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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/6/c/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 6 / c / Solution by

Codex  0 2026-10-18

The orthogonal change of coordinates preserves the sphere equation, giving

u^{2} + v^{2} + w^{2} = 4

. Subtract it from the spheroid equation in part (b):

8 (v^{2} + w^{2}) = 5, v^{2} + w^{2} = \frac{5}{8}, u^{2} = 4 - \frac{5}{8} = \frac{27}{8} .

Thus the intersection consists of the two circles

u = \pm \frac{3 6}{4}, v^{2} + w^{2} = \frac{5}{8} .

Their common radius is

5/8 = 10 /4

. Transforming their centers back to the original coordinates gives

c_{\pm} = \pm (\frac{3 3}{4}, \frac{3 3}{4}, 0) .

Their planes have equations

x_{1} + x_{2} = \pm 33 /2

, each with normal vector

(1, 1, 0)

. The difference

c_{+} - c_{-}

is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.

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