- Identify a conic in polar form.
- Graph the polar equations of conics.
- Define conics in terms of a focus and a directrix.
A conic section in polar form with focus at the origin has the equation [latex]r = \frac{ep}{1 \pm e\cos\theta}[/latex] or [latex]r = \frac{ep}{1 \pm e\sin\theta}[/latex], where [latex]e[/latex] is the eccentricity and [latex]p[/latex] is the distance to the directrix (a fixed reference line). The eccentricity determines the shape:
- If [latex]0 \leq e < 1[/latex]: ellipse
- If [latex]e = 1[/latex]: parabola
- If [latex]e > 1[/latex]: hyperbola
Orbital Mechanics
Satellites, planets, and comets follow conic paths around celestial bodies. The shape of an orbit depends on the object’s velocity and distance from the body it orbits. Understanding these polar equations helps scientists predict orbital behavior and plan space missions.
A satellite’s orbital path is described by [latex]r = \frac{6}{3 + 2\sin\theta}[/latex], where [latex]r[/latex] is measured in thousands of kilometers. Identify the type of orbit, the eccentricity, and the directrix.
A space probe follows the path [latex]r = \frac{7}{2 - 2\sin\theta}[/latex]. Identify the orbit characteristics.
A comet’s path is given by [latex]r = \frac{12}{4 + 5\cos\theta}[/latex]. Identify the type of conic, the eccentricity, and the directrix.
Type of conic: [response area]
Eccentricity [latex]e =[/latex] [response area]
Directrix: [response area]
Correct answer:
- Type: hyperbola
- Eccentricity: [latex]\frac{5}{4}[/latex]
- Directrix: [latex]x = \frac{12}{5}[/latex] or [latex]x = 2.4[/latex]
Feedback for correct answer: Excellent! You rewrote the equation as [latex]r = \frac{3}{1 + \frac{5}{4}\cos\theta}[/latex], identified [latex]e = \frac{5}{4} > 1[/latex] (hyperbola), and solved [latex]\frac{5}{4}p = 3[/latex] to get [latex]p = \frac{12}{5}[/latex]. Since [latex]\cos\theta[/latex] appears with a plus sign, the directrix is [latex]x = \frac{12}{5}[/latex]. This hyperbolic path means the comet will escape the gravitational pull!
Feedback for incorrect answer: Start by multiplying numerator and denominator by [latex]\frac{1}{4}[/latex] to get standard form. The coefficient of [latex]\cos\theta[/latex] in the denominator is your eccentricity [latex]e[/latex]. Compare [latex]e[/latex] to 1 to determine the conic type. Since [latex]\cos\theta[/latex] appears, the directrix has form [latex]x = p[/latex].
Identify the conic type and eccentricity for [latex]r = \frac{2}{3 - \cos\theta}[/latex].
Type: [response area]
Eccentricity [latex]e =[/latex] [response area]
Correct answer:
- Type: ellipse
- Eccentricity: [latex]\frac{1}{3}[/latex]
Feedback for correct answer: Perfect! You converted to standard form [latex]r = \frac{\frac{2}{3}}{1 - \frac{1}{3}\cos\theta}[/latex] and identified [latex]e = \frac{1}{3} < 1[/latex], which means this is an ellipse. The relatively small eccentricity indicates a nearly circular orbit.
Feedback for incorrect answer: Multiply both numerator and denominator by [latex]\frac{1}{3}[/latex] to rewrite with 1 in the denominator. The coefficient of [latex]\cos\theta[/latex] after this conversion is your eccentricity. Since this value is less than 1, what type of conic is it?
Mission control needs to place a satellite in an elliptical orbit with focus at Earth’s center, eccentricity [latex]e = \frac{3}{5}[/latex], and directrix at [latex]x = 4[/latex] (in thousands of km). Write the polar equation for this orbit.
Solution:
The directrix [latex]x = 4[/latex] tells us:
- Use [latex]\cos\theta[/latex] in the denominator (vertical directrix)
- Since [latex]4 > 0[/latex], use the [latex]+[/latex] sign
- [latex]p = 4[/latex]
The standard form is:
[latex]r = \frac{ep}{1 + e\cos\theta}[/latex]
Substituting [latex]e = \frac{3}{5}[/latex] and [latex]p = 4[/latex]:
[latex]\begin{aligned} r &= \frac{\frac{3}{5} \cdot 4}{1 + \frac{3}{5}\cos\theta} \ &= \frac{\frac{12}{5}}{1 + \frac{3}{5}\cos\theta} \end{aligned}[/latex]
To eliminate the fraction in the numerator, multiply by [latex]\frac{5}{5}[/latex]:
[latex]\begin{aligned} r &= \frac{\frac{12}{5}}{\frac{5}{5} + \frac{3}{5}\cos\theta} \ &= \frac{12}{5} \cdot \frac{5}{5 + 3\cos\theta} \ &= \frac{12}{5 + 3\cos\theta} \end{aligned}[/latex]
The orbital equation is [latex]r = \frac{12}{5 + 3\cos\theta}[/latex].
You are a mission planner designing trajectories for three different space missions. Each mission requires a specific orbital path with a focus at Earth’s center. Choose one mission and write its polar equation.
Mission A: Communications Satellite
- Orbit type: Elliptical (for stable, repeating coverage)
- Eccentricity: [latex]e = \frac{2}{3}[/latex]
- Directrix: [latex]y = 6[/latex] thousand km
Mission B: Interstellar Probe
- Orbit type: Hyperbolic (to escape the solar system)
- Eccentricity: [latex]e = \frac{5}{2}[/latex]
- Directrix: [latex]x = -4[/latex] thousand km
Mission C: Solar Observation Probe
- Orbit type: Parabolic (at exact escape velocity)
- Eccentricity: [latex]e = 1[/latex]
- Directrix: [latex]y = -3[/latex] thousand km
Choose your mission and write the polar equation: [latex]r =[/latex] [response area]
Correct answer for Mission A: [latex]r = \frac{4}{1 + \frac{2}{3}\sin\theta}[/latex] or [latex]r = \frac{12}{3 + 2\sin\theta}[/latex]
Feedback for Mission A correct: Excellent mission planning! You identified that directrix [latex]y = 6[/latex] requires [latex]\sin\theta[/latex] with a plus sign. With [latex]e = \frac{2}{3}[/latex] and [latex]p = 6[/latex], you calculated [latex]ep = 4[/latex]. Your equation [latex]r = \frac{4}{1 + \frac{2}{3}\sin\theta}[/latex] creates a stable elliptical orbit perfect for consistent satellite coverage. The eccentricity of [latex]\frac{2}{3}[/latex] provides an elongated but closed orbit.
Feedback for Mission A incorrect: For directrix [latex]y = 6[/latex], use the form [latex]r = \frac{ep}{1 + e\sin\theta}[/latex] (positive [latex]p[/latex] means plus sign). Calculate [latex]ep = \frac{2}{3} \cdot 6 = 4[/latex] for the numerator.
Correct answer for Mission B: [latex]r = \frac{10}{1 - \frac{5}{2}\cos\theta}[/latex] or [latex]r = \frac{20}{2 - 5\cos\theta}[/latex]
Feedback for Mission B correct: Outstanding! You recognized that directrix [latex]x = -4[/latex] requires [latex]\cos\theta[/latex] with a minus sign. With [latex]e = \frac{5}{2}[/latex] and [latex]p = 4[/latex], you found [latex]ep = 10[/latex]. Your hyperbolic trajectory [latex]r = \frac{10}{1 - \frac{5}{2}\cos\theta}[/latex] will successfully launch the probe out of the solar system. Since [latex]e > 1[/latex], this open orbit provides the escape path needed!
Feedback for Mission B incorrect: For directrix [latex]x = -4[/latex], use the form [latex]r = \frac{ep}{1 - e\cos\theta}[/latex] (negative [latex]p[/latex] means minus sign, but use [latex]|p| = 4[/latex]). Calculate [latex]ep = \frac{5}{2} \cdot 4 = 10[/latex] for the numerator.
Correct answer for Mission C: [latex]r = \frac{3}{1 - \sin\theta}[/latex]
Feedback for Mission C correct: Perfect trajectory design! You correctly identified that directrix [latex]y = -3[/latex] requires [latex]\sin\theta[/latex] with a minus sign. For a parabola, [latex]e = 1[/latex], so [latex]ep = (1)(3) = 3[/latex]. Your equation [latex]r = \frac{3}{1 - \sin\theta}[/latex] places the probe at exactly escape velocity—the boundary between returning to Earth and escaping forever. This is ideal for a solar observation mission that needs to break free of Earth’s orbit!
Feedback for Mission C incorrect: A parabola has [latex]e = 1[/latex]. For directrix [latex]y = -3[/latex], use [latex]r = \frac{ep}{1 - e\sin\theta}[/latex]. With [latex]e = 1[/latex] and [latex]|p| = 3[/latex], calculate [latex]ep[/latex] for the numerator.