 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/10b/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

r = \frac{r _{0}}{1 + e c o s θ}, r_{0} = \frac{l ^{2}}{k} .

Consequently

l = k r_{0}, E = \frac{mk ( e ^{2} - 1 )}{2 r _{0}} .

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