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  1. Preface
  2. 1 Prerequisites
    1. Introduction to Prerequisites
    2. 1.1 Real Numbers: Algebra Essentials
    3. 1.2 Exponents and Scientific Notation
    4. 1.3 Radicals and Rational Exponents
    5. 1.4 Polynomials
    6. 1.5 Factoring Polynomials
    7. 1.6 Rational Expressions
    8. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    9. Exercises
      1. Review Exercises
      2. Practice Test
  3. 2 Equations and Inequalities
    1. Introduction to Equations and Inequalities
    2. 2.1 The Rectangular Coordinate Systems and Graphs
    3. 2.2 Linear Equations in One Variable
    4. 2.3 Models and Applications
    5. 2.4 Complex Numbers
    6. 2.5 Quadratic Equations
    7. 2.6 Other Types of Equations
    8. 2.7 Linear Inequalities and Absolute Value Inequalities
    9. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  4. 3 Functions
    1. Introduction to Functions
    2. 3.1 Functions and Function Notation
    3. 3.2 Domain and Range
    4. 3.3 Rates of Change and Behavior of Graphs
    5. 3.4 Composition of Functions
    6. 3.5 Transformation of Functions
    7. 3.6 Absolute Value Functions
    8. 3.7 Inverse Functions
    9. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  5. 4 Linear Functions
    1. Introduction to Linear Functions
    2. 4.1 Linear Functions
    3. 4.2 Modeling with Linear Functions
    4. 4.3 Fitting Linear Models to Data
    5. Chapter Review
      1. Key Terms
      2. Key Concepts
    6. Exercises
      1. Review Exercises
      2. Practice Test
  6. 5 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 5.1 Quadratic Functions
    3. 5.2 Power Functions and Polynomial Functions
    4. 5.3 Graphs of Polynomial Functions
    5. 5.4 Dividing Polynomials
    6. 5.5 Zeros of Polynomial Functions
    7. 5.6 Rational Functions
    8. 5.7 Inverses and Radical Functions
    9. 5.8 Modeling Using Variation
    10. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  7. 6 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 6.1 Exponential Functions
    3. 6.2 Graphs of Exponential Functions
    4. 6.3 Logarithmic Functions
    5. 6.4 Graphs of Logarithmic Functions
    6. 6.5 Logarithmic Properties
    7. 6.6 Exponential and Logarithmic Equations
    8. 6.7 Exponential and Logarithmic Models
    9. 6.8 Fitting Exponential Models to Data
    10. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  8. 7 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 7.1 Systems of Linear Equations: Two Variables
    3. 7.2 Systems of Linear Equations: Three Variables
    4. 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 7.4 Partial Fractions
    6. 7.5 Matrices and Matrix Operations
    7. 7.6 Solving Systems with Gaussian Elimination
    8. 7.7 Solving Systems with Inverses
    9. 7.8 Solving Systems with Cramer's Rule
    10. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  9. 8 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 8.1 The Ellipse
    3. 8.2 The Hyperbola
    4. 8.3 The Parabola
    5. 8.4 Rotation of Axes
    6. 8.5 Conic Sections in Polar Coordinates
    7. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    8. Exercises
      1. Review Exercises
      2. Practice Test
  10. 9 Sequences, Probability, and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 9.1 Sequences and Their Notations
    3. 9.2 Arithmetic Sequences
    4. 9.3 Geometric Sequences
    5. 9.4 Series and Their Notations
    6. 9.5 Counting Principles
    7. 9.6 Binomial Theorem
    8. 9.7 Probability
    9. Chapter Review
      1. Key Terms
      2. Key Equations
      3. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
  12. Index

Learning Objectives

In this section, you will:

  • Locate a hyperbola’s vertices and foci.
  • Write equations of hyperbolas in standard form.
  • Graph hyperbolas centered at the origin.
  • Graph hyperbolas not centered at the origin.
  • Solve applied problems involving hyperbolas.

Corequisite Skills

Learning Objectives

  • Use the Distance Formula. (IA 11.1.1)
  • Graph a hyperbola with center at (0,0). (IA 11.4.1)

Objective 1: Use the Distance Formula. (IA 11.1.1)

Distance Formula

Distance Formula: The distance d between two points x1, y1x1, y1 and x2, y2x2, y2 is d=x2-x12+y2-y12 d=x2-x12+y2-y12 .

distance formula graph

Example 1

Use the Distance Formula

Use the distance formula to find the distance between the points (−5, −3) and (7,2).

Practice Makes Perfect

Use the Distance Formula.

1.

Use the Distance Formula to find the distance between the points (−2,−5) and (−14,−10).

2.

Use the Distance Formula to find the distance between the points (10, −4) and (−1,5). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Objective 2: Graph a hyperbola with center at (0,0). (IA 11.4.1)

A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The line through the foci is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

hyperbola definitional label graph
Equation x2a2-y2b2=1x2a2-y2b2=1 y2a2-x2b2=1y2a2-x2b2=1
Orientation Transverse axis is horizontal.
Opens right and left.
Transverse axis is vertical.
Opens up and down.
Vertices (-a, 0), (a, 0) (0, -a), (0, a)
x-intercepts (-a, 0), (a, 0) none)
y-intercepts none (0, -a), (0, a)
Rectangle Use (±a,0), (0,±b) Use (0,±a), (±b,0)
Asymptotes ( y=±baxy=±bax y=±abxy=±abx
Table 2 The standard form of the equation of a hyperbola with center (0,0)
hyperbola graph

Notice that, unlike the equation of an ellipse, the denominator of x2x2 is not always a2a2 and the denominator of y2y2 is not always b2b2 .

Notice that when the x2x2 term is positive, the transverse axis is on the x-axis. When the y2y2 term is positive, the transverse axis is on the y-axis.

How To

How To Graph a hyperbola with center at (0, 0).

  1. Step 1. Write the equation in standard form.
  2. Step 2. Determine if the transverse axis is horizontal or vertical.
  3. Step 3. Find the vertices.
  4. Step 4. Sketch the rectangle, entered at the origin, intersecting one axis at ±a and the other at ±b.
  5. Step 5. Sketch the asymptotes – the lines through the diagonals of the rectangle.
  6. Step 6. Draw the two branches of the hyperbola. Start at the vertex and use the asymptotes as a guide.

Example 2

Graph a hyperbola with center at (0,0).

Graph x 2 25 y 2 4 = 1 . x 2 25 y 2 4 = 1 .

Example 3

Graph 4 y 2 16 x 2 = 64 . 4 y 2 16 x 2 = 64 .

Practice Makes Perfect

Graph a hyperbola with center at (0,0).

3.

Graph x29-y216=1x29-y216=1 .

4.

Graph 25y2-9x2=22525y2-9x2=225 .

What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See Figure 1.

Figure 1 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom.

Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom.

Locating the Vertices and Foci of a Hyperbola

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See Figure 2.

Figure 2 A hyperbola

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points ( x,y ) ( x,y ) in a plane such that the difference of the distances between ( x,y ) ( x,y ) and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See Figure 3.

Figure 3 Key features of the hyperbola

In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Deriving the Equation of a Hyperbola Centered at the Origin

Let ( c,0 ) ( c,0 ) and ( c,0 ) ( c,0 ) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points ( x,y ) ( x,y ) such that the difference of the distances from ( x,y ) ( x,y ) to the foci is constant. See Figure 4.

A horizontal hyperbola in the x y coordinate system centered at (0, 0) with Vertices at (negative a, 0) and (a, 0) and Foci at (negative c, 0) and (c, 0), with lines of length d1 and d2 connecting a point on the right branch of the hyperbola to the foci.
Figure 4

If ( a,0 ) ( a,0 ) is a vertex of the hyperbola, the distance from ( c,0 ) ( c,0 ) to ( a,0 ) ( a,0 ) is a( c )=a+c. a( c )=a+c. The distance from ( c,0 ) ( c,0 ) to ( a,0 ) ( a,0 ) is ca. ca. The difference of the distances from the foci to the vertex is

( a+c )( ca )=2a ( a+c )( ca )=2a

If ( x,y ) ( x,y ) is a point on the hyperbola, we can define the following variables:

d 2 =the distance from ( c,0 )to ( x,y ) d 1 =the distance from ( c,0 )to ( x,y ) d 2 =the distance from ( c,0 )to ( x,y ) d 1 =the distance from ( c,0 )to ( x,y )

By definition of a hyperbola, d 2 d 1 d 2 d 1 is constant for any point ( x,y ) ( x,y ) on the hyperbola. We know that the difference of these distances is 2a 2a for the vertex (a,0). (a,0). It follows that d 2 d 1 =2a d 2 d 1 =2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

                                      d 2 d 1 = (x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 =2a Distance Formula (x+c) 2 + y 2 (xc) 2 + y 2 =2a Simplify expressions.                            (x+c) 2 + y 2 =2a+ (xc) 2 + y 2 Move radical to opposite side.                              (x+c) 2 + y 2 = ( 2a+ (xc) 2 + y 2 ) 2 Square both sides.                     x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (xc) 2 + y 2 + (xc) 2 + y 2 Expand the squares.                     x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (xc) 2 + y 2 + x 2 2cx+ c 2 + y 2 Expand remaining square.                                             2cx=4 a 2 +4a (xc) 2 + y 2 2cx Combine like terms.                                  4cx4 a 2 =4a (xc) 2 + y 2 Isolate the radical.                                      cx a 2 =a (xc) 2 + y 2 Divide by 4.                                   ( cx a 2 ) 2 = a 2 ( (xc) 2 + y 2 ) 2 Square both sides.                     c 2 x 2 2 a 2 cx+ a 4 = a 2 ( x 2 2cx+ c 2 + y 2 ) Expand the squares.                    c 2 x 2 2 a 2 cx+ a 4 = a 2 x 2 2 a 2 cx+ a 2 c 2 + a 2 y 2 Distribute  a 2 .                                   a 4 + c 2 x 2 = a 2 x 2 + a 2 c 2 + a 2 y 2 Combine like terms.                  c 2 x 2 a 2 x 2 a 2 y 2 = a 2 c 2 a 4 Rearrange terms.                    x 2 ( c 2 a 2 ) a 2 y 2 = a 2 ( c 2 a 2 ) Factor common terms.                              x 2 b 2 a 2 y 2 = a 2 b 2 Set  b 2 = c 2 a 2 .                             x 2 b 2 a 2 b 2 a 2 y 2 a 2 b 2 = a 2 b 2 a 2 b 2 Divide both sides by  a 2 b 2                                     x 2 a 2 y 2 b 2 =1                                       d 2 d 1 = (x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 =2a Distance Formula (x+c) 2 + y 2 (xc) 2 + y 2 =2a Simplify expressions.                            (x+c) 2 + y 2 =2a+ (xc) 2 + y 2 Move radical to opposite side.                              (x+c) 2 + y 2 = ( 2a+ (xc) 2 + y 2 ) 2 Square both sides.                     x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (xc) 2 + y 2 + (xc) 2 + y 2 Expand the squares.                     x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (xc) 2 + y 2 + x 2 2cx+ c 2 + y 2 Expand remaining square.                                             2cx=4 a 2 +4a (xc) 2 + y 2 2cx Combine like terms.                                  4cx4 a 2 =4a (xc) 2 + y 2 Isolate the radical.                                      cx a 2 =a (xc) 2 + y 2 Divide by 4.                                   ( cx a 2 ) 2 = a 2 ( (xc) 2 + y 2 ) 2 Square both sides.                     c 2 x 2 2 a 2 cx+ a 4 = a 2 ( x 2 2cx+ c 2 + y 2 ) Expand the squares.                    c 2 x 2 2 a 2 cx+ a 4 = a 2 x 2 2 a 2 cx+ a 2 c 2 + a 2 y 2 Distribute  a 2 .                                   a 4 + c 2 x 2 = a 2 x 2 + a 2 c 2 + a 2 y 2 Combine like terms.                  c 2 x 2 a 2 x 2 a 2 y 2 = a 2 c 2 a 4 Rearrange terms.                    x 2 ( c 2 a 2 ) a 2 y 2 = a 2 ( c 2 a 2 ) Factor common terms.                              x 2 b 2 a 2 y 2 = a 2 b 2 Set  b 2 = c 2 a 2 .                             x 2 b 2 a 2 b 2 a 2 y 2 a 2 b 2 = a 2 b 2 a 2 b 2 Divide both sides by  a 2 b 2                                     x 2 a 2 y 2 b 2 =1

This equation defines a hyperbola centered at the origin with vertices ( ±a,0 ) ( ±a,0 ) and co-vertices ( 0±b ). ( 0±b ).

Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center ( 0,0 ) ( 0,0 ) and transverse axis on the x-axis is

x 2 a 2 y 2 b 2 =1 x 2 a 2 y 2 b 2 =1

where

  • the length of the transverse axis is 2a 2a
  • the coordinates of the vertices are ( ±a,0 ) ( ±a,0 )
  • the length of the conjugate axis is 2b 2b
  • the coordinates of the co-vertices are ( 0,±b ) ( 0,±b )
  • the distance between the foci is 2c, 2c, where c 2 = a 2 + b 2 c 2 = a 2 + b 2
  • the coordinates of the foci are ( ±c,0 ) ( ±c,0 )
  • the equations of the asymptotes are y=± b a x y=± b a x

See Figure 5a.

The standard form of the equation of a hyperbola with center ( 0,0 ) ( 0,0 ) and transverse axis on the y-axis is

y 2 a 2 x 2 b 2 =1 y 2 a 2 x 2 b 2 =1

where

  • the length of the transverse axis is 2a 2a
  • the coordinates of the vertices are ( 0,±a ) ( 0,±a )
  • the length of the conjugate axis is 2b 2b
  • the coordinates of the co-vertices are ( ±b,0 ) ( ±b,0 )
  • the distance between the foci is 2c, 2c, where c 2 = a 2 + b 2 c 2 = a 2 + b 2
  • the coordinates of the foci are ( 0,±c ) ( 0,±c )
  • the equations of the asymptotes are y=± a b x y=± a b x

See Figure 5b.

Note that the vertices, co-vertices, and foci are related by the equation c 2 = a 2 + b 2 . c 2 = a 2 + b 2 . When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

Figure 5 (a) Horizontal hyperbola with center ( 0,0 ) ( 0,0 ) (b) Vertical hyperbola with center ( 0,0 ) ( 0,0 )

How To

Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a 2 a 2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
    1. If the equation has the form x 2 a 2 y 2 b 2 =1, x 2 a 2 y 2 b 2 =1, then the transverse axis lies on the x-axis. The vertices are located at (±a,0), (±a,0), and the foci are located at ( ±c,0 ). ( ±c,0 ).
    2. If the equation has the form y 2 a 2 x 2 b 2 =1, y 2 a 2 x 2 b 2 =1, then the transverse axis lies on the y-axis. The vertices are located at (0,±a), (0,±a), and the foci are located at ( 0,±c ). ( 0,±c ).
  2. Solve for a a using the equation a= a 2 . a= a 2 .
  3. Solve for c c using the equation c= a 2 + b 2 . c= a 2 + b 2 .

Example 1

Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation y 2 49 x 2 32 =1. y 2 49 x 2 32 =1.

Try It #1

Identify the vertices and foci of the hyperbola with equation x 2 9 y 2 25 =1. x 2 9 y 2 25 =1.

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at ( 0,0 ), ( 0,0 ), we see that the vertices, co-vertices, and foci are related by the equation c 2 = a 2 + b 2 . c 2 = a 2 + b 2 . Note that this equation can also be rewritten as b 2 = c 2 a 2 . b 2 = c 2 a 2 . This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How To

Given the vertices and foci of a hyperbola centered at ( 0,0 ), ( 0,0 ), write its equation in standard form.

  1. Determine whether the transverse axis lies on the x- or y-axis.
    1. If the given coordinates of the vertices and foci have the form ( ±a,0 ) ( ±a,0 ) and ( ±c,0 ), ( ±c,0 ), respectively, then the transverse axis is the x-axis. Use the standard form x 2 a 2 y 2 b 2 =1. x 2 a 2 y 2 b 2 =1.
    2. If the given coordinates of the vertices and foci have the form ( 0,±a ) ( 0,±a ) and ( 0,±c ), ( 0,±c ), respectively, then the transverse axis is the y-axis. Use the standard form y 2 a 2 x 2 b 2 =1. y 2 a 2 x 2 b 2 =1.
  2. Find b 2 b 2 using the equation b 2 = c 2 a 2 . b 2 = c 2 a 2 .
  3. Substitute the values for a 2 a 2 and b 2 b 2 into the standard form of the equation determined in Step 1.

Example 2

Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices ( ±6,0 ) ( ±6,0 ) and foci ( ±2 10 ,0 )? ( ±2 10 ,0 )?

Try It #2

What is the standard form equation of the hyperbola that has vertices ( 0,±2 ) ( 0,±2 ) and foci ( 0,±2 5 )? ( 0,±2 5 )?

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated h h units horizontally and k k units vertically, the center of the hyperbola will be ( h,k ). ( h,k ). This translation results in the standard form of the equation we saw previously, with x x replaced by ( xh ) ( xh ) and y y replaced by ( yk ). ( yk ).

Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center ( h,k ) ( h,k ) and transverse axis parallel to the x-axis is

( xh ) 2 a 2 ( yk ) 2 b 2 =1 ( xh ) 2 a 2 ( yk ) 2 b 2 =1

where

  • the length of the transverse axis is 2a 2a
  • the coordinates of the vertices are ( h±a,k ) ( h±a,k )
  • the length of the conjugate axis is 2b 2b
  • the coordinates of the co-vertices are ( h,k±b ) ( h,k±b )
  • the distance between the foci is 2c, 2c, where c 2 = a 2 + b 2 c 2 = a 2 + b 2
  • the coordinates of the foci are ( h±c,k ) ( h±c,k )

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is 2a 2a and its width is 2b. 2b. The slopes of the diagonals are ± b a , ± b a , and each diagonal passes through the center ( h,k ). ( h,k ). Using the point-slope formula, it is simple to show that the equations of the asymptotes are y=± b a ( xh )+k. y=± b a ( xh )+k. See Figure 7a

The standard form of the equation of a hyperbola with center ( h,k ) ( h,k ) and transverse axis parallel to the y-axis is

( yk ) 2 a 2 ( xh ) 2 b 2 =1 ( yk ) 2 a 2 ( xh ) 2 b 2 =1

where

  • the length of the transverse axis is 2a 2a
  • the coordinates of the vertices are ( h,k±a ) ( h,k±a )
  • the length of the conjugate axis is 2b 2b
  • the coordinates of the co-vertices are ( h±b,k ) ( h±b,k )
  • the distance between the foci is 2c, 2c, where c 2 = a 2 + b 2 c 2 = a 2 + b 2
  • the coordinates of the foci are ( h,k±c ) ( h,k±c )

Using the reasoning above, the equations of the asymptotes are y=± a b ( xh )+k. y=± a b ( xh )+k. See Figure 7b.

This is a horizontal parabola opening to the right with Vertex (0, 0), Focus (6, 0), and Directrix x = negative 6. The Latus Rectum is shown, a vertical line passing through the Focus and terminating on the parabola at (6, 12) and (6, negative 12).
Figure 7 (a) Horizontal hyperbola with center ( h,k ) ( h,k ) (b) Vertical hyperbola with center ( h,k ) ( h,k )

Like hyperbolas centered at the origin, hyperbolas centered at a point ( h,k ) ( h,k ) have vertices, co-vertices, and foci that are related by the equation c 2 = a 2 + b 2 . c 2 = a 2 + b 2 . We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How To

Given the vertices and foci of a hyperbola centered at ( h,k ), ( h,k ), write its equation in standard form.

  1. Determine whether the transverse axis is parallel to the x- or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form ( xh ) 2 a 2 ( yk ) 2 b 2 =1. ( xh ) 2 a 2 ( yk ) 2 b 2 =1.
    2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form ( yk ) 2 a 2 ( xh ) 2 b 2 =1. ( yk ) 2 a 2 ( xh ) 2 b 2 =1.
  2. Identify the center of the hyperbola, ( h,k ), ( h,k ), using the midpoint formula and the given coordinates for the vertices.
  3. Find a 2 a 2 by solving for the length of the transverse axis, 2a 2a , which is the distance between the given vertices.
  4. Find c 2 c 2 using h h and k k found in Step 2 along with the given coordinates for the foci.
  5. Solve for b 2 b 2 using the equation b 2 = c 2 a 2 . b 2 = c 2 a 2 .
  6. Substitute the values for h, k, a 2 , h, k, a 2 , and b 2 b 2 into the standard form of the equation determined in Step 1.

Example 3

Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at (0,−2) (0,−2) and (6,−2) (6,−2) and foci at (−2,−2) (−2,−2) and (8,−2)? (8,−2)?

Try It #3

What is the standard form equation of the hyperbola that has vertices ( 1,−2 ) ( 1,−2 ) and ( 1,8 ) ( 1,8 ) and foci ( 1,−10 ) ( 1,−10 ) and ( 1,16 )? ( 1,16 )?

Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form x 2 a 2 y 2 b 2 =1 x 2 a 2 y 2 b 2 =1 for horizontal hyperbolas and the standard form y 2 a 2 x 2 b 2 =1 y 2 a 2 x 2 b 2 =1 for vertical hyperbolas.

How To

Given a standard form equation for a hyperbola centered at ( 0,0 ), ( 0,0 ), sketch the graph.

  1. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
    1. If the equation is in the form x 2 a 2 y 2 b 2 =1, x 2 a 2 y 2 b 2 =1, then
      • the transverse axis is on the x-axis
      • the coordinates of the vertices are ( ±a,0 ) ( ±a,0 )
      • the coordinates of the co-vertices are ( 0,±b ) ( 0,±b )
      • the coordinates of the foci are ( ±c,0 ) ( ±c,0 )
      • the equations of the asymptotes are y=± b a x y=± b a x
    2. If the equation is in the form y 2 a 2 x 2 b 2 =1, y 2 a 2 x 2 b 2 =1, then
      • the transverse axis is on the y-axis
      • the coordinates of the vertices are ( 0,±a ) ( 0,±a )
      • the coordinates of the co-vertices are ( ±b,0 ) ( ±b,0 )
      • the coordinates of the foci are ( 0,±c ) ( 0,±c )
      • the equations of the asymptotes are y=± a b x y=± a b x
  3. Solve for the coordinates of the foci using the equation c=± a 2 + b 2 . c=± a 2 + b 2 .
  4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

Example 4

Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form

Graph the hyperbola given by the equation y 2 64 x 2 36 =1. y 2 64 x 2 36 =1. Identify and label the vertices, co-vertices, foci, and asymptotes.

Try It #4

Graph the hyperbola given by the equation x 2 144 y 2 81 =1. x 2 144 y 2 81 =1. Identify and label the vertices, co-vertices, foci, and asymptotes.

Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point ( h,k ) ( h,k ) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms ( xh ) 2 a 2 ( yk ) 2 b 2 =1 ( xh ) 2 a 2 ( yk ) 2 b 2 =1 for horizontal hyperbolas, and ( yk ) 2 a 2 ( xh ) 2 b 2 =1 ( yk ) 2 a 2 ( xh ) 2 b 2 =1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

How To

Given a general form for a hyperbola centered at ( h, k ), ( h, k ), sketch the graph.

  1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
    1. If the equation is in the form ( xh ) 2 a 2 ( yk ) 2 b 2 =1, ( xh ) 2 a 2 ( yk ) 2 b 2 =1, then
      • the transverse axis is parallel to the x-axis
      • the center is ( h,k ) ( h,k )
      • the coordinates of the vertices are ( h±a,k ) ( h±a,k )
      • the coordinates of the co-vertices are ( h,k±b ) ( h,k±b )
      • the coordinates of the foci are ( h±c,k ) ( h±c,k )
      • the equations of the asymptotes are y=± b a ( xh )+k y=± b a ( xh )+k
    2. If the equation is in the form ( yk ) 2 a 2 ( xh ) 2 b 2 =1, ( yk ) 2 a 2 ( xh ) 2 b 2 =1, then
      • the transverse axis is parallel to the y-axis
      • the center is ( h,k ) ( h,k )
      • the coordinates of the vertices are ( h,k±a ) ( h,k±a )
      • the coordinates of the co-vertices are ( h±b,k ) ( h±b,k )
      • the coordinates of the foci are ( h,k±c ) ( h,k±c )
      • the equations of the asymptotes are y=± a b ( xh )+k y=± a b ( xh )+k
  3. Solve for the coordinates of the foci using the equation c=± a 2 + b 2 . c=± a 2 + b 2 .
  4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

Example 5

Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form

Graph the hyperbola given by the equation 9 x 2 4 y 2 36x40y388=0. 9 x 2 4 y 2 36x40y388=0. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Try It #5

Graph the hyperbola given by the standard form of an equation ( y+4 ) 2 100 ( x3 ) 2 64 =1. ( y+4 ) 2 100 ( x3 ) 2 64 =1. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See Figure 10. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

Figure 10 Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example 6

Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.

Figure 11 Project design for a natural draft cooling tower

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Try It #6

A design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Project design for a natural draft cooling tower. The overall height is 167.082 meters. The diameter at the top is 60 meters, and at their closest, 79.6 meters from the top, the sides are 60 meters apart.
Figure 12

8.2 Section Exercises

Verbal

1.

Define a hyperbola in terms of its foci.

2.

What can we conclude about a hyperbola if its asymptotes intersect at the origin?

3.

What must be true of the foci of a hyperbola?

4.

If the transverse axis of a hyperbola is vertical, what do we know about the graph?

5.

Where must the center of hyperbola be relative to its foci?

Algebraic

For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form.

6.

3 y 2 +2x=6 3 y 2 +2x=6

7.

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

8.

5 y 2 +4 x 2 =6x 5 y 2 +4 x 2 =6x

9.

25 x 2 16 y 2 =400 25 x 2 16 y 2 =400

10.

9 x 2 +18x+ y 2 +4y14=0 9 x 2 +18x+ y 2 +4y14=0

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes.

11.

x 2 25 y 2 36 =1 x 2 25 y 2 36 =1

12.

x 2 100 y 2 9 =1 x 2 100 y 2 9 =1

13.

y 2 4 x 2 81 =1 y 2 4 x 2 81 =1

14.

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

15.

( x1 ) 2 9 ( y2 ) 2 16 =1 ( x1 ) 2 9 ( y2 ) 2 16 =1

16.

( y6 ) 2 36 ( x+1 ) 2 16 =1 ( y6 ) 2 36 ( x+1 ) 2 16 =1

17.

( x2 ) 2 49 ( y+7 ) 2 49 =1 ( x2 ) 2 49 ( y+7 ) 2 49 =1

18.

4 x 2 8x9 y 2 72y+112=0 4 x 2 8x9 y 2 72y+112=0

19.

9 x 2 54x+9 y 2 54y+81=0 9 x 2 54x+9 y 2 54y+81=0

20.

4 x 2 24x36 y 2 360y+864=0 4 x 2 24x36 y 2 360y+864=0

21.

4 x 2 +24x+16 y 2 128y+156=0 4 x 2 +24x+16 y 2 128y+156=0

22.

4 x 2 +40x+25 y 2 100y+100=0 4 x 2 +40x+25 y 2 100y+100=0

23.

x 2 +2x100 y 2 1000y+2401=0 x 2 +2x100 y 2 1000y+2401=0

24.

9 x 2 +72x+16 y 2 +16y+4=0 9 x 2 +72x+16 y 2 +16y+4=0

25.

4 x 2 +24x25 y 2 +200y464=0 4 x 2 +24x25 y 2 +200y464=0

For the following exercises, find the equations of the asymptotes for each hyperbola.

26.

y 2 3 2 x 2 3 2 =1 y 2 3 2 x 2 3 2 =1

27.

( x3 ) 2 5 2 ( y+4 ) 2 2 2 =1 ( x3 ) 2 5 2 ( y+4 ) 2 2 2 =1

28.

( y3 ) 2 3 2 ( x+5 ) 2 6 2 =1 ( y3 ) 2 3 2 ( x+5 ) 2 6 2 =1

29.

9 x 2 18x16 y 2 +32y151=0 9 x 2 18x16 y 2 +32y151=0

30.

16 y 2 +96y4 x 2 +16x+112=0 16 y 2 +96y4 x 2 +16x+112=0

Graphical

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

31.

x 2 49 y 2 16 =1 x 2 49 y 2 16 =1

32.

x 2 64 y 2 4 =1 x 2 64 y 2 4 =1

33.

y 2 9 x 2 25 =1 y 2 9 x 2 25 =1

34.

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

35.

( y+5 ) 2 9 ( x4 ) 2 25 =1 ( y+5 ) 2 9 ( x4 ) 2 25 =1

36.

( x2 ) 2 8 ( y+3 ) 2 27 =1 ( x2 ) 2 8 ( y+3 ) 2 27 =1

37.

( y3 ) 2 9 ( x3 ) 2 9 =1 ( y3 ) 2 9 ( x3 ) 2 9 =1

38.

4 x 2 8x+16 y 2 32y52=0 4 x 2 8x+16 y 2 32y52=0

39.

x 2 8x25 y 2 100y109=0 x 2 8x25 y 2 100y109=0

40.

x 2 +8x+4 y 2 40y+88=0 x 2 +8x+4 y 2 40y+88=0

41.

64 x 2 +128x9 y 2 72y656=0 64 x 2 +128x9 y 2 72y656=0

42.

16 x 2 +64x4 y 2 8y4=0 16 x 2 +64x4 y 2 8y4=0

43.

100 x 2 +1000x+ y 2 10y2575=0 100 x 2 +1000x+ y 2 10y2575=0

44.

4 x 2 +16x4 y 2 +16y+16=0 4 x 2 +16x4 y 2 +16y+16=0

For the following exercises, given information about the graph of the hyperbola, find its equation.

45.

Vertices at ( 3,0 ) ( 3,0 ) and ( −3,0 ) ( −3,0 ) and one focus at ( 5,0 ). ( 5,0 ).

46.

Vertices at ( 0,6 ) ( 0,6 ) and ( 0,−6 ) ( 0,−6 ) and one focus at ( 0,−8 ). ( 0,−8 ).

47.

Vertices at ( 1,1 ) ( 1,1 ) and ( 11,1 ) ( 11,1 ) and one focus at ( 12,1 ). ( 12,1 ).

48.

Center: ( 0,0 ); ( 0,0 ); vertex: ( 0,−13 ); ( 0,−13 ); one focus: ( 0, 313 ). ( 0, 313 ).

49.

Center: ( 4,2 ); ( 4,2 ); vertex: ( 9,2 ); ( 9,2 ); one focus: ( 4+ 26 ,2 ). ( 4+ 26 ,2 ).

50.

Center: ( 3,5 ); ( 3,5 ); vertex: ( 3,11 ); ( 3,11 ); one focus: ( 3,5+2 10 ). ( 3,5+2 10 ).

For the following exercises, given the graph of the hyperbola, find its equation.

51.
A vertical hyperbola centered at (0, 0) with vertices at (0, negative 4) and (0, 4). The slant asymptotes are shown but not labeled.
52.
A horizontal hyperbola centered at (1, 1) with  vertices at (1 minus square root of 2, 1) and (1 + square root of 2, 1) and foci at (1 minus square root of 5, 1) and (1 + square root of 5, 1)
53.
A vertical hyperbola centered at (negative 1, 0) with vertices at (negative 1, negative 3) and (negative 1, 3) and foci at (negative 1, negative 3 square root of 2) and (negative 1, 3 square root of 2).
54.
A vertical hyperbola centered at (3, 1) with vertices at (3, 1 minus square root of 2) and (3, 1 + square root of 2) and foci at (3, 1 minus square root of 7) and (3, 1 + square root of 7).
55.
A horizontal hyperbola centered at (negative 3, negative 3) with vertices at (negative 8, negative 3) and (2, negative 3) and foci at (negative 3 minus 5 square root of 2, negative 3) and (negative 3 + 5 square root of 2, negative 3).

Extensions

For the following exercises, express the equation for the hyperbola as two functions, with y y as a function of x. x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes.

56.

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

57.

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

58.

( x2 ) 2 16 ( y+3 ) 2 25 =1 ( x2 ) 2 16 ( y+3 ) 2 25 =1

59.

4 x 2 16x+ y 2 2y19=0 4 x 2 16x+ y 2 2y19=0

60.

4 x 2 24x y 2 4y+16=0 4 x 2 24x y 2 4y+16=0

Real-World Applications

For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph.

61.

The hedge will follow the asymptotes y=xand y=x, y=xand y=x, and its closest distance to the center fountain is 5 yards.

62.

The hedge will follow the asymptotes y=2xand y=−2x, y=2xand y=−2x, and its closest distance to the center fountain is 6 yards.

63.

The hedge will follow the asymptotes y= 1 2 x y= 1 2 x and y= 1 2 x, y= 1 2 x, and its closest distance to the center fountain is 10 yards.

64.

The hedge will follow the asymptotes y= 2 3 x y= 2 3 x and y= 2 3 x, y= 2 3 x, and its closest distance to the center fountain is 12 yards.

65.

The hedge will follow the asymptotes y= 3 4 x y= 3 4 x and y= 3 4 x, y= 3 4 x, and its closest distance to the center fountain is 20 yards.

For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information.

66.

The object enters along a path approximated by the line y=x2 y=x2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y=x+2. y=x+2.

67.

The object enters along a path approximated by the line y=2x2 y=2x2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y=−2x+2. y=−2x+2.

68.

The object enters along a path approximated by the line y=0.5x+2 y=0.5x+2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y=−0.5x2. y=−0.5x2.

69.

The object enters along a path approximated by the line y= 1 3 x1 y= 1 3 x1 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y= 1 3 x+1. y= 1 3 x+1.

70.

The object enters along a path approximated by the line y=3x9 y=3x9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y=−3x+9. y=−3x+9.

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