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Conics, Parametric Equations, and Polar Coordinates

10.6 Polar Equations of Conics and Kepler’s Laws

Focus on eccentricity and conic type, focus and directrix, kepler’s laws.

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Ideas in this lesson

Idea 1 of 3

Eccentricity and conic type

In a standard polar conic, e<1e < 1 gives an ellipse, e=1e = 1 a parabola, and e>1e > 1 a hyperbola.

Worked example 1

Classify r=21+0.5cos⁡θr = \dfrac{2}{1 + 0.5 \cos \theta }.
  1. r=21+0.5cos⁡θr = \dfrac{2}{1 + 0.5 \cos \theta }; e=0.5e = 0.5
  2. e<1  ⟹  e < 1 \implies Ellipse

Answer: Ellipse

Worked example 2

Classify r=31+1cos⁡θr = \dfrac{3}{1 + 1 \cos \theta }.
  1. r=31+1cos⁡θr = \dfrac{3}{1 + 1 \cos \theta }; e=1e = 1
  2. e=1  ⟹  e = 1 \implies Parabola

Answer: Parabola

Idea 2 of 3

Focus and directrix

The numerator in r=ed1±ecos⁡θr = \frac{ed}{1 \pm e \cos \theta } is eded, not dd alone. Recover the directrix distance by dividing by ee.

Worked example 1

For r=21+0.5cos⁡θr = \dfrac{2}{1 + 0.5 \cos \theta }, find the positive distance from the focus to the directrix.
  1. ed=2ed = 2; e=0.5e = 0.5
  2. d=20.5=4d = \dfrac{2}{0.5} = 4

Answer: 44

Worked example 2

For r=31+1cos⁡θr = \dfrac{3}{1 + 1 \cos \theta }, find the positive distance from the focus to the directrix.
  1. ed=3ed = 3; e=1e = 1
  2. d=31=3d = \dfrac{3}{1} = 3

Answer: 33

Idea 3 of 3

Kepler’s Laws

For bodies orbiting the same central mass, T2a3\frac{T^{2}}{a^{3}} is constant, where aa is semimajor axis. Equal areas are swept in equal times.

Worked example 1

Two planets orbit the same star. Planet AA has semimajor axis 11 and period 11. Planet BB has semimajor axis 22. Find its period.
  1. T2a3=1213=1\dfrac{T^{2}}{a^{3}} = \dfrac{1^{2}}{1^{3}} = 1
  2. TB2=23=8T_B^{2} = 2^{3} = 8
  3. TB=8=8≈2.83T_B = \sqrt{8} = \sqrt{8} \approx 2.83

Answer: 8≈2.83\sqrt{8} \approx 2.83

Worked example 2

Two planets orbit the same star. Planet AA has semimajor axis 11 and period 11. Planet BB has semimajor axis 33. Find its period squared.
  1. T2a3=1213=1\dfrac{T^{2}}{a^{3}} = \dfrac{1^{2}}{1^{3}} = 1
  2. TB2=33=27T_B^{2} = 3^{3} = 27
  3. TB2=27T_B^{2} = 27

Answer: 2727

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §10.6,p10.6, p. 734734. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.