fundamental forms, with the coefficients giving:

$$E = a^2\cos^2v+b^2\sin^2v+4u^2, \>\>F = u\cos v\sin v(b^2-a^2), \>\> G = u^2(a^2\sin^2v+b^2\cos^2v)$$Also: $$L = \frac{2ab}{(*)}, \>\> M = 0, \>\> N = \frac{2abu^2}{(*)}$$ Where $$(*) = \sqrt{4b^2u^2\cos^2v+4a^2u^2\sin^2v+a^2b^2}$$ Now I am trying to find the principal curvatures by finding the matrices $F_{I}, F_{II}$, where:$$F_{I} = \begin{pmatrix} E & F\\ F & G\end{pmatrix}, \>\>F_{II} = \begin{pmatrix} L & M \\ M & N\end{pmatrix}$$ and solving for: $$\det(F_{II}-\kappa F_{I})$$ Where the two unique solutions for this term would give the principal curvatures. But the thing is the math gets way too complicated and messy. I also tried to find the Principal Curvatures by finding the Gaussian Curvature and Mean Curvature, and using the formula:$$\kappa^2-2H\kappa+K = 0$$ where $K$ is the Gaussian Curvature, $H$ is the Mean Curvature, and finding the roots of the equation. But once again, this is too complicated. Is this the only way for me to find the principal curvatures? Or is there a simpler way to approach it? Thank you for your time.

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    What makes you think this question has a "nice" solution? – 2017-02-20
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    @TedShifrin I am aware that the solution will not be pretty... however following my approach there is no way for me to simplify it enough to solve the equations in terms of $\kappa$. – 2017-02-20
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    Unless I am missing something that allows for further simplification... which may be the case. – 2017-02-20
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    I don't think it helps much — i.e., it's just rephrasing your calculation — but you can try finding the eigenvalues of $F_I^{-1}F_{II}$ using CAS. This is a calculation I've never tried doing in my long differential geometry life. I wouldn't be surprised if it's not explicitly doable. – 2017-02-20
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    @TedShifrin it's worth a shot. Thanks! – 2017-02-20
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    @TedShifrin Do you think I could just find the matrix $F_{I}^{-1}F_{II} = A$, and then use the formula:$$\lambda = \frac{\text{tr}(A)\pm\sqrt{\text{tr}^2(A) - 4\det(A)}}{2}$$ to find the eigenvalues? – 2017-02-20
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    Sure, in theory. That's what I suggested. But I doubt it's going to be anything manageable. Let me know :) – 2017-02-20

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