Figure 1. Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)
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 parabolax=2+y2 shown in Figure 2.
Figure 2
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,θ) 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=PDPF 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≤e<1, the conic is an ellipse
if e=1, the conic is a parabola
if e>1, the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, x=±p, the eccentricity e, and the angle θ. Thus, each conic may be written as a polar equation, an equation written in terms of r and θ.
A General Note: The Polar Equation for a Conic
For a conic with a focus at the origin, if the directrix is x=±p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation
r=1±ecosθep
For a conic with a focus at the origin, if the directrix is y=±p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation
r=1±esinθep
How To: Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.
Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
Identify the eccentricity e as the coefficient of the trigonometric function in the denominator.
Compare e with 1 to determine the shape of the conic.
Determine the directrix as x=p if cosine is in the denominator and y=p if sine is in the denominator. Set ep equal to the numerator in standard form to solve for x or y.
Example 1: Identifying a Conic Given the Polar Form
For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
r=3+2sinθ6
r=4+5cosθ12
r=2−2sinθ7
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, c1, where c is that constant.
Because sine is in the denominator, the directrix is y=−p. Comparing to standard form, e=1. Therefore, from the numerator,
27=ep27=(1)p27=p
Because e=1, the conic is a parabola. The eccentricity is e=1 and the directrix is y=−27=−3.5.
Try It 1
Identify the conic with focus at the origin, the directrix, and the eccentricity for r=3−cosθ2.
Solution
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 θ and solve for r to plot a few key points. Setting θ equal to 0,2π,π, and 23π provides the vertices so we can create a rough sketch of the graph.
Example 2: Graphing a Parabola in Polar Form
Graph r=3+3cosθ5.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 31.
Because e=1, we will graph a parabola with a focus at the origin. The function has a cosθ, and there is an addition sign in the denominator, so the directrix is x=p.
35=ep35=(1)p35=p
The directrix is x=35.
Plotting a few key points as in the table below will enable us to see the vertices.
A
B
C
D
θ
0
2π
π
23π
r=3+3cosθ5
65≈0.83
35≈1.67
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35≈1.67
Figure 3
Analysis of the Solution
We can check our result with a graphing utility.
Figure 4
Example 3: Graphing a Hyperbola in Polar Form
Graph r=2−3sinθ8.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 21.
Because e=23,e>1, so we will graph a hyperbola with a focus at the origin. The function has a sinθ term and there is a subtraction sign in the denominator, so the directrix is y=−p.
4=ep4=(23)p4(32)=p38=p
The directrix is y=−38.
Plotting a few key points as in the table below will enable us to see the vertices.
A
B
C
D
θ
0
2π
π
23π
r=2−3sinθ8
4
−8
4
58=1.6
Figure 5
Example 4: Graphing an Ellipse in Polar Form
Graph r=5−4cosθ10.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 51.
Because e=54,e<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ, and there is a subtraction sign in the denominator, so the directrix is x=−p.
2=ep2=(54)p2(45)=p25=p
The directrix is x=−25.
Plotting a few key points as in the table below will enable us to see the vertices.
A
B
C
D
θ
0
2π
π
23π
r=5−4cosθ10
10
2
910≈1.1
2
Figure 6
Analysis of the Solution
We can check our result using a graphing utility.
Figure 7.r=5−4cosθ10 graphed on a viewing window of [−3,12,1] by [−4,4,1],θmin =0 and θmax =2π.
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.
How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, we use the general polar form in terms of sine. If the directrix is given in terms of x, we use the general polar form in terms of cosine.
Determine the sign in the denominator. If p<0, use subtraction. If p>0, use addition.
Write the coefficient of the trigonometric function as the given eccentricity.
Write the absolute value of p in the numerator, and simplify the equation.
Example 5: Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of the conic given a focus at the origin, e=3 and directrixy=−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
r=1−esinθep
and e=3 and ∣−2∣=2=p.
Therefore,
r=1−3sinθ(3)(2)r=1−3sinθ6
Example 6: Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of a conic given a focus at the origin, e=53, and directrixx=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
Find the polar form of the conic given a focus at the origin, e=1, and directrix x=−1.
Solution
Example 5: Converting a Conic in Polar Form to Rectangular Form
Convert the conic r=5−5sinθ1 to rectangular form.
Solution
We will rearrange the formula to use the identities r=x2+y2,x=rcosθ,and y=rsinθ.
r=5−5sinθ1r⋅(5−5sinθ)=5−5sinθ1⋅(5−5sinθ)5r−5rsinθ=15r=1+5rsinθ25r2=(1+5rsinθ)225(x2+y2)=(1+5y)225x2+25y2=1+10y+25y225x2−10y=1Eliminate the fraction.Distribute.Isolate 5r.Square both sides.Substitute r=x2+y2 and y=rsinθ.Distribute and use FOIL.Rearrange terms and set equal to 1.
Try It 4
Convert the conic r=1+2cosθ2 to rectangular form.
Solution
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,θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis.
A conic is the set of all points e=PDPF, 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=x2+y2,x=rcosθ, and y=rsinθ to convert the equation for a conic from polar to rectangular form.
Glossary
eccentricity
the ratio of the distances from a point P on the graph to the focus F and to the directrix D represented by e=PDPF, where e is a positive real number
polar equation
an equation of a curve in polar coordinates r and θ
Section Exercises
1. Explain how eccentricity determines which conic section is given.
2. If a conic section is written as a polar equation, what must be true of the denominator?
3. If a conic section is written as a polar equation, and the denominator involves sinθ, what conclusion can be drawn about the directrix?
4. If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?
5. What do we know about the focus/foci of a conic section if it is written as a polar equation?
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.
6. r=1−2cosθ6
7. r=4−4sinθ3
8. r=4−3cosθ8
9. r=1+2sinθ5
10. r=4+3cosθ16
11. r=10+10cosθ3
12. r=1−cosθ2
13. r=7+2cosθ4
14. r(1−cosθ)=3
15. r(3+5sinθ)=11
16. r(4−5sinθ)=1
17. r(7+8cosθ)=7
For the following exercises, convert the polar equation of a conic section to a rectangular equation.
18. r=1+3sinθ4
19. r=5−3sinθ2
20. r=3−2cosθ8
21. r=2+5cosθ3
22. r=2+2sinθ4
23. r=8−8cosθ3
24. r=6+7cosθ2
25. r=5−11sinθ5
26. r(5+2cosθ)=6
27. r(2−cosθ)=1
28. r(2.5−2.5sinθ)=5
29. r=−2+3secθ6secθ
30. r=3+2cscθ6cscθ
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.
31. r=2+cosθ5
32. r=3+3sinθ2
33. r=5−4sinθ10
34. r=1+2cosθ3
35. r=4−5cosθ8
36. r=4−4cosθ3
37. r=1−sinθ2
38. r=3+2sinθ6
39. r(1+cosθ)=5
40. r(3−4sinθ)=9
41. r(3−2sinθ)=6
42. r(6−4cosθ)=5
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.
43. Directrix: x=4;e=51
44. Directrix: x=−4;e=5
45. Directrix: y=2;e=2
46. Directrix: y=−2;e=21
47. Directrix: x=1;e=1
48. Directrix: x=−1;e=1
49. Directrix: x=−41;e=27
50. Directrix: y=52;e=27
51. Directrix: y=4;e=23
52. Directrix: x=−2;e=38
53. Directrix: x=−5;e=43
54. Directrix: y=2;e=2.5
55. Directrix: x=−3;e=31
Equations of conics with an xy term have rotated graphs. For the following exercises, express each equation in polar form with r as a function of θ.
56. xy=2
57. x2+xy+y2=4
58. 2x2+4xy+2y2=9
59. 16x2+24xy+9y2=4
60. 2xy+y=1