Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 4 / ii / Solution

With the ellipse center as origin and its major axis horizontal, the auxiliary-circle construction gives particle coordinates $X=a\cos E$, $Y=a\sqrt{1-e^2}\sin E$, while the stellar focus is at $(ae,0)$. ThereforeSquaring and adding gives $r^2=a^2(1-e\cos E{)}^2$. The Kepler orbit radius and the anomaly conversion are consequentlyEquivalently,The sine and cosine formulas, or a quadrant-aware angle function, fix the correct branch when using the half-angle relation.