 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2008/a/2/ii/solution

Expanding the characteristic determinant gives

(EG - F^{2}) κ^{2} - (L G + NE - 2 MF) κ + (L N - M^{2}) = 0.

The product and sum of its two roots therefore give

κ_{1} κ_{2} = \frac{L N - M ^{2}}{EG - F ^{2}} = K, H = \frac{κ _{1} + κ _{2}}{2} = \frac{L G + NE - 2 MF}{2 ( EG - F ^{2} )} .

Reversing the normal changes both principal curvatures and the mean curvature in sign, while leaving their product unchanged.

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