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According to my calculation,
\text {Semi major axis, a = }\frac {31.5 + 0.5}{2} = 16 \text { AU}
Period, T^2 = a^3 \implies T = 16^{1.5} = 64 \text { years}
\text {Distance between foci and centre, } c = 16-0.5 = 15.5 AU
\therefore \text {Semi minor axis, } b = \sqrt{a^2-c^2} = \sqrt{16^2-15.5^2} \approx 3.9686 \text { (5 s.f.)}
Area of ellipse is therefore = \pi a b = \pi(16)(3.9686) \\ \implies \text {Area} \approx 199.4835, this is for the whole 64 years.
\therefore, \text {For one year, area swept = } \frac {1}{64}(199.4835) = \boxed {3.1169}
However, this answer is incorrect. Can anyone care to explain why this is and what would be the correct approach? Any thoughtful comment is greatly appreciated.