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Precalculus

10.5 Conic Sections in Polar Coordinates

Precalculus10.5 Conic Sections in Polar Coordinates

Learning Objectives

In this section, you will:

  • Identify a conic in polar form.
  • Graph the polar equations of conics.
  • Define conics in terms of a focus and a directrix.
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 parabola x=2+ y 2 x=2+ y 2 shown in Figure 2.

A horizontal parabola, labeled x = 2 + y squared, opening to the right is shown. The Focus is labeled Focus @ pole and is on the horizontal Polar Axis. The vertical Directrix is shown. A point on the upper side of the parabola is labeled P times (r, theta) and two lines of equal length r are drawn from it, one to the Focus and the other to the Directrix and perpendicular to it. The line to the Focus makes an angle theta with the Polar Axis.
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) P(r,θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis.

If F F is a fixed point, the focus, and D D is a fixed line, the directrix, then we can let e 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 P such that e= PF PD e= PF PD is a conic. In other words, we can define a conic as the set of all points P P with the property that the ratio of the distance from P P to F F to the distance from P P to D D is equal to the constant e. e.

For a conic with eccentricity e, e,

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

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

The Polar Equation for a Conic

For a conic with a focus at the origin, if the directrix is x=±p, x=±p, where p p is a positive real number, and the eccentricity is a positive real number e, e, the conic has a polar equation

r= ep 1±ecosθ r= ep 1±ecosθ

For a conic with a focus at the origin, if the directrix is y=±p, y=±p, where p p is a positive real number, and the eccentricity is a positive real number e, e, the conic has a polar equation

r= ep 1±esinθ r= ep 1±esinθ

How To

Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.

  1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
  2. Identify the eccentricity e e as the coefficient of the trigonometric function in the denominator.
  3. Compare e e with 1 to determine the shape of the conic.
  4. Determine the directrix as x=p x=p if cosine is in the denominator and y=p y=p if sine is in the denominator. Set ep ep equal to the numerator in standard form to solve for x x or y. 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.

  1. r= 6 3+2sinθ r= 6 3+2sinθ
  2. r= 12 4+5cosθ r= 12 4+5cosθ
  3. r= 7 22sinθ r= 7 22sinθ

Try It #1

Identify the conic with focus at the origin, the directrix, and the eccentricity for r= 2 3cosθ . r= 2 3cosθ .

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 e and, therefore, the shape of the curve. The next step is to substitute values for θ θ and solve for r r to plot a few key points. Setting θ θ equal to 0, π 2 ,π, 0, π 2 ,π, and 3π 2 3π 2 provides the vertices so we can create a rough sketch of the graph.

Example 2

Graphing a Parabola in Polar Form

Graph r= 5 3+3cosθ . r= 5 3+3cosθ .

Analysis

We can check our result with a graphing utility. See Figure 4.

A horizontal parabola opening left is shown in a polar coordinate system. The Vertex is on the Polar Axis at r = 1. The Polar Axis tick marks are labeled 2, 3, 4, 5.
Figure 4

Example 3

Graphing a Hyperbola in Polar Form

Graph r= 8 23sinθ . r= 8 23sinθ .

Example 4

Graphing an Ellipse in Polar Form

Graph r= 10 54cosθ . r= 10 54cosθ .

Analysis

We can check our result using a graphing utility. See Figure 7.

Figure 7 r= 10 54cosθ r= 10 54cosθ graphed on a viewing window of [ –3,12,1 ] [ –3,12,1 ] by [4,4,1],θmin =0 [4,4,1],θmin =0 and θmax =2π. θmax =2π.

Try It #2

Graph r= 2 4cosθ . r= 2 4cosθ .

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.

  1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, y, we use the general polar form in terms of sine. If the directrix is given in terms of x, x, we use the general polar form in terms of cosine.
  2. Determine the sign in the denominator. If p<0, p<0, use subtraction. If p>0, p>0, use addition.
  3. Write the coefficient of the trigonometric function as the given eccentricity.
  4. Write the absolute value of p 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 e=3 and directrix y=2. y=2.

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= 3 5 , e= 3 5 , and directrix x=4. x=4.

Try It #3

Find the polar form of the conic given a focus at the origin, e=1, e=1, and directrix x=−1. x=−1.

Example 7

Converting a Conic in Polar Form to Rectangular Form

Convert the conic r= 1 55sinθ r= 1 55sinθ to rectangular form.

Try It #4

Convert the conic r= 2 1+2cosθ r= 2 1+2cosθ to rectangular form.

Media

Access these online resources for additional instruction and practice with conics in polar coordinates.

10.5 Section Exercises

Verbal

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, 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?

Algebraic

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.

6.

r= 6 12cosθ r= 6 12cosθ

7.

r= 3 44sinθ r= 3 44sinθ

8.

r= 8 43cosθ r= 8 43cosθ

9.

r= 5 1+2sinθ r= 5 1+2sinθ

10.

r= 16 4+3cosθ r= 16 4+3cosθ

11.

r= 3 10+10cosθ r= 3 10+10cosθ

12.

r= 2 1cosθ r= 2 1cosθ

13.

r= 4 7+2cosθ r= 4 7+2cosθ

14.

r(1cosθ)=3 r(1cosθ)=3

15.

r(3+5sinθ)=11 r(3+5sinθ)=11

16.

r(45sinθ)=1 r(45sinθ)=1

17.

r(7+8cosθ)=7 r(7+8cosθ)=7

For the following exercises, convert the polar equation of a conic section to a rectangular equation.

18.

r= 4 1+3sinθ r= 4 1+3sinθ

19.

r= 2 53sinθ r= 2 53sinθ

20.

r= 8 32cosθ r= 8 32cosθ

21.

r= 3 2+5cosθ r= 3 2+5cosθ

22.

r= 4 2+2sinθ r= 4 2+2sinθ

23.

r= 3 88cosθ r= 3 88cosθ

24.

r= 2 6+7cosθ r= 2 6+7cosθ

25.

r= 5 511sinθ r= 5 511sinθ

26.

r(5+2cosθ)=6 r(5+2cosθ)=6

27.

r(2cosθ)=1 r(2cosθ)=1

28.

r(2.52.5sinθ)=5 r(2.52.5sinθ)=5

29.

r= 6secθ 2+3secθ r= 6secθ 2+3secθ

30.

r= 6cscθ 3+2cscθ r= 6cscθ 3+2cscθ

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= 5 2+cosθ r= 5 2+cosθ

32.

r= 2 3+3sinθ r= 2 3+3sinθ

33.

r= 10 54sinθ r= 10 54sinθ

34.

r= 3 1+2cosθ r= 3 1+2cosθ

35.

r= 8 45cosθ r= 8 45cosθ

36.

r= 3 44cosθ r= 3 44cosθ

37.

r= 2 1sinθ r= 2 1sinθ

38.

r= 6 3+2sinθ r= 6 3+2sinθ

39.

r(1+cosθ)=5 r(1+cosθ)=5

40.

r(34sinθ)=9 r(34sinθ)=9

41.

r(32sinθ)=6 r(32sinθ)=6

42.

r(64cosθ)=5 r(64cosθ)=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= 1 5 x=4;e= 1 5

44.

Directrix: x=4;e=5 x=4;e=5

45.

Directrix: y=2;e=2 y=2;e=2

46.

Directrix: y=2;e= 1 2 y=2;e= 1 2

47.

Directrix: x=1;e=1 x=1;e=1

48.

Directrix: x=1;e=1 x=1;e=1

49.

Directrix: x= 1 4 ;e= 7 2 x= 1 4 ;e= 7 2

50.

Directrix: y= 2 5 ;e= 7 2 y= 2 5 ;e= 7 2

51.

Directrix: y=4;e= 3 2 y=4;e= 3 2

52.

Directrix: x=−2;e= 8 3 x=−2;e= 8 3

53.

Directrix: x=−5;e= 3 4 x=−5;e= 3 4

54.

Directrix: y=2;e=2.5 y=2;e=2.5

55.

Directrix: x=−3;e= 1 3 x=−3;e= 1 3

Extensions

Recall from Rotation of Axes that equations of conics with an xy xy term have rotated graphs. For the following exercises, express each equation in polar form with r r as a function of θ. θ.

56.

xy=2 xy=2

57.

x 2 +xy+ y 2 =4 x 2 +xy+ y 2 =4

58.

2 x 2 +4xy+2 y 2 =9 2 x 2 +4xy+2 y 2 =9

59.

16 x 2 +24xy+9 y 2 =4 16 x 2 +24xy+9 y 2 =4

60.

2xy+y=1 2xy+y=1

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