 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/10c/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

r_{1}

r_{2}

, the given Kepler orbit satisfies

r_{1} = \frac{h ^{2}}{k ( 1 + e )}, r_{2} = \frac{h ^{2}}{k ( 1 - e )} .

Hence

e = \frac{r _{2} - r _{1}}{r _{1} + r _{2}}, h^{2} = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}} .

Therefore

h = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}}, C = 2 .

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