Conic Sections in Polar Coordinates

Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.

In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.

Identifying a Conic in Polar Form

Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola shown below, whose focus is at the pole.

In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus $P(r,\theta)$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.

If $F$ is a fixed point, the focus, and $D$ is a fixed line, the directrix, then we can let $e$ be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points $P$ such that $e=\tfrac{PF}{PD}$ is a conic. In other words, we can define a conic as the set of all points $P$ with the property that the ratio of the distance from $P$ to $F$ to the distance from $P$ to $D$ is equal to the constant $e$.

For a conic with eccentricity $e$,

• if $0\le e<1$, the conic is an ellipse
• if $e=1$, the conic is a parabola
• if $e>1$, the conic is a hyperbola

With this definition, we may now define a conic in terms of the directrix, $x=\pm p$, the eccentricity $e$, and the angle $\theta$. Thus, each conic may be written as a polar equation, an equation written in terms of $r$ and $\theta$.

Example. For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.

a. $r=\tfrac{6}{3+2\sin\theta}$ b. $r=\tfrac{12}{4+5\cos\theta}$ c. $r=\tfrac{7}{2-2\sin\theta}$

Solution. For each of the three conics, we will rewrite the equation in standard form. Standard form has a $1$ as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, $\tfrac1c$, where $c$ is that constant.

a. Multiply the numerator and denominator by $\tfrac13$.

Because $\sin\theta$ is in the denominator, the directrix is $y=p$. Comparing to standard form, note that $e=\tfrac23$. Therefore, from the numerator,

Since $e<1$, the conic is an ellipse. The eccentricity is $e=\tfrac23$ and the directrix is $y=3$.

b. Multiply the numerator and denominator by $\tfrac14$.

Because $\cos\theta$ is in the denominator, the directrix is $x=p$. Comparing to standard form, $e=\tfrac54$. Therefore, from the numerator,

Since $e>1$, the conic is a hyperbola. The eccentricity is $e=\tfrac54$ and the directrix is $x=\tfrac{12}{5}=2.4$.

c. Multiply the numerator and denominator by $\tfrac12$.

Because sine is in the denominator, the directrix is $y=-p$. Comparing to standard form, $e=1$. Therefore, from the numerator,

Because $e=1$, the conic is a parabola. The eccentricity is $e=1$ and the directrix is $y=-\tfrac72=-3.5$.

Graphing the Polar Equations of Conics

When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with $1$. This enables us to determine $e$ and, therefore, the shape of the curve. The next step is to substitute values for $\theta$ and solve for $r$ to plot a few key points. Setting $\theta$ equal to $0,\tfrac{\pi}{2},\pi,$ and $\tfrac{3\pi}{2}$ provides the vertices so we can create a rough sketch of the graph.

Example. Graph $r=\tfrac{5}{3+3\cos\theta}$.

Solution. First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of $3$, which is $\tfrac13$.

Because $e=1$, we will graph a parabola with a focus at the origin. The function has a $\cos\theta$, and there is an addition sign in the denominator, so the directrix is $x=p$.

The directrix is $x=\tfrac53$.

Plotting a few key points as in the table below will enable us to see the vertices. See the figure below.

Analysis. We can check our result with a graphing utility. See the figure below.

Example. Graph $r=\tfrac{8}{2-3\sin\theta}$.

Solution. First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of $2$, which is $\tfrac12$.

Because $e=\tfrac32,e>1$, so we will graph a hyperbola with a focus at the origin. The function has a $\sin\theta$ term and there is a subtraction sign in the denominator, so the directrix is $y=-p$.

The directrix is $y=-\tfrac83$.

Plotting a few key points as in the table below will enable us to see the vertices. See the figure below.

Example. Graph $r=\tfrac{10}{5-4\cos\theta}$.

Solution. First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of $5$, which is $\tfrac15$.

Because $e=\tfrac45,e<1$, so we will graph an ellipse with a focus at the origin. The function has a $\cos\theta$, and there is a subtraction sign in the denominator, so the directrix is $x=-p$.

The directrix is $x=-\tfrac52$.

Plotting a few key points as in the table below will enable us to see the vertices. See the figure below.

Analysis. We can check our result using a graphing utility. See the figure below.

Defining Conics in Terms of a Focus and a Directrix

So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.

Example. Find the polar form of the conic given a focus at the origin, $e=3$, and directrix $y=-2$.

Solution. The directrix is $y=-p$, so we know the trigonometric function in the denominator is sine.

Because $y=-2,-2<0$, so we know there is a subtraction sign in the denominator. We use the standard form of

and $e=3$ and $|-2|=2=p$.

Therefore,

Example. Find the polar form of a conic given a focus at the origin, $e=\tfrac35$, and directrix $x=4$.

Solution. Because the directrix is $x=p$, we know the function in the denominator is cosine. Because $x=4,4>0$, so we know there is an addition sign in the denominator. We use the standard form of

and $e=\tfrac35$ and $|4|=4=p$.

Therefore,

Example. Convert the conic $r=\tfrac{1}{5-5\sin\theta}$ to rectangular form.

Solution. We will rearrange the formula to use the identities $r=\sqrt{x^2+y^2},x=r\cos\theta,$ and $y=r\sin\theta$.

Key concepts

• Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus $P(r,\theta)$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.
• A conic is the set of all points $e=\tfrac{PF}{PD}$, where eccentricity $e$ is a positive real number. Each conic may be written in terms of its polar equation.
• The polar equations of conics can be graphed.
• Conics can be defined in terms of a focus, a directrix, and eccentricity.
• We can use the identities $r=\sqrt{x^2+y^2},x=r\cos\theta,$ and $y=r\sin\theta$ to convert the equation for a conic from polar to rectangular form.

Practice

Identify a conic in polar form

Graph the polar equations of conics

Define conics in terms of a focus and a directrix

This section is adapted from Precalculus 2e, Section 10.5: Conic Sections in Polar Coordinates by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: this module (unlike m49438–m49441 earlier in the chapter) carries no coreq-skills prelude, and its own Section Exercises are followed by the chapter’s Review Exercises and Practice Test appended in the same file; those chapter-level exercises were not transcribed and no Practice item was drawn from them. Omitted the credit photograph of the solar system (NASA Blueshift, Flickr), keeping the two paragraphs that introduce it. Kept the Media callout’s introductory sentence but omitted its three external video links, matching house precedent elsewhere in this book. Recreated all six instructional figures as accessible spec-first SVGs on ordinary Cartesian axes with tick labels, rather than the source’s polar-grid background of concentric circles and radial spokes — matching this book’s own §8.4 convention for polar-equation graphs, since the figure engine has no polar-grid primitive: the introductory focus/directrix/angle schematic, drawn as the exact parabola $r=\tfrac{2}{1-\cos\theta}$ (focus at the pole, directrix $x=-2$, a point $P$ at $\theta=70^\circ$ with its equal distances to the focus and to the directrix marked) — the source labels its figure $x=2+y^2$ while placing the focus at the pole, which that parabola’s focus $(\tfrac94,0)$ is not, so the local figure and the sentence introducing it name no equation other than the polar one it actually plots (a source defect, logged in the errata); the three worked-example graphs (a parabola, a hyperbola, and an ellipse), each sampled as polylines from the exact equation with points A–D placed at the printed table’s $\theta=0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}$ values and the hyperbola’s two branches split where the denominator changes sign; and the two “Analysis” graphing-utility check figures (bare re-renders of the same two curves, unlabeled, matching the source’s own unlabeled check art). The grader proves two polar equations of the same conic equal, so “find the polar equation” answers carry no answerForm — a printed-subject retype is not a hazard here because the prompt states only the focus, eccentricity, and directrix in prose, never an equation to retype — and are keyed in whichever of the standard or fraction-cleared integer form the source itself prints. Every “identify the conic” ask is a multiplechoice over ellipse/parabola/hyperbola; eccentricity is keyed as a bare number or fraction, never the letter $e$; a directrix is keyed as an equation ($x=-2$, $y=\tfrac52$) with the question saying “as an equation.” The module’s second in-page Try It (“Graph $r=\tfrac{2}{4-\cos\theta}$”, solution a figure only) is a graph-mode multiple choice over four spec-first options that vary orientation, mirror sign, and eccentricity — following the 9.3 precedent — since the grader cannot take a drawn curve as a submitted answer; the section’s other in-page Try Its keep their source form (identify/eccentricity/directrix fill-ins and multiple choice, or a safe polar-to-rectangular re-expression fill-in). In the closing Practice block, three end-of-section “graph the conic” exercises (Algebraic #39, #33, and #35 in this book’s local numbering, for the vertex, both-vertices, and graph-recognition items respectively) are likewise adapted into a computed fill-in on the vertex or vertices the sketch needs, or a graph-recognition multiple choice, each appearing once (not duplicated with the in-page set). Nine selected end-of-section Algebraic exercises were adapted into the interactive components of the closing Practice block, one Practice group per objective, every one independently re-derived — including by running the arithmetic in Node — rather than read off the source key. One suspected source defect (Algebraic #55: directrix $x=-3,e=\tfrac13$) prints a key, $r=\tfrac{3}{3-3\cos\theta}$, that simplifies to $e=1$ and contradicts its own stated eccentricity of $\tfrac13$; the correctly derived $r=\tfrac{3}{3-\cos\theta}$ was independently verified but the exercise itself was not used on this page, since equivalent, unaffected exercises (#47, #51) were available.