The Hyperbola - OpenStax CNX · hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides

OpenStax-CNX module: m49439 1

The Hyperbola∗

OpenStax College

This work is produced by OpenStax-CNX and licensed under the

Creative Commons Attribution License 4.0†

Abstract

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.

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

∗Version 1.5: Mar 16, 2015 8:28 am -0500†http://creativecommons.org/licenses/by/4.0/

http://https://legacy.cnx.org/content/m49439/1.5/


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 breakingthe sound barrier long before the �rst supersonic �ight. The crack of a whip occurs because the tip isexceeding the speed of sound. The bullets shot from many �rearms also break the sound barrier, althoughthe bang of the gun usually supersedes the sound of the sonic boom.

1 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 aplane at an angle such that both halves of the cone are intersected. This intersection produces two separateunbounded curves that are mirror images of each other. See Figure 2.



Figure 2: A hyperbola

Like the ellipse, the hyperbola can also be de�ned as a set of points in the coordinate plane. A hyperbolais the set of all points (x, y) in a plane such that the di�erence of the distances between (x, y) and the fociis a positive constant.

Notice that the de�nition of a hyperbola is very similar to that of an ellipse. The distinction is that thehyperbola is de�ned in terms of the di�erence of two distances, whereas the ellipse is de�ned in terms of thesum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segmentthat passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line thatcontains 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 ahyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of thehyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful toolfor graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketchand 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 horizontallyin the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider twocases: those that are centered at the origin, and those that are centered at a point other than the origin.

1.1 Deriving the Equation of an Ellipse Centered at the Origin

Let (−c, 0) and (c, 0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of allpoints (x, y) such that the di�erence of the distances from (x, y) to the foci is constant. See Figure 4.



Figure 4

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

(a + c)− (c− a) = 2a (1)

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

d2 = the distance from (−c, 0) to (x, y)

d1 = the distance from (c, 0) to (x, y)(2)

By de�nition of a hyperbola, d2 − d1 is constant for any point (x, y) on the hyperbola. We know thatthe di�erence of these distances is 2a for the vertex (a, 0) . It follows that d2 − d1 = 2a for any point on thehyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distanceformula. The rest of the derivation is algebraic. Compare this derivation with the one from the previoussection for ellipses.



d2 − d1 =√

(x− (−c))2 + (y − 0)2 −√

(x− c)2 + (y − 0)2 = 2a Distance Formula√(x + c)2 + y2 −

√(x− c)2 + y2 = 2a Simplify expressions.√

(x + c)2 + y2 = 2a +√

(x− c)2 + y2 Move radical to opposite side.

(x + c)2 + y2 =

(2a +

√(x− c)2 + y2

)2

Square both sides.

x2 + 2cx + c2 + y2 = 4a2 + 4a√

(x− c)2 + y2 + (x− c)2 + y2 Expand the squares.

x2 + 2cx + c2 + y2 = 4a2 + 4a√

(x− c)2 + y2 + x2 − 2cx + c2 + y2 Expand remaining square.

2cx = 4a2 + 4a√

(x− c)2 + y2 − 2cx Combine like terms.

4cx− 4a2 = 4a√

(x− c)2 + y2 Isolate the radical.

cx− a2 = a√

(x− c)2 + y2 Divide by 4.

(cx− a2)2

= a2

[√(x− c)2 + y2

]2

Square both sides.

c2x2 − 2a2cx + a4 = a2 (x2 − 2cx + c2 + y2) Expand the squares.

c2x2 − 2a2cx + a4 = a2x2 − 2a2cx + a2c2 + a2y2 Distribute a2.

a4 + c2x2 = a2x2 + a2c2 + a2y2 Combine like terms.

c2x2 − a2x2 − a2y2 = a2c2 − a4 Rearrange terms.

x2 (c2 − a2)− a2y2 = a2 (c2 − a2) Factor common terms.

x2b2 − a2y2 = a2b2 Set b2 = c2 − a2.x2b2

a2b2− a2y2

a2b2= a2b2

a2b2Divide both sides by a2b2

x2

a2 − y2

b2= 1

(3)

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

A General Note: The standard form of the equation of a hyperbola with center (0, 0) andtransverse axis on the x-axis is

x2

a2− y2

b2= 1 (4)

where

• the length of the transverse axis is 2a• the coordinates of the vertices are (±a, 0)• the length of the conjugate axis is 2b• the coordinates of the co-vertices are (0,±b)• the distance between the foci is 2c, where c2 = a2 + b2

• the coordinates of the foci are (±c, 0)



• the equations of the asymptotes are y = ± bax

See Figure 5a.

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

y2

a2− x2

b2= 1 (5)

where

• the length of the transverse axis is 2a• the coordinates of the vertices are (0,±a)• the length of the conjugate axis is 2b• the coordinates of the co-vertices are (±b, 0)• the distance between the foci is 2c, where c2 = a2 + b2

• the coordinates of the foci are (0,±c)• the equations of the asymptotes are y = ±a

b x

See Figure 5b.

Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2.When we aregiven 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) (b) Vertical hyperbola with center (0, 0)

How To: Given the equation of a hyperbola in standard form, locate its vertices andfoci.



1.Determine whether the transverse axis lies on the x- or y-axis. Notice that a2 is always underthe variable with the positive coe�cient. So, if you set the other variable equal to zero, youcan easily �nd the intercepts. In the case where the hyperbola is centered at the origin, theintercepts coincide with the vertices.

a.If the equation has the form x2

a2 − y2

b2 = 1, then the transverse axis lies on the x-axis. Thevertices are located at (±a, 0) , and the foci are located at (±c, 0) .

b.If the equation has the form y2

a2 − x2

b2 = 1, then the transverse axis lies on the y-axis. Thevertices are located at (0,±a) , and the foci are located at (0,±c) .

2.Solve for a using the equation a =√

a2.3.Solve for c using the equation c =

√a2 + b2.

Example 1Locating a Hyperbola's Vertices and Foci

Identify the vertices and foci of the hyperbola with equation y2

49 −x2

32 = 1.

SolutionThe equation has the form y2

a2 − x2

b2 = 1, so the transverse axis lies on the y-axis. The hyperbola iscentered at the origin, so the vertices serve as the y-intercepts of the graph. To �nd the vertices,setx = 0, and solve for y.

1 = y2

49 −x2

32

1 = y2

49 −02

32

1 = y2

49

y2 = 49

y = ±√

49 = ±7

(6)

The foci are located at (0,±c) . Solving for c,

c =√

a2 + b2 =√

49 + 32 =√

81 = 9 (7)

Therefore, the vertices are located at (0,±7) , and the foci are located at (0, 9) .

Try It:

Exercise 2 (Solution on p. 32.)

Identify the vertices and foci of the hyperbola with equation x2

9 −y2

25 = 1.

2 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 keyfeatures: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverseand conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We beginby �nding standard equations for hyperbolas centered at the origin. Then we will turn our attention to�nding standard equations for hyperbolas centered at some point other than the origin.



2.1 Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at (0, 0) ,we see that the vertices, co-vertices, andfoci are related by the equation c2 = a2+b2.Note that this equation can also be rewritten as b2 = c2−a2.Thisrelationship 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) , write itsequation in standard form.

1.Determine whether the transverse axis lies on the x- or y-axis.

a.If the given coordinates of the vertices and foci have the form (±a, 0) and (±c, 0) , respectively,

then the transverse axis is the x-axis. Use the standard form x2

a2 − y2

b2 = 1.b.If the given coordinates of the vertices and foci have the form (0,±a) and (0,±c) , respectively,

then the transverse axis is the y-axis. Use the standard form y2

a2 − x2

b2 = 1.

2.Find b2 using the equation b2 = c2 − a2.3.Substitute the values for a2 and b2 into the standard form of the equation determined in Step1.

Example 2Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and VerticesWhat is the standard form equation of the hyperbola that has vertices (±6, 0) and foci

(±2√

10, 0)?

SolutionThe vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the

form x2

a2 − y2

b2 = 1.The vertices are (±6, 0) , so a = 6 and a2 = 36.The foci are

(±2√

10, 0), so c = 2

√10 and c2 = 40.

Solving for b2, we have

b2 = c2 − a2

b2 = 40− 36 Substitute for c2 and a2.

b2 = 4 Subtract.

(8)

Finally, we substitute a2 = 36 and b2 = 4 into the standard form of the equation, x2

a2 − y2

b2 = 1. The

equation of the hyperbola is x2

36 −y2

4 = 1, as shown in Figure 6.



Figure 6

Try It:


What is the standard form equation of the hyperbola that has vertices (0,±2) andfoci

(0,±2

√5)?

2.2 Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translatedh unitshorizontally and k units vertically, the center of the hyperbola will be (h, k) .This translation results in thestandard form of the equation we saw previously, withx replaced by (x− h) and y replaced by (y − k) .

A General Note: The standard form of the equation of a hyperbola with center (h, k) andtransverse axis parallel to the x-axis is

(x− h)2

a2− (y − k)2

b2= 1 (9)

where



• the length of the transverse axis is 2a• the coordinates of the vertices are (h± a, k)• the length of the conjugate axis is 2b• the coordinates of the co-vertices are (h, k ± b)• the distance between the foci is 2c, where c2 = a2 + b2

• the coordinates of the foci are (h± c, k)

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length ofthe rectangle is 2a and its width is 2b.The slopes of the diagonals are ± b

a ,and each diagonal passesthrough the center (h, k) .Using the point-slope formula, it is simple to show that the equationsof the asymptotes are y = ± b

a (x− h) + k. See Figure 7a

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

(y − k)2

a2− (x− h)2

b2= 1 (10)

where

�the length of the transverse axis is 2a�the coordinates of the vertices are (h, k ± a)�the length of the conjugate axis is 2b�the coordinates of the co-vertices are (h± b, k)�the distance between the foci is 2c, where c2 = a2 + b2

�the coordinates of the foci are (h, k ± c)

Using the reasoning above, the equations of the asymptotes are y = ±ab (x− h) + k. See Figure 7b.

Figure 7: (a) Horizontal hyperbola with center (h, k) (b) Vertical hyperbola with center (h, k)



Like hyperbolas centered at the origin, hyperbolas centered at a point (h, k) have vertices, co-vertices, andfoci that are related by the equation c2 = a2 + b2.We can use this relationship along with the midpoint anddistance formulas to �nd 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) ,write its equationin standard form.

1.Determine whether the transverse axis is parallel to the x- or y-axis.

a.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 (x−h)2

a2 − (y−k)2

b2 = 1.b.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 (y−k)2

a2 − (x−h)2

b2 = 1.

2.Identify the center of the hyperbola, (h, k) ,using the midpoint formula and the given coordi-nates for the vertices.

3.Find a2 by solving for the length of the transverse axis, 2a, which is the distance between thegiven vertices.

4.Find c2 usingh and k found in Step 2 along with the given coordinates for the foci.5.Solve for b2 using the equation b2 = c2 − a2.6.Substitute the values forh, k, a2, and b2 into the standard form of the equation determined inStep 1.

Example 3Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and VerticesWhat is the standard form equation of the hyperbola that has vertices at(0,−2)and(6,−2)andfoci at(−2,−2)and(8,−2)?

SolutionThe y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to thex-axis. Thus, the equation of the hyperbola will have the form

(x− h)2

a2− (y − k)2

b2= 1 (11)

First, we identify the center, (h, k) .The center is halfway between the vertices (0,−2) and (6,−2) .Applyingthe midpoint formula, we have

(h, k) =(

0 + 62

,−2 + (−2)

2

)= (3,−2) (12)

Next, we �nd a2.The length of the transverse axis, 2a,is bounded by the vertices. So, we can�nd a2 by �nding the distance between the x-coordinates of the vertices.

2a = |0− 6|2a = 6

a = 3

a2 = 9

(13)

Now we need to �nd c2.The coordinates of the foci are (h± c, k) . So (h− c, k) = (−2,−2) and(h + c, k) =(8,−2) .We can use the x-coordinate from either of these points to solve for c.Using the point(8,−2) , andsubstitutingh = 3,



h + c = 8

3 + c = 8

c = 5

c2 = 25

(14)

Next, solve for b2 using the equation b2 = c2 − a2 :

b2 = c2 − a2

= 25− 9

= 16

(15)

Finally, substitute the values found forh, k, a2,and b2 into the standard form of the equation.

(x− 3)2

9− (y + 2)2

16= 1 (16)

Try It:


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

3 Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret itsparts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengthsand positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use

the standard form x2

a2 − y2

b2 = 1 for horizontal hyperbolas and the standard form y2

a2 − x2

b2 = 1 for verticalhyperbolas.

How To: Given a standard form equation for a hyperbola centered at (0, 0) , sketchthe graph.

1.Determine which of the standard forms applies to the given equation.2.Use the standard form identi�ed 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.

a.If the equation is in the form x2

a2 − y2

b2 = 1, then

• the transverse axis is on the x-axis• the coordinates of the vertices are (±a, 0)• the coordinates of the co-vertices are (0,±b)• the coordinates of the foci are (±c, 0)• the equations of the asymptotes are y = ± b

ax

b.If the equation is in the form y2

a2 − x2

b2 = 1, then



• the transverse axis is on the y-axis• the coordinates of the vertices are (0,±a)• the coordinates of the co-vertices are (±b, 0)• the coordinates of the foci are (0,±c)• the equations of the asymptotes are y = ±a

b x

3.Solve for the coordinates of the foci using the equation c = ±√

a2 + b2.4.Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smoothcurve to form the hyperbola.

Example 4Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form

Graph the hyperbola given by the equation y2

64−x2

36 = 1. Identify and label the vertices, co-vertices,foci, and asymptotes.

SolutionThe standard form that applies to the given equation is y2

a2 − x2

b2 = 1. Thus, the transverse axis ison the y-axis

The coordinates of the vertices are (0,±a) =(0,±√

64)

= (0,±8)The coordinates of the co-vertices are (±b, 0) =

(±√

36, 0)

= (±6, 0)The coordinates of the foci are (0,±c) , where c = ±

√a2 + b2. Solving for c, we have

c = ±√

a2 + b2 = ±√

64 + 36 = ±√

100 = ±10 (17)

Therefore, the coordinates of the foci are (0,±10)The equations of the asymptotes are y = ±a

b x = ± 86x = ± 4

3xPlot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the

rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extendthe diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotesprovide the framework needed to sketch an accurate graph of the hyperbola. Label the foci andasymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 8.



Figure 8

Try It:


Graph the hyperbola given by the equation x2

144 −y2

81 = 1. Identify and label the vertices,co-vertices, foci, and asymptotes.

4 Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point (h, k)other than the origin is similar to graphing ellipses centered

at a point other than the origin. We use the standard forms (x−h)2

a2 − (y−k)2

b2 = 1 for horizontal hyperbolas,



and (y−k)2

a2 − (x−h)2

b2 = 1 for vertical hyperbolas. From these standard form equations we can easily calculateand plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equationsof its asymptotes; and the positions of the transverse and conjugate axes.

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

1.Convert the general form to that standard form. Determine which of the standard formsapplies to the given equation.

2.Use the standard form identi�ed in Step 1 to determine the position of the transverse axis;coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.

a.If the equation is in the form (x−h)2

a2 − (y−k)2

b2 = 1, then

• the transverse axis is parallel to the x-axis• the center is (h, k)• the coordinates of the vertices are (h± a, k)• the coordinates of the co-vertices are (h, k ± b)• the coordinates of the foci are (h± c, k)• the equations of the asymptotes are y = ± b

a (x− h) + k

b.If the equation is in the form (y−k)2

a2 − (x−h)2

b2 = 1, then

• the transverse axis is parallel to the y-axis• the center is (h, k)• the coordinates of the vertices are (h, k ± a)• the coordinates of the co-vertices are (h± b, k)• the coordinates of the foci are (h, k ± c)• the equations of the asymptotes are y = ±a

b (x− h) + k

3.Solve for the coordinates of the foci using the equation c = ±√

a2 + b2.4.Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw asmooth curve to form the hyperbola.

Example 5Graphing a Hyperbola Centered at (h, k) Given an Equation in General FormGraph the hyperbola given by the equation 9x2 − 4y2 − 36x − 40y − 388 = 0. Identify and labelthe center, vertices, co-vertices, foci, and asymptotes.

SolutionStart by expressing the equation in standard form. Group terms that contain the same variable,and move the constant to the opposite side of the equation.

(9x2 − 36x

)−(4y2 + 40y

)= 388 (18)

Factor the leading coe�cient of each expression.

9(x2 − 4x

)− 4

(y2 + 10y

)= 388 (19)

Complete the square twice. Remember to balance the equation by adding the same constants toeach side.

9(x2 − 4x + 4

)− 4

(y2 + 10y + 25

)= 388 + 36− 100 (20)

Rewrite as perfect squares.

9(x− 2)2 − 4(y + 5)2 = 324 (21)



Divide both sides by the constant term to place the equation in standard form.

(x− 2)2

36− (y + 5)2

81= 1 (22)

The standard form that applies to the given equation is (x−h)2

a2 − (y−k)2

b2 = 1, where a2 = 36 and b2 =81,or a = 6 and b = 9.Thus, the transverse axis is parallel to the x-axis. It follows that:

• the center of the ellipse is (h, k) = (2,−5)• the coordinates of the vertices are (h± a, k) = (2± 6,−5) , or (−4,−5) and (8,−5)• the coordinates of the co-vertices are (h, k ± b) = (2,−5± 9) , or (2,−14) and (2, 4)• the coordinates of the foci are (h± c, k) , where c = ±

√a2 + b2. Solving for c,we have

c = ±√

36 + 81 = ±√

117 = ±3√

13 (23)

Therefore, the coordinates of the foci are(2− 3

√13,−5

)and

(2 + 3

√13,−5

).

The equations of the asymptotes are y = ± ba (x− h) + k = ± 3

2 (x− 2)− 5.Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth

curves to form the hyperbola, as shown in Figure 9.



Figure 9

Try It:


Graph the hyperbola given by the standard form of an equation (y+4)2

100 − (x−3)2

64 =1. Identify and label the center, vertices, co-vertices, foci, and asymptotes.



5 Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many �elds,such as astronomy, physics, engineering, and architecture. The design e�ciency of hyperbolic cooling towersis particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are oftentouted for their ability to generate power e�ciently. Because of their hyperbolic form, these structures areable 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 incheswide!

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

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



Example 6Solving Applied Problems Involving HyperbolasThe 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 thatthe center of the hyperbola�indicated by the intersection of dashed perpendicular lines in the�gure�is the origin of the coordinate plane. Round �nal values to four decimal places.

SolutionWe are assuming the center of the tower is at the origin, so we can use the standard form of a

horizontal hyperbola centered at the origin: x2

a2 − y2

b2 = 1,where the branches of the hyperbola formthe sides of the cooling tower. We must �nd the values of a2 and b2to complete the model.

First, we �nd a2.Recall that the length of the transverse axis of a hyperbola is 2a.This lengthis represented by the distance where the sides are closest, which is given as 65.3 meters. So, 2a =



60.Therefore, a = 30 and a2 = 900.To solve for b2,we need to substitute forx and y in our equation using a known point. To do this,

we can use the dimensions of the tower to �nd some point (x, y) that lies on the hyperbola. Wewill use the top right corner of the tower to represent that point. Since the y-axis bisects the tower,our x-value can be represented by the radius of the top, or 36 meters. The y-value is representedby the distance from the origin to the top, which is given as 79.6 meters. Therefore,

x2

a2 − y2

b2 = 1 Standard form of horizontal hyperbola.

b2 = y2

x2

a2−1Isolate b2

= (79.6)2

(36)2900 −1

Substitute for a2, x, and y

≈ 14400.3636 Round to four decimal places

(24)

The sides of the tower can be modeled by the hyperbolic equation

x2

900− y2

14400.3636= 1, or

x2

302 −y2

120.00152 = 1 (25)

Try It:


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



Figure 12

Media: Access these online resources for additional instruction and practice with hyperbolas.

• Conic Sections: The Hyperbola Part 1 of 21

• Conic Sections: The Hyperbola Part 2 of 22

• Graph a Hyperbola with Center at Origin3

• Graph a Hyperbola with Center not at Origin4

1http://openstaxcollege.org/l/hyperbola12http://openstaxcollege.org/l/hyperbola23http://openstaxcollege.org/l/hyperbolaorigin4http://openstaxcollege.org/l/hbnotorigin



6 Key Equations

Hyperbola, center at origin, transverse axis on x-axis x2

a2 − y2

b2 = 1

Hyperbola, center at origin, transverse axis on y-axis y2

a2 − x2

b2 = 1

Hyperbola, center at (h, k) ,transverse axis parallel to x-axis (x−h)2

a2 − (y−k)2

b2 = 1

Hyperbola, center at (h, k) ,transverse axis parallel to y-axis (y−k)2

a2 − (x−h)2

b2 = 1

Table 1

7 Key Concepts

• A hyperbola is the set of all points (x, y) in a plane such that the di�erence of the distances between (x, y) andthe foci is a positive constant.

• The standard form of a hyperbola can be used to locate its vertices and foci. See Example 1.• When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the

hyperbola in standard form. See Example 2 and Example 3.• When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes,

and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. SeeExample 4 and Example 5.

• Real-world situations can be modeled using the standard equations of hyperbolas. For instance, giventhe dimensions of a natural draft cooling tower, we can �nd a hyperbolic equation that models itssides. See Example 6.

8 Section Exercises

8.1 Verbal


De�ne a hyperbola in terms of its foci.

Exercise 14What can we conclude about a hyperbola if its asymptotes intersect at the origin?


What must be true of the foci of a hyperbola?

Exercise 16If the transverse axis of a hyperbola is vertical, what do we know about the graph?


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

8.2 Algebraic

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

Exercise 183y2 + 2x = 6

Exercise 19 (Solution on p. 33.)x2

36 −y2

9 = 1



Exercise 205y2 + 4x2 = 6x


25x2 − 16y2 = 400Exercise 22−9x2 + 18x + y2 + 4y − 14 = 0

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


25 −y2

36 = 1Exercise 24

x2

100 −y2

9 = 1Exercise 25 (Solution on p. 34.)

y2

4 −x2

81 = 1Exercise 26

9y2 − 4x2 = 1Exercise 27 (Solution on p. 34.)

(x−1)2

9 − (y−2)2

16 = 1Exercise 28

(y−6)2

36 − (x+1)2


(x−2)2

49 − (y+7)2

49 = 1Exercise 30

4x2 − 8x− 9y2 − 72y + 112 = 0Exercise 31 (Solution on p. 34.)

−9x2 − 54x + 9y2 − 54y + 81 = 0Exercise 32

4x2 − 24x− 36y2 − 360y + 864 = 0Exercise 33 (Solution on p. 34.)

−4x2 + 24x + 16y2 − 128y + 156 = 0Exercise 34−4x2 + 40x + 25y2 − 100y + 100 = 0


x2 + 2x− 100y2 − 1000y + 2401 = 0Exercise 36−9x2 + 72x + 16y2 + 16y + 4 = 0


4x2 + 24x− 25y2 + 200y − 464 = 0For the following exercises, �nd the equations of the asymptotes for each hyperbola.

Exercise 38y2

32 − x2


(x−3)2

52 − (y+4)2

22 = 1



Exercise 40(y−3)2

32 − (x+5)2


9x2 − 18x− 16y2 + 32y − 151 = 0Exercise 42

16y2 + 96y − 4x2 + 16x + 112 = 0

8.3 Graphical

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


49 −y2

16 = 1Exercise 44

x2

64 −y2


y2

9 −x2

25 = 1Exercise 46

81x2 − 9y2 = 1Exercise 47 (Solution on p. 36.)

(y+5)2

9 − (x−4)2

25 = 1Exercise 48

(x−2)2

8 − (y+3)2


(y−3)2

9 − (x−3)2

9 = 1Exercise 50−4x2 − 8x + 16y2 − 32y − 52 = 0


x2 − 8x− 25y2 − 100y − 109 = 0Exercise 52−x2 + 8x + 4y2 − 40y + 88 = 0


64x2 + 128x− 9y2 − 72y − 656 = 0Exercise 54

16x2 + 64x− 4y2 − 8y − 4 = 0Exercise 55 (Solution on p. 40.)

−100x2 + 1000x + y2 − 10y − 2575 = 0Exercise 56

4x2 + 16x− 4y2 + 16y + 16 = 0For the following exercises, given information about the graph of the hyperbola, �nd its equation.


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

Exercise 58Vertices at (0, 6) and (0,−6) and one focus at (0,−8) .




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

Exercise 60Center: (0, 0) ;vertex: (0,−13) ;one focus:

(0,√

313).


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

26, 2).

Exercise 62Center: (3, 5) ; vertex: (3, 11) ; one focus:

(3, 5 + 2

√10).

For the following exercises, given the graph of the hyperbola, �nd its equation.Exercise 63 (Solution on p. 41.)



Exercise 64






Exercise 66




8.4 Extensions

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

Exercise 68x2

4 −y2


y2

9 −x2

1 = 1Exercise 70

(x−2)2

16 − (y+3)2


−4x2 − 16x + y2 − 2y − 19 = 0Exercise 72

4x2 − 24x− y2 − 4y + 16 = 0



8.5 Real-World Applications

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


The hedge will follow the asymptotes y = x and y = −x, and its closest distance to the centerfountain is 5 yards.

Exercise 74The hedge will follow the asymptotes y = 2x and y = −2x, and its closest distance to the centerfountain is 6 yards.


The hedge will follow the asymptotes y = 12x and y = − 1

2x, and its closest distance to the centerfountain is 10 yards.

Exercise 76The hedge will follow the asymptotes y = 2

3x and y = − 23x, and its closest distance to the center

fountain is 12 yards.


The hedge will follow the asymptotes y = 34x and y = − 3

4x, and its closest distance to the centerfountain is 20 yards.

For the following exercises, assume an object enters our solar system and we want to graph its path on acoordinate 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 �ight path of each object using the given information.

Exercise 78The object enters along a path approximated by the line y = x− 2 and passes within 1 au (astro-nomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. Itthen departs the solar system along a path approximated by the line y = −x + 2.


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

Exercise 80The object enters along a path approximated by the line y = 0.5x + 2 and passes within 1 au ofthe sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solarsystem along a path approximated by the line y = −0.5x− 2.


The object enters along a path approximated by the line y = 13x − 1 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 solarsystem along a path approximated by the line y = − 1

3x + 1.

Exercise 82The object It enters along a path approximated by the line y = 3x − 9 and passes within 1 au ofthe sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solarsystem along a path approximated by the line y = −3x + 9.



Solutions to Exercises in this Module

Solution to Exercise (p. 8)Vertices: (±3, 0) ; Foci:

(±√

34, 0)

Solution to Exercise (p. 10)y2

4 −x2

16 = 1Solution to Exercise (p. 13)(y−3)2

25 + (x−1)2

144 = 1Solution to Exercise (p. 15)vertices: (±12, 0) ; co-vertices: (0,±9) ; foci: (±15, 0) ; asymptotes: y = ± 3

4x;

So-lution to Exercise (p. 18)center: (3,−4) ; vertices: (3,−14) and (3, 6) ; co-vertices: (−5,−4) ; and (11,−4) ; foci:

(3,−4− 2

√41)and

(3,−4 + 2

√41); asymptotes: y =

± 54 (x− 3)− 4



Solution to Exercise (p. 21)

The sides of the tower can be modeled by the hyperbolic equation. x2

400 −y2

3600 = 1or x2

202 − y2

602 = 1.Solution to Exercise (p. 23)A hyperbola is the set of points in a plane the di�erence of whose distances from two �xed points (foci) isa positive constant.Solution to Exercise (p. 23)The foci must lie on the transverse axis and be in the interior of the hyperbola.Solution to Exercise (p. 23)The center must be the midpoint of the line segment joining the foci.Solution to Exercise (p. 23)

yes x2

62 − y2

32 = 1




yes x2

42 − y2

52 = 1Solution to Exercise (p. 24)x2

52 − y2

62 = 1; vertices: (5, 0) , (−5, 0) ; foci:(√

61, 0),(−√

61, 0); asymptotes: y = 6

5x, y = − 65x

Solution to Exercise (p. 24)y2

22 − x2

92 = 1; vertices: (0, 2) , (0,−2) ; foci:(0,√

85),(0,−√

85); asymptotes: y = 2

9x, y = − 29x

Solution to Exercise (p. 24)(x−1)2

32 − (y−2)2

42 = 1; vertices: (4, 2) , (−2, 2) ; foci: (6, 2) , (−4, 2) ; asymptotes: y = 43 (x− 1)+2, y = − 4

3 (x− 1)+2Solution to Exercise (p. 24)(x−2)2

72 − (y+7)2

72 = 1; vertices: (9,−7) , (−5,−7) ; foci:(2 + 7

√2,−7

),(2− 7

√2,−7

); asymptotes: y = x −

9, y = −x− 5Solution to Exercise (p. 24)(x+3)2

32 − (y−3)2

32 = 1; vertices: (0, 3) , (−6, 3) ; foci:(−3 + 3

√2, 1),(−3− 3

√2, 1); asymptotes: y = x+6, y =

−xSolution to Exercise (p. 24)

(y−4)2

22 − (x−3)2

42 = 1; vertices: (3, 6) , (3, 2) ; foci:(3, 4 + 2

√5),(3, 4− 2

√5); asymptotes: y = 1

2 (x− 3) +4, y = − 1

2 (x− 3) + 4Solution to Exercise (p. 24)(y+5)2

72 − (x+1)2

702 = 1; vertices: (−1, 2) , (−1,−12) ; foci:(−1,−5 + 7

√101),(−1,−5− 7

√101); asymptotes: y =

110 (x + 1)− 5, y = − 1

10 (x + 1)− 5Solution to Exercise (p. 24)(x+3)2

52 − (y−4)2

22 = 1; vertices: (2, 4) , (−8, 4) ; foci:(−3 +

√29, 4

),(−3−

√29, 4

); asymptotes: y = 2

5 (x + 3)+4, y = − 2

5 (x + 3) + 4Solution to Exercise (p. 24)y = 2

5 (x− 3)− 4, y = − 25 (x− 3)− 4

Solution to Exercise (p. 25)y = 3

4 (x− 1) + 1, y = − 34 (x− 1) + 1






















Solution to Exercise (p. 25)x2

9 −y2

16 = 1Solution to Exercise (p. 25)(x−6)2

25 − (y−1)2

11 = 1Solution to Exercise (p. 26)(x−4)2

25 − (y−2)2

1 = 1Solution to Exercise (p. 26)y2

16 −x2

25 = 1Solution to Exercise (p. 28)y2

9 −(x+1)2

9 = 1Solution to Exercise (p. 30)(x+3)2

25 − (y+3)2

25 = 1Solution to Exercise (p. 30)y (x) = 3

√x2 + 1, y (x) = −3

√x2 + 1



Solution to Exercise (p. 30)y (x) = 1 + 2

√x2 + 4x + 5, y (x) = 1− 2

√x2 + 4x + 5




25 −y2

25 = 1



So-lution to Exercise (p. 31)

x2

100 −y2

25 = 1




400 −y2

225 = 1


0.25 −y2

0.75 = 1




4 − y2

5 = 1

Glossary

De�nition 1: center of a hyperbolathe midpoint of both the transverse and conjugate axes of a hyperbola

De�nition 2: conjugate axisthe axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as itsendpoints

De�nition 3: hyperbolathe set of all points (x, y) in a plane such that the di�erence of the distances between (x, y) andthe foci is a positive constant

De�nition 4: transverse axisthe axis of a hyperbola that includes the foci and has the vertices as its endpoints


Documents

The Hyperbola - OpenStax CNX · hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides