Conics in Polar Form
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.5: “Conic Sections” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-5-conic-sections |
| Textbook used in class | Stewart, Calculus, Section 10.6: “Conic Sections in Polar Coordinates” (Examples 1, 2, 3, 4) |
Opening Scenario
The equations for ellipses, hyperbolas, and parabolas in Cartesian coordinates look completely different from one another. In polar coordinates, with the focus placed at the origin, all three conics fit a single formula. That unified formula is the natural language for planetary orbits: Kepler derived his laws of planetary motion in this form, and NASA uses it today to calculate trajectories.
Quick Reference
Theorem. A conic section with one focus at the pole, corresponding directrix at distance $d$ from that focus, and eccentricity $e > 0$ has polar equation:
\[ r = \frac{ed}{1 + e\cos\theta} \quad \text{(directrix to the right, vertical)} \]
or one of the following variants depending on the directrix orientation:
| Equation | Directrix location | Axis |
|---|---|---|
| $r = \dfrac{ed}{1 + e\cos\theta}$ | right of pole ($x = d$) | horizontal |
| $r = \dfrac{ed}{1 - e\cos\theta}$ | left of pole ($x = -d$) | horizontal |
| $r = \dfrac{ed}{1 + e\sin\theta}$ | above pole ($y = d$) | vertical |
| $r = \dfrac{ed}{1 - e\sin\theta}$ | below pole ($y = -d$) | vertical |
Conic type from $e$:
- $e < 1$: ellipse
- $e = 1$: parabola
- $e > 1$: hyperbola
Important: $d > 0$ and $e > 0$. The numerator $ed$ is a single positive constant.
Key Concepts
1. Why One Formula Covers All Three Conics
The focus-directrix definition unifies all conics: a conic is the set of all points $P$ where \[ \frac{|PF|}{|P\ell|} = e, \] with $|PF|$ the distance to the focus and $|P\ell|$ the distance to the directrix. When $e = 1$ this gives the parabola definition directly. For $e \neq 1$ it turns out to produce an ellipse ($e < 1$) or a hyperbola ($e > 1$).
Placing the focus at the pole (the origin of polar coordinates) and the directrix perpendicular to the polar axis at $x = d$, the condition $|PF|/|P\ell| = e$ translates directly into the polar equation $r = ed/(1 + e\cos\theta)$. The algebra: $|PF| = r$ and $|P\ell| = d - r\cos\theta$ (horizontal distance from the point to the vertical line $x = d$), so $r/(d - r\cos\theta) = e$, giving $r = e(d - r\cos\theta)$, then $r(1 + e\cos\theta) = ed$, hence $r = ed/(1 + e\cos\theta)$.
2. Identifying the Conic and Its Orientation
To analyze $r = f(\theta)$:
- Divide numerator and denominator by the constant in front of $\cos\theta$ or $\sin\theta$ so the constant term in the denominator is $1$.
- Read off $e$ as the coefficient of the trig function.
- Read off $d = \text{numerator}/e$.
- Use $e$ to identify the conic type.
- Use the trig function and the sign to identify the directrix.
Example 1. Identify the conic $r = \dfrac{8}{2 + \sin\theta}$.
Goal. Rewrite so the denominator constant is $1$, then read $e$. (This is Stewart 10.6, Example 1.)
Divide numerator and denominator by $2$: \[ r = \frac{4}{1 + \frac{1}{2}\sin\theta}. \]
Here $e = \dfrac{1}{2} < 1$, so the conic is an ellipse. The denominator uses $\sin\theta$ with a $+$ sign, so the directrix is above the pole at $y = d$. From $ed = 4$ and $e = 1/2$, we get $d = 8$.
Boxed answer: Ellipse with $e = 1/2$; directrix $y = 8$ (above the pole).
3. Matching the Equation to a Conic’s Shape
Example 2. Identify the conic and sketch its key features for $r = \dfrac{10}{3 - 2\cos\theta}$. (This is Stewart 10.6, Example 2.)
Divide by $3$: \[ r = \frac{10/3}{1 - \frac{2}{3}\cos\theta}. \]
Here $e = 2/3 < 1$: ellipse. The denominator uses $-\cos\theta$, so the directrix is to the left.
The parameter: $ed = 10/3$, so $d = (10/3)/(2/3) = 5$. Directrix: $x = -5$.
Key $r$ values: at $\theta = 0$, $r = 10/(3-2) = 10$ (rightmost point); at $\theta = \pi$, $r = 10/(3+2) = 2$ (leftmost point, closest to the left directrix). The semi-major axis runs between $r = 10$ and $r = 2$, so $2a = 12$ and $a = 6$. Center is at distance $a - r_{\min} = 6 - 2 = 4$ from the focus.
Boxed answer: Ellipse with $e = 2/3$, directrix $x = -5$; $r$ ranges from $2$ to $10$.
4. Writing a Polar Equation From Given Data
Example 3. Find a polar equation for the ellipse with eccentricity $1/3$ and directrix $x = 4$ (to the right of the focus). (This is Stewart 10.6, Example 3.)
Goal. Use the form $r = ed/(1 + e\cos\theta)$ with $e = 1/3$ and $d = 4$.
\[ r = \frac{(1/3)(4)}{1 + (1/3)\cos\theta} = \frac{4/3}{1 + (1/3)\cos\theta}. \]
Multiply numerator and denominator by $3$: \[ r = \frac{4}{3 + \cos\theta}. \]
Boxed answer: $r = \dfrac{4}{3 + \cos\theta}$.
Example 4. Find a polar equation for the parabola with focus at the origin and directrix $y = -3$ (below the pole). (This is Stewart 10.6, Example 4 type.)
Goal. Parabola means $e = 1$. The directrix is below at $y = -d$ with $d = 3$, so use $r = d/(1 - \sin\theta)$ (minus sign for below, $\sin$ for vertical directrix).
\[ r = \frac{(1)(3)}{1 - \sin\theta} = \frac{3}{1 - \sin\theta}. \]
Boxed answer: $r = \dfrac{3}{1 - \sin\theta}$.
treating $ed$ as the $r$-value at the vertex. The numerator $ed$ is not the $r$-value at any particular point. The vertex (closest point to the focus) occurs where the denominator is largest. For $r = ed/(1 + e\cos\theta)$, the denominator is largest at $\theta = 0$, giving $r_{\min} = ed/(1 + e)$. Students often confuse $ed$ with this vertex value. Always rewrite the equation in standard form first, read $e$, then compute $d = \text{numerator}/e$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Not dividing to get $1$ in the denominator | reading $e = 2$ from $r = 8/(2 + \sin\theta)$ | Divide first: $r = 4/(1 + \frac{1}{2}\sin\theta)$, so $e = 1/2$ |
| Confusing $+\cos$ with directrix on the left | placing $x = -d$ when the denominator is $1 + e\cos\theta$ | Denominator $1 + e\cos\theta$ goes with directrix to the right ($x = +d$) |
| Confusing $+\sin$ with directrix below | placing $y = -d$ when the denominator is $1 + e\sin\theta$ | Denominator $1 + e\sin\theta$ goes with directrix above ($y = +d$) |
| Forgetting $e = 1$ for a parabola | choosing $e$ other than $1$ for a parabola | Parabola always has $e = 1$; this is the defining eccentricity |
Common Misconceptions
the eccentricity $e$ can be read directly from the denominator of $r = f(\theta)$ without first dividing to make the constant term equal to $1$.
This is the concept-image-conflicts-definition error. The standard polar form requires the constant term in the denominator to be exactly $1$. In $r = 8/(2 + \sin\theta)$, dividing numerator and denominator by $2$ gives $r = 4/(1 + \frac{1}{2}\sin\theta)$, revealing $e = 1/2$. Reading $e = 2$ from the un-standardized denominator would classify an ellipse as a hyperbola.
the numerator $ed$ in $r = ed/(1 + e\cos\theta)$ equals the $r$-value at the vertex.
This is the limit-equals-function-value error applied to polar conics. The vertex occurs where the denominator is largest, giving $r_{\min} = ed/(1 + e)$, not $r = ed$. The numerator $ed$ is simply the product of eccentricity and directrix distance; it matches an $r$-value only at a specific angle, not at the vertex.
Leveled Practice
Level 1 -- Identify and Read
Problem 1. Identify the type of conic: $r = \dfrac{12}{4 + 3\cos\theta}$.
Show answer
Divide by $4$: $r = \dfrac{3}{1 + \frac{3}{4}\cos\theta}$.
$e = 3/4 < 1$: ellipse. Directrix to the right at $x = d$, where $d = 3/(3/4) = 4$.
Problem 2. Identify the conic: $r = \dfrac{5}{1 - 2\sin\theta}$.
Show answer
The denominator constant is already $1$. $e = 2 > 1$: hyperbola. The $-\sin\theta$ means the directrix is below the pole at $y = -d$, where $d = 5/2$.
Level 2 -- Write the Equation
Problem 3. Write a polar equation for the hyperbola with eccentricity $3$ and directrix $y = 2$ (above the focus at the origin).
Show answer
Above means $+\sin\theta$, parabola would use $e = 1$ but here $e = 3$, $d = 2$: \[ r = \frac{(3)(2)}{1 + 3\sin\theta} = \frac{6}{1 + 3\sin\theta}. \]
Problem 4. The planet Mercury has an approximately elliptical orbit with eccentricity $e = 0.206$ and perihelion distance (closest approach) $r_{\min} = 4.60 \times 10^7$ km. Find a polar equation for the orbit with the Sun at the focus.
Show answer
With perihelion at $\theta = 0$ and the directrix to the right, $r = ed/(1 + e\cos\theta)$. At $\theta = 0$, $r_{\min} = ed/(1 + e)$, so \[ ed = r_{\min}(1 + e) = 4.60 \times 10^7 \times 1.206 \approx 5.55 \times 10^7. \]
Boxed answer: $r = \dfrac{5.55 \times 10^7}{1 + 0.206\cos\theta}$ (km).
Mastery Checklist
Mental Model
Think of eccentricity as a dial. Turn it from $0$ toward $1$ and the ellipse stretches from a circle toward a long, flat oval. At exactly $1$ you pass through the parabola, where one branch escapes to infinity. Turn past $1$ and the two branches of a hyperbola open up, spread farther apart as $e$ grows. The polar formula $r = ed/(1 + e\cos\theta)$ is the same equation throughout this dial-turning; only $e$ changes.
The denominator $1 + e\cos\theta$ becomes zero at $\cos\theta = -1/e$ when $e > 1$. That is where $r \to \infty$, which corresponds to the asymptotic directions of the hyperbola. For $e < 1$ the denominator is always positive, so $r$ is always finite: the closed ellipse.