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# Conic Sections in Polar Coordinates

Module by: First Last. E-mail the author

Summary: 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.

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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.

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.

### A General Note: 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±e cos θ r= ep 1±e cos θ

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±e sin θ r= ep 1±e sin θ

### 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

#### Problem 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+2 sin θ r= 6 3+2 sin θ
2. r= 12 4+5 cos θ r= 12 4+5 cos θ
3. r= 7 22 sin θ r= 7 22 sin θ
##### 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, 1 c , 1 c , where c c is that constant.

1. Multiply the numerator and denominator by 1 3 . 1 3 .
r= 6 3+2sin θ ( 1 3 ) ( 1 3 ) = 6( 1 3 ) 3( 1 3 )+2( 1 3 )sin θ = 2 1+ 2 3  sin θ r= 6 3+2sin θ ( 1 3 ) ( 1 3 ) = 6( 1 3 ) 3( 1 3 )+2( 1 3 )sin θ = 2 1+ 2 3  sin θ

Because sin θ sin θ is in the denominator, the directrix is y=p. y=p. Comparing to standard form, note that e= 2 3 . e= 2 3 . Therefore, from the numerator,

2=ep      2= 2 3 p ( 3 2 )2=( 3 2 ) 2 3 p      3=p      2=ep      2= 2 3 p ( 3 2 )2=( 3 2 ) 2 3 p      3=p

Since e<1, e<1, the conic is an ellipse. The eccentricity is e= 2 3 e= 2 3 and the directrix is y=3. y=3.

2. Multiply the numerator and denominator by 1 4 . 1 4 .
r= 12 4+5 cos θ ( 1 4 ) ( 1 4 ) r= 12( 1 4 ) 4( 1 4 )+5( 1 4 )cos θ r= 3 1+ 5 4  cos θ r= 12 4+5 cos θ ( 1 4 ) ( 1 4 ) r= 12( 1 4 ) 4( 1 4 )+5( 1 4 )cos θ r= 3 1+ 5 4  cos θ

Because  cosθ   cosθ  is in the denominator, the directrix is x=p. x=p. Comparing to standard form, e= 5 4 . e= 5 4 . Therefore, from the numerator,

3=ep       3= 5 4 p ( 4 5 )3=( 4 5 ) 5 4 p     12 5 =p       3=ep       3= 5 4 p ( 4 5 )3=( 4 5 ) 5 4 p     12 5 =p

Since e>1, e>1, the conic is a hyperbola. The eccentricity is e= 5 4 e= 5 4 and the directrix is x= 12 5 =2.4. x= 12 5 =2.4.

3. Multiply the numerator and denominator by 1 2 . 1 2 .
r= 7 22 sin θ ( 1 2 ) ( 1 2 ) r= 7( 1 2 ) 2( 1 2 )2( 1 2 ) sin θ r= 7 2 1sin θ r= 7 22 sin θ ( 1 2 ) ( 1 2 ) r= 7( 1 2 ) 2( 1 2 )2( 1 2 ) sin θ r= 7 2 1sin θ

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

7 2 =ep 7 2 =( 1 )p 7 2 =p 7 2 =ep 7 2 =( 1 )p 7 2 =p

Because e=1, e=1, the conic is a parabola. The eccentricity is e=1 e=1 and the directrix is y= 7 2 =−3.5. y= 7 2 =−3.5.

### Try It:

#### Exercise 1

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

##### Solution

ellipse; e= 1 3 ;x=2 e= 1 3 ;x=2

## 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

#### Problem 1

##### Graphing a Parabola in Polar Form

Graph r= 5 3+3 cos θ . r= 5 3+3 cos θ .

##### Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 1 3 . 1 3 .

r= 5 3+3 cos θ = 5( 1 3 ) 3( 1 3 )+3( 1 3 )cos θ r= 5 3 1+cos θ r= 5 3+3 cos θ = 5( 1 3 ) 3( 1 3 )+3( 1 3 )cos θ r= 5 3 1+cos θ

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

5 3 =ep 5 3 =(1)p 5 3 =p 5 3 =ep 5 3 =(1)p 5 3 =p

The directrix is x= 5 3 . x= 5 3 .

Plotting a few key points as in Table 1 will enable us to see the vertices. See Figure 3.

Table 1
A B C D
θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
r= 5 3+3 cos θ r= 5 3+3 cos θ 5 6 0.83 5 6 0.83 5 3 1.67 5 3 1.67 undefined 5 3 1.67 5 3 1.67
##### Analysis

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

### Example 3

#### Problem 1

##### Graphing a Hyperbola in Polar Form

Graph r= 8 23 sin θ . r= 8 23 sin θ .

##### Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 1 2 . 1 2 .

r= 8 23sin θ = 8( 1 2 ) 2( 1 2 )3( 1 2 )sin θ r= 4 1 3 2  sin θ r= 8 23sin θ = 8( 1 2 ) 2( 1 2 )3( 1 2 )sin θ r= 4 1 3 2  sin θ

Because e= 3 2 ,e>1, e= 3 2 ,e>1, so we will graph a hyperbola with a focus at the origin. The function has a sin θ sin θ term and there is a subtraction sign in the denominator, so the directrix is y=p. y=p.

4=ep       4=( 3 2 )p 4( 2 3 )=p       8 3 =p       4=ep       4=( 3 2 )p 4( 2 3 )=p       8 3 =p

The directrix is y= 8 3 . y= 8 3 .

Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5.

Table 2
A B C D
θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
r= 8 23sinθ r= 8 23sinθ 4 4 8 8 4 4 8 5 =1.6 8 5 =1.6

### Example 4

#### Problem 1

##### Graphing an Ellipse in Polar Form

Graph r= 10 54 cos θ . r= 10 54 cos θ .

##### Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 1 5 . 1 5 .

r= 10 54cos θ = 10( 1 5 ) 5( 1 5 )4( 1 5 )cos θ r= 2 1 4 5  cos θ r= 10 54cos θ = 10( 1 5 ) 5( 1 5 )4( 1 5 )cos θ r= 2 1 4 5  cos θ

Because e= 4 5 ,e<1, e= 4 5 ,e<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ, cosθ, and there is a subtraction sign in the denominator, so the directrix is x=p. x=p.

2=ep       2=( 4 5 )p 2( 5 4 )=p       5 2 =p       2=ep       2=( 4 5 )p 2( 5 4 )=p       5 2 =p

The directrix is x= 5 2 . x= 5 2 .

Plotting a few key points as in Table 3 will enable us to see the vertices. See Figure 6.

Table 3
A B C D
θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
r= 10 54 cos θ r= 10 54 cos θ 10 10 2 2 10 9 1.1 10 9 1.1 2 2
##### Analysis

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

### Try It:

#### Exercise 2

Graph r= 2 4cos θ . r= 2 4cos θ .

## 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

#### Problem 1

##### 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.

##### Solution

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

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

r= ep 1e sin θ r= ep 1e sin θ

and e=3 e=3 and | −2 |=2=p. | −2 |=2=p.

Therefore,

r= (3)(2) 13 sin θ r= 6 13 sin θ r= (3)(2) 13 sin θ r= 6 13 sin θ

### Example 6

#### Problem 1

##### 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.

##### Solution

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

r= ep 1+e cos θ r= ep 1+e cos θ

and e= 3 5 e= 3 5 and | 4 |=4=p. | 4 |=4=p.

Therefore,

r= ( 3 5 )(4) 1+ 3 5 cosθ r= 12 5 1+ 3 5 cosθ r= 12 5 1( 5 5 )+ 3 5 cosθ r= 12 5 5 5 + 3 5 cosθ r= 12 5 5 5+3cosθ r= 12 5+3cosθ r= ( 3 5 )(4) 1+ 3 5 cosθ r= 12 5 1+ 3 5 cosθ r= 12 5 1( 5 5 )+ 3 5 cosθ r= 12 5 5 5 + 3 5 cosθ r= 12 5 5 5+3cosθ r= 12 5+3cosθ

### Try It:

#### Exercise 3

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

##### Solution

r= 1 1cosθ r= 1 1cosθ

### Example 7

#### Problem 1

##### Converting a Conic in Polar Form to Rectangular Form

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

##### Solution

We will rearrange the formula to use the identities  r= x 2 + y 2 ,x=rcosθ,and y=rsinθ.  r= x 2 + y 2 ,x=rcosθ,and y=rsinθ.

r= 1 55sinθ  r(55sinθ)= 1 55sinθ (55sinθ) Eliminate the fraction.        5r5rsinθ=1 Distribute.                         5r=1+5rsinθ Isolate 5r.                     25 r 2 = (1+5rsinθ) 2 Square both sides.          25( x 2 + y 2 )= (1+5y) 2 Substitute r= x 2 + y 2  and y=rsinθ.        25 x 2 +25 y 2 =1+10y+25 y 2 Distribute and use FOIL.          25 x 2 10y=1 Rearrange terms and set equal to 1.                           r= 1 55sinθ  r(55sinθ)= 1 55sinθ (55sinθ) Eliminate the fraction.        5r5rsinθ=1 Distribute.                         5r=1+5rsinθ Isolate 5r.                     25 r 2 = (1+5rsinθ) 2 Square both sides.          25( x 2 + y 2 )= (1+5y) 2 Substitute r= x 2 + y 2  and y=rsinθ.        25 x 2 +25 y 2 =1+10y+25 y 2 Distribute and use FOIL.          25 x 2 10y=1 Rearrange terms and set equal to 1.

### Try It:

#### Exercise 4

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

##### Solution

48x+3 x 2 y 2 =0 48x+3 x 2 y 2 =0

### Media:

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

Visit this website for additional practice questions from Learningpod.

## 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) 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= PF PD , e= PF PD , where eccentricity e e is a positive real number. Each conic may be written in terms of its polar equation. See Example 1.
• The polar equations of conics can be graphed. See Example 2, Example 3, and Example 4.
• Conics can be defined in terms of a focus, a directrix, and eccentricity. See Example 5 and Example 6.
• We can use the identities r= x 2 + y 2 ,x=r cos θ, r= x 2 + y 2 ,x=r cos θ, and y=r sin θ y=r sin θ to convert the equation for a conic from polar to rectangular form. See Example 7.

## Section Exercises

### Verbal

#### Exercise 5

Explain how eccentricity determines which conic section is given.

##### Solution

If eccentricity is less than 1, it is an ellipse. If eccentricity is equal to 1, it is a parabola. If eccentricity is greater than 1, it is a hyperbola.

#### Exercise 6

If a conic section is written as a polar equation, what must be true of the denominator?

#### Exercise 7

If a conic section is written as a polar equation, and the denominator involves sin θ, sin θ, what conclusion can be drawn about the directrix?

##### Solution

The directrix will be parallel to the polar axis.

#### Exercise 8

If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?

#### Exercise 9

What do we know about the focus/foci of a conic section if it is written as a polar equation?

##### Solution

One of the foci will be located at the origin.

### Algebraic

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

#### Exercise 10

r= 6 12 cos θ r= 6 12 cos θ

#### Exercise 11

r= 3 44 sin θ r= 3 44 sin θ

##### Solution

Parabola with e=1 e=1 and directrix 3 4 3 4 units below the pole.

#### Exercise 12

r= 8 43 cos θ r= 8 43 cos θ

#### Exercise 13

r= 5 1+2 sin θ r= 5 1+2 sin θ

##### Solution

Hyperbola with e=2 e=2 and directrix 5 2 5 2 units above the pole.

#### Exercise 14

r= 16 4+3 cos θ r= 16 4+3 cos θ

#### Exercise 15

r= 3 10+10 cos θ r= 3 10+10 cos θ

##### Solution

Parabola with e=1 e=1 and directrix 3 10 3 10 units to the right of the pole.

#### Exercise 16

r= 2 1cos θ r= 2 1cos θ

#### Exercise 17

r= 4 7+2 cos θ r= 4 7+2 cos θ

##### Solution

Ellipse with e= 2 7 e= 2 7 and directrix 2 2 units to the right of the pole.

#### Exercise 18

r(1cos θ)=3 r(1cos θ)=3

#### Exercise 19

r(3+5sin θ)=11 r(3+5sin θ)=11

##### Solution

Hyperbola with e= 5 3 e= 5 3 and directrix 11 5 11 5 units above the pole.

#### Exercise 20

r(45sin θ)=1 r(45sin θ)=1

#### Exercise 21

r(7+8cos θ)=7 r(7+8cos θ)=7

##### Solution

Hyperbola with e= 8 7 e= 8 7 and directrix 7 8 7 8 units to the right of the pole.

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

#### Exercise 22

r= 4 1+3 sin θ r= 4 1+3 sin θ

#### Exercise 23

r= 2 53 sin θ r= 2 53 sin θ

##### Solution

25 x 2 +16 y 2 12y4=0 25 x 2 +16 y 2 12y4=0

#### Exercise 24

r= 8 32 cos θ r= 8 32 cos θ

#### Exercise 25

r= 3 2+5 cos θ r= 3 2+5 cos θ

##### Solution

21 x 2 4 y 2 30x+9=0 21 x 2 4 y 2 30x+9=0

#### Exercise 26

r= 4 2+2 sin θ r= 4 2+2 sin θ

#### Exercise 27

r= 3 88 cos θ r= 3 88 cos θ

##### Solution

64 y 2 =48x+9 64 y 2 =48x+9

#### Exercise 28

r= 2 6+7 cos θ r= 2 6+7 cos θ

#### Exercise 29

r= 5 511 sin θ r= 5 511 sin θ

##### Solution

96 y 2 25 x 2 +110y+25=0 96 y 2 25 x 2 +110y+25=0

#### Exercise 30

r(5+2 cos θ)=6 r(5+2 cos θ)=6

#### Exercise 31

r(2cos θ)=1 r(2cos θ)=1

##### Solution

3 x 2 +4 y 2 2x1=0 3 x 2 +4 y 2 2x1=0

#### Exercise 32

r(2.52.5 sin θ)=5 r(2.52.5 sin θ)=5

#### Exercise 33

r= 6sec θ 2+3 sec θ r= 6sec θ 2+3 sec θ

##### Solution

5 x 2 +9 y 2 24x36=0 5 x 2 +9 y 2 24x36=0

#### Exercise 34

r= 6csc θ 3+2 csc θ r= 6csc θ 3+2 csc θ

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.

#### Exercise 35

r= 5 2+cos θ r= 5 2+cos θ

#### Exercise 36

r= 2 3+3 sin θ r= 2 3+3 sin θ

#### Exercise 37

r= 10 54 sin θ r= 10 54 sin θ

#### Exercise 38

r= 3 1+2 cos θ r= 3 1+2 cos θ

#### Exercise 39

r= 8 45 cos θ r= 8 45 cos θ

#### Exercise 40

r= 3 44 cos θ r= 3 44 cos θ

#### Exercise 41

r= 2 1sin θ r= 2 1sin θ

#### Exercise 42

r= 6 3+2 sin θ r= 6 3+2 sin θ

#### Exercise 43

r(1+cos θ)=5 r(1+cos θ)=5

#### Exercise 44

r(34sin θ)=9 r(34sin θ)=9

#### Exercise 45

r(32sin θ)=6 r(32sin θ)=6

#### Exercise 46

r(64cos θ)=5 r(64cos θ)=5

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.

#### Exercise 47

Directrix: x=4;e= 1 5 x=4;e= 1 5

##### Solution

r= 4 5+cosθ r= 4 5+cosθ

#### Exercise 48

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

#### Exercise 49

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

##### Solution

r= 4 1+2sinθ r= 4 1+2sinθ

#### Exercise 50

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

#### Exercise 51

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

##### Solution

r= 1 1+cosθ r= 1 1+cosθ

#### Exercise 52

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

#### Exercise 53

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

##### Solution

r= 7 828cosθ r= 7 828cosθ

#### Exercise 54

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

#### Exercise 55

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

##### Solution

r= 12 2+3sinθ r= 12 2+3sinθ

#### Exercise 56

Directrix: x=−2;e= 8 3 x=−2;e= 8 3

#### Exercise 57

Directrix: x=−5;e= 3 4 x=−5;e= 3 4

##### Solution

r= 15 43cosθ r= 15 43cosθ

#### Exercise 58

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

#### Exercise 59

Directrix: x=−3;e= 1 3 x=−3;e= 1 3

##### Solution

r= 3 33cosθ r= 3 33cosθ

### 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.

xy=2 xy=2

#### Exercise 61

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

##### Solution

r=± 2 1+sinθcosθ r=± 2 1+sinθcosθ

#### Exercise 62

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

#### Exercise 63

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

##### Solution

r=± 2 4cosθ+3sinθ r=± 2 4cosθ+3sinθ

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

## Chapter Review Exercises

### The Ellipse

For the following exercises, write the equation of the ellipse in standard form. Then identify the center, vertices, and foci.

#### Exercise 65

x 2 25 + y 2 64 =1 x 2 25 + y 2 64 =1

##### Solution

x 2 5 2 + y 2 8 2 =1; x 2 5 2 + y 2 8 2 =1; center: ( 0,0 ); ( 0,0 ); vertices: ( 5,0 ),( −5,0 ),( 0,8 ),( 0,8 ); ( 5,0 ),( −5,0 ),( 0,8 ),( 0,8 ); foci: ( 0, 39 ),( 0, 39 ) ( 0, 39 ),( 0, 39 )

#### Exercise 66

(x2) 2 100 + ( y+3 ) 2 36 =1 (x2) 2 100 + ( y+3 ) 2 36 =1

#### Exercise 67

9 x 2 + y 2 +54x4y+76=0 9 x 2 + y 2 +54x4y+76=0

##### Solution

(x+3) 2 1 2 + (y2) 2 3 2 =1(3,2);(2,2),(4,2),(3,5),(3,1);( 3,2+2 2 ),( 3,22 2 ) (x+3) 2 1 2 + (y2) 2 3 2 =1(3,2);(2,2),(4,2),(3,5),(3,1);( 3,2+2 2 ),( 3,22 2 )

#### Exercise 68

9 x 2 +36 y 2 36x+72y+36=0 9 x 2 +36 y 2 36x+72y+36=0

For the following exercises, graph the ellipse, noting center, vertices, and foci.

#### Exercise 69

x 2 36 + y 2 9 =1 x 2 36 + y 2 9 =1

##### Solution

center: ( 0,0 ); ( 0,0 ); vertices: ( 6,0 ),( −6,0 ),( 0,3 ),( 0,−3 ); ( 6,0 ),( −6,0 ),( 0,3 ),( 0,−3 ); foci: ( 3 3 ,0 ),( 3 3 ,0 ) ( 3 3 ,0 ),( 3 3 ,0 )

#### Exercise 70

(x4) 2 25 + ( y+3 ) 2 49 =1 (x4) 2 25 + ( y+3 ) 2 49 =1

#### Exercise 71

4 x 2 + y 2 +16x+4y44=0 4 x 2 + y 2 +16x+4y44=0

##### Solution

center: ( −2,−2 ); ( −2,−2 ); vertices: ( 2,−2 ),( −6,−2 ),( −2,6 ),( −2,−10 ); ( 2,−2 ),( −6,−2 ),( −2,6 ),( −2,−10 ); foci: ( −2,−2+4 3 , ),( −2,−2−4 3 ) ( −2,−2+4 3 , ),( −2,−2−4 3 )

#### Exercise 72

2 x 2 +3 y 2 20x+12y+38=0 2 x 2 +3 y 2 20x+12y+38=0

For the following exercises, use the given information to find the equation for the ellipse.

#### Exercise 73

Center at ( 0,0 ), ( 0,0 ), focus at ( 3,0 ), ( 3,0 ), vertex at ( −5,0 ) ( −5,0 )

##### Solution

x 2 25 + y 2 16 =1 x 2 25 + y 2 16 =1

#### Exercise 74

Center at ( 2,−2 ), ( 2,−2 ), vertex at ( 7,−2 ), ( 7,−2 ), focus at ( 4,−2 ) ( 4,−2 )

#### Exercise 75

A whispering gallery is to be constructed such that the foci are located 35 feet from the center. If the length of the gallery is to be 100 feet, what should the height of the ceiling be?

##### Solution

Approximately 35.71 feet

### The Hyperbola

For the following exercises, write the equation of the hyperbola in standard form. Then give the center, vertices, and foci.

#### Exercise 76

x 2 81 y 2 9 =1 x 2 81 y 2 9 =1

#### Exercise 77

( y+1 ) 2 16 ( x4 ) 2 36 =1 ( y+1 ) 2 16 ( x4 ) 2 36 =1

##### Solution

( y+1 ) 2 4 2 ( x4 ) 2 6 2 =1; ( y+1 ) 2 4 2 ( x4 ) 2 6 2 =1; center: ( 4,−1 ); ( 4,−1 ); vertices: ( 4,3 ),( 4,−5 ); ( 4,3 ),( 4,−5 ); foci: ( 4,−1+2 13 ),( 4,−12 13 ) ( 4,−1+2 13 ),( 4,−12 13 )

#### Exercise 78

9 y 2 4 x 2 +54y16x+29=0 9 y 2 4 x 2 +54y16x+29=0

#### Exercise 79

3 x 2 y 2 12x6y9=0 3 x 2 y 2 12x6y9=0

##### Solution

( x2 ) 2 2 2 ( y+3 ) 2 ( 2 3 ) 2 =1; ( x2 ) 2 2 2 ( y+3 ) 2 ( 2 3 ) 2 =1; center: ( 2,−3 ); ( 2,−3 ); vertices: ( 4,−3 ),( 0,−3 ); ( 4,−3 ),( 0,−3 ); foci: ( 6,−3 ),( −2,−3 ) ( 6,−3 ),( −2,−3 )

For the following exercises, graph the hyperbola, labeling vertices and foci.

#### Exercise 80

x 2 9 y 2 16 =1 x 2 9 y 2 16 =1

#### Exercise 81

( y1 ) 2 49 ( x+1 ) 2 4 =1 ( y1 ) 2 49 ( x+1 ) 2 4 =1

#### Exercise 82

x 2 4 y 2 +6x+32y91=0 x 2 4 y 2 +6x+32y91=0

#### Exercise 83

2 y 2 x 2 12y6=0 2 y 2 x 2 12y6=0

##### Solution

For the following exercises, find the equation of the hyperbola.

#### Exercise 84

Center at ( 0,0 ), ( 0,0 ), vertex at ( 0,4 ), ( 0,4 ), focus at ( 0,−6 ) ( 0,−6 )

#### Exercise 85

Foci at ( 3,7 ) ( 3,7 ) and ( 7,7 ), ( 7,7 ), vertex at ( 6,7 ) ( 6,7 )

##### Solution

( x5 ) 2 1 ( y7 ) 2 3 =1 ( x5 ) 2 1 ( y7 ) 2 3 =1

### The Parabola

For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix.

#### Exercise 86

y 2 =12x y 2 =12x

#### Exercise 87

( x+2 ) 2 = 1 2 ( y1 ) ( x+2 ) 2 = 1 2 ( y1 )

##### Solution

( x+2 ) 2 = 1 2 ( y1 ); ( x+2 ) 2 = 1 2 ( y1 ); vertex: ( −2,1 ); ( −2,1 ); focus: ( −2, 9 8 ); ( −2, 9 8 ); directrix: y= 7 8 y= 7 8

#### Exercise 88

y 2 6y6x3=0 y 2 6y6x3=0

#### Exercise 89

x 2 +10xy+23=0 x 2 +10xy+23=0

##### Solution

( x+5 ) 2 =( y+2 ); ( x+5 ) 2 =( y+2 ); vertex: ( 5,2 ); ( 5,2 ); focus: ( 5, 7 4 ); ( 5, 7 4 ); directrix: y= 9 4 y= 9 4

For the following exercises, graph the parabola, labeling vertex, focus, and directrix.

#### Exercise 90

x 2 +4y=0 x 2 +4y=0

#### Exercise 91

( y1 ) 2 = 1 2 ( x+3 ) ( y1 ) 2 = 1 2 ( x+3 )

#### Exercise 92

x 2 8x10y+46=0 x 2 8x10y+46=0

#### Exercise 93

2 y 2 +12y+6x+15=0 2 y 2 +12y+6x+15=0

##### Solution

For the following exercises, write the equation of the parabola using the given information.

#### Exercise 94

Focus at ( −4,0 ); ( −4,0 ); directrix is x=4 x=4

#### Exercise 95

Focus at ( 2, 9 8 ); ( 2, 9 8 ); directrix is y= 7 8 y= 7 8

##### Solution

( x2 ) 2 =( 1 2 )( y1 ) ( x2 ) 2 =( 1 2 )( y1 )

#### Exercise 96

A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep.

### Rotation of Axes

For the following exercises, determine which of the conic sections is represented.

#### Exercise 97

16 x 2 +24xy+9 y 2 +24x60y60=0 16 x 2 +24xy+9 y 2 +24x60y60=0

##### Solution

B 2 4AC=0, B 2 4AC=0, parabola

#### Exercise 98

4 x 2 +14xy+5 y 2 +18x6y+30=0 4 x 2 +14xy+5 y 2 +18x6y+30=0

#### Exercise 99

4 x 2 +xy+2 y 2 +8x26y+9=0 4 x 2 +xy+2 y 2 +8x26y+9=0

##### Solution

B 2 4AC=31<0, B 2 4AC=31<0, ellipse

For the following exercises, determine the angle θ θ that will eliminate the xy xy term, and write the corresponding equation without the xy xy term.

#### Exercise 100

x 2 +4xy2 y 2 6=0 x 2 +4xy2 y 2 6=0

#### Exercise 101

x 2 xy+ y 2 6=0 x 2 xy+ y 2 6=0

##### Solution

θ= 45 , x 2 +3 y 2 12=0 θ= 45 , x 2 +3 y 2 12=0

For the following exercises, graph the equation relative to the x y x y system in which the equation has no x y x y term.

#### Exercise 102

9 x 2 24xy+16 y 2 80x60y+100=0 9 x 2 24xy+16 y 2 80x60y+100=0

#### Exercise 103

x 2 xy+ y 2 2=0 x 2 xy+ y 2 2=0

θ= 45 θ= 45

#### Exercise 104

6 x 2 +24xy y 2 12x+26y+11=0 6 x 2 +24xy y 2 12x+26y+11=0

### Conic Sections in Polar Coordinates

For the following exercises, given the polar equation of the conic with focus at the origin, identify the eccentricity and directrix.

#### Exercise 105

r= 10 15 cos θ r= 10 15 cos θ

##### Solution

Hyperbola with e=5 e=5 and directrix 2 2 units to the left of the pole.

#### Exercise 106

r= 6 3+2 cos θ r= 6 3+2 cos θ

#### Exercise 107

r= 1 4+3 sin θ r= 1 4+3 sin θ

##### Solution

Ellipse with e= 3 4 e= 3 4 and directrix 1 3 1 3 unit above the pole.

#### Exercise 108

r= 3 55 sin θ r= 3 55 sin θ

For the following exercises, graph the conic given in polar form. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci.

#### Exercise 109

r= 3 1sin θ r= 3 1sin θ

#### Exercise 110

r= 8 4+3 sin θ r= 8 4+3 sin θ

#### Exercise 111

r= 10 4+5 cos θ r= 10 4+5 cos θ

#### Exercise 112

r= 9 36 cos θ r= 9 36 cos θ

For the following exercises, given information about the graph of a conic with focus at the origin, find the equation in polar form.

#### Exercise 113

Directrix is x=3 x=3 and eccentricity e=1 e=1

##### Solution

r= 3 1+cos  θ r= 3 1+cos  θ

#### Exercise 114

Directrix is y=−2 y=−2 and eccentricity e=4 e=4

## Practice Test

For the following exercises, write the equation in standard form and state the center, vertices, and foci.

### Exercise 115

x 2 9 + y 2 4 =1 x 2 9 + y 2 4 =1

#### Solution

x 2 3 2 + y 2 2 2 =1; x 2 3 2 + y 2 2 2 =1; center: ( 0,0 ); ( 0,0 ); vertices: ( 3,0 ),( –3,0 ),( 0,2 ),( 0,−2 ); ( 3,0 ),( –3,0 ),( 0,2 ),( 0,−2 ); foci: ( 5 ,0 ),( 5 ,0 ) ( 5 ,0 ),( 5 ,0 )

### Exercise 116

9 y 2 +16 x 2 36y+32x92=0 9 y 2 +16 x 2 36y+32x92=0

For the following exercises, sketch the graph, identifying the center, vertices, and foci.

### Exercise 117

( x3 ) 2 64 + ( y2 ) 2 36 =1 ( x3 ) 2 64 + ( y2 ) 2 36 =1

#### Solution

center: ( 3,2 ); ( 3,2 ); vertices: ( 11,2 ),( −5,2 ),( 3,8 ),( 3,−4 ); ( 11,2 ),( −5,2 ),( 3,8 ),( 3,−4 ); foci: ( 3+2 7 ,2 ),( 32 7 ,2 ) ( 3+2 7 ,2 ),( 32 7 ,2 )

### Exercise 118

2 x 2 + y 2 +8x6y7=0 2 x 2 + y 2 +8x6y7=0

### Exercise 119

Write the standard form equation of an ellipse with a center at ( 1,2 ), ( 1,2 ), vertex at ( 7,2 ), ( 7,2 ), and focus at ( 4,2 ). ( 4,2 ).

#### Solution

( x1 ) 2 36 + ( y2 ) 2 27 =1 ( x1 ) 2 36 + ( y2 ) 2 27 =1

### Exercise 120

A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be?

For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes.

### Exercise 121

x 2 49 y 2 81 =1 x 2 49 y 2 81 =1

#### Solution

x 2 7 2 y 2 9 2 =1; x 2 7 2 y 2 9 2 =1; center: ( 0,0 ); ( 0,0 ); vertices ( 7,0 ),( −7,0 ); ( 7,0 ),( −7,0 ); foci: ( 130 ,0 ),( 130 ,0 ); ( 130 ,0 ),( 130 ,0 ); asymptotes: y=± 9 7 x y=± 9 7 x

### Exercise 122

16 y 2 9 x 2 +128y+112=0 16 y 2 9 x 2 +128y+112=0

For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes.

### Exercise 123

( x3 ) 2 25 ( y+3 ) 2 1 =1 ( x3 ) 2 25 ( y+3 ) 2 1 =1

#### Solution

center: ( 3,−3 ); ( 3,−3 ); vertices: ( 8,−3 ),( −2,−3 ); ( 8,−3 ),( −2,−3 ); foci: ( 3+ 26 ,−3 ),( 3 26 ,−3 ); ( 3+ 26 ,−3 ),( 3 26 ,−3 ); asymptotes: y=± 1 5 (x3)3 y=± 1 5 (x3)3

### Exercise 124

y 2 x 2 +4y4x18=0 y 2 x 2 +4y4x18=0

### Exercise 125

Write the standard form equation of a hyperbola with foci at ( 1,0 ) ( 1,0 ) and ( 1,6 ), ( 1,6 ), and a vertex at ( 1,2 ). ( 1,2 ).

#### Solution

( y3 ) 2 1 ( x1 ) 2 8 =1 ( y3 ) 2 1 ( x1 ) 2 8 =1

For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix.

### Exercise 126

y 2 +10x=0 y 2 +10x=0

### Exercise 127

3 x 2 12xy+11=0 3 x 2 12xy+11=0

#### Solution

( x2 ) 2 = 1 3 ( y+1 ); ( x2 ) 2 = 1 3 ( y+1 ); vertex: ( 2,−1 ); ( 2,−1 ); focus: ( 2, 11 12 ); ( 2, 11 12 ); directrix: y= 13 12 y= 13 12

For the following exercises, graph the parabola, labeling the vertex, focus, and directrix.

### Exercise 128

( x1 ) 2 =−4( y+3 ) ( x1 ) 2 =−4( y+3 )

### Exercise 129

y 2 +8x8y+40=0 y 2 +8x8y+40=0

### Exercise 130

Write the equation of a parabola with a focus at ( 2,3 ) ( 2,3 ) and directrix y=−1. y=−1.

### Exercise 131

A searchlight is shaped like a paraboloid of revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be?

#### Solution

Approximately 8.49 8.49 feet

For the following exercises, determine which conic section is represented by the given equation, and then determine the angle θ θ that will eliminate the xy xy term.

### Exercise 132

3 x 2 2xy+3 y 2 =4 3 x 2 2xy+3 y 2 =4

### Exercise 133

x 2 +4xy+4 y 2 +6x8y=0 x 2 +4xy+4 y 2 +6x8y=0

#### Solution

parabola; θ 63.4 θ 63.4

For the following exercises, rewrite in the x y x y system without the x y x y term, and graph the rotated graph.

### Exercise 134

11 x 2 +10 3 xy+ y 2 =4 11 x 2 +10 3 xy+ y 2 =4

### Exercise 135

16 x 2 +24xy+9 y 2 125x=0 16 x 2 +24xy+9 y 2 125x=0

#### Solution

x 2 4 x +3 y =0 x 2 4 x +3 y =0

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

### Exercise 136

r= 3 2sin θ r= 3 2sin θ

### Exercise 137

r= 5 4+6 cos θ r= 5 4+6 cos θ

#### Solution

Hyperbola with e= 3 2 , e= 3 2 , and directrix 5 6 5 6 units to the right of the pole.

For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci.

### Exercise 138

r= 12 48 sin θ r= 12 48 sin θ

### Exercise 139

r= 2 4+4 sin θ r= 2 4+4 sin θ

### Exercise 140

Find a polar equation of the conic with focus at the origin, eccentricity of e=2, e=2, and directrix: x=3. x=3.

## Glossary

eccentricity:
the ratio of the distances from a point P P on the graph to the focus F F and to the directrix D D represented by e= PF PD , e= PF PD , where e e is a positive real number
polar equation:
an equation of a curve in polar coordinates r r and θ θ

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