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
The trigonometric function indicates directrix orientation: [latex]\cos\theta[/latex] means a vertical directrix ([latex]x = \pm p[/latex]), while [latex]\sin\theta[/latex] means a horizontal directrix ([latex]y = \pm p[/latex]).
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.
First, rewrite in standard form by multiplying numerator and denominator by [latex]\frac{1}{3}[/latex]: [latex]\begin{aligned} r &= \frac{6}{3 + 2\sin\theta} \cdot \frac{\frac{1}{3}}{\frac{1}{3}} \ &= \frac{6 \cdot \frac{1}{3}}{3 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3}\sin\theta} \ &= \frac{2}{1 + \frac{2}{3}\sin\theta} \end{aligned}[/latex] Now identify the characteristics: The eccentricity is [latex]e = \frac{2}{3}[/latex]. Since [latex]e < 1[/latex], this is an ellipse. Since [latex]\sin\theta[/latex] appears in the denominator with a plus sign, the directrix is [latex]y = p[/latex] where [latex]p > 0[/latex]. To find [latex]p[/latex], use [latex]ep = 2[/latex]: [latex]\begin{aligned} \frac{2}{3} \cdot p &= 2 \ p &= 2 \cdot \frac{3}{2} \ p &= 3 \end{aligned}[/latex] The satellite follows an elliptical orbit with eccentricity [latex]e = \frac{2}{3}[/latex] and directrix [latex]y = 3[/latex] (3,000 km above the origin).
A space probe follows the path [latex]r = \frac{7}{2 - 2\sin\theta}[/latex]. Identify the orbit characteristics.
Multiply numerator and denominator by [latex]\frac{1}{2}[/latex]: [latex]\begin{aligned} r &= \frac{7}{2 - 2\sin\theta} \cdot \frac{\frac{1}{2}}{\frac{1}{2}} \ &= \frac{\frac{7}{2}}{1 - \sin\theta} \end{aligned}[/latex] The eccentricity is [latex]e = 1[/latex], so this is a parabola. Since [latex]\sin\theta[/latex] appears with a minus sign, the directrix is [latex]y = -p[/latex] where [latex]p > 0[/latex]. From [latex]ep = \frac{7}{2}[/latex]: [latex]\begin{aligned} (1) \cdot p &= \frac{7}{2} \ p &= \frac{7}{2} \end{aligned}[/latex] The directrix is [latex]y = -\frac{7}{2} = -3.5[/latex].
A parabolic orbit ([latex]e = 1[/latex]) represents the boundary between closed orbits (ellipses) and open orbits (hyperbolas). Objects at exactly escape velocity follow parabolic paths!The general forms are [latex]r = \frac{ep}{1 \pm e\cos\theta}[/latex] or [latex]r = \frac{ep}{1 \pm e\sin\theta}[/latex]. Choose [latex]\cos\theta[/latex] for a vertical directrix ([latex]x = \pm p[/latex]) and [latex]\sin\theta[/latex] for a horizontal directrix ([latex]y = \pm p[/latex]). Use [latex]+[/latex] when [latex]p > 0[/latex] and [latex]-[/latex] when [latex]p < 0[/latex].
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]: