Polar Form of Conics — Mathematics Knowledge System

Overview

Every conic section (other than a circle centered at the origin) can be written in polar coordinates with one focus at the pole, using a single formula involving the eccentricity e and the semi-latus rectum. This polar form unifies the ellipse, parabola, and hyperbola into one equation, distinguished only by the value of e, and is especially natural for orbital mechanics, since Kepler's laws describe planetary orbits as conics with the Sun at one focus — exactly the pole of the polar coordinate system.

Intuition

The polar form comes directly from the focus-directrix definition of a conic: the distance from any point on the curve to the focus equals e times its distance to the directrix. Placing the focus at the pole makes the distance to a point simply r, and the distance to a (vertical) directrix a distance d away works out to d − rcosθ in polar terms; substituting into |PF| = e·d(P, directrix) and solving for r produces the polar formula directly, with no need for the messier Cartesian derivation involving square roots.

Formal Definition

Definition

With the focus at the pole and directrix a distance d from it, the polar equation of a conic with eccentricity e is:

r = \frac{ed}{1 + e\cos\theta}

Polar conic, directrix perpendicular to polar axis (to the right)

r = \frac{ed}{1 - e\cos\theta}

Directrix to the left

r = \frac{ed}{1 \pm e\sin\theta}

Directrix perpendicular to the vertical axis

0 \le e < 1: \text{ellipse}, \quad e = 1: \text{parabola}, \quad e > 1: \text{hyperbola}

Classification by eccentricity

Worked Examples

Match to r = ed/(1+ecosθ): e=0.5, and ed=2 so d=4. Since 0<e<1, this is an ellipse.
$e = 0.5 \ (\text{ellipse})$
Evaluate r at θ=0°.
$r(0°) = \frac{2}{1+0.5(1)} = \frac{2}{1.5} = \frac{4}{3}$
Evaluate r at θ=90° and θ=180°.
$r(90°) = \frac{2}{1+0.5(0)} = 2, \qquad r(180°) = \frac{2}{1+0.5(-1)} = \frac{2}{0.5} = 4$

Answer: e = 0.5 (ellipse); r(0°) = 4/3, r(90°) = 2, r(180°) = 4

Practice Problems

Difficulty 5/10

Classify the conic r = 4/(1 − 2cosθ) by its eccentricity.

Difficulty 6/10

For r = 2/(1+0.5cosθ) (the ellipse from the worked example), find r at θ = 60°, given cos(60°)=0.5.

Difficulty 7/10

A comet's orbit is modeled by r = 5/(1+0.8cosθ) (AU), with the Sun at the pole. Find the comet's closest approach (perihelion, θ=0°) and farthest distance within one orbit (aphelion, θ=180°), and classify the orbit.

Quiz

Summary

Every conic with a focus at the pole satisfies r = ed/(1±ecosθ) or r = ed/(1±esinθ), depending on directrix orientation.
The eccentricity e alone classifies the conic: 0≤e<1 ellipse, e=1 parabola, e>1 hyperbola.
This polar form directly models orbital mechanics — Kepler's laws describe planetary and cometary orbits as conics with the Sun at a focus.

References

WebsiteWikipedia — Conic section (polar coordinates)