The Hyperbola

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.

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.

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 c−a.𝑐−𝑎. The difference of the distances from the foci to the vertex is

(a+c)−(c−a)=2a(𝑎+𝑐)−(𝑐−𝑎)=2𝑎

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

d2=the distance from (−c,0)to (x,y)d1=the distance from (c,0)to (x,y)𝑑2=the distance from (−𝑐,0)to (𝑥,𝑦)𝑑1=the distance from (𝑐,0)to (𝑥,𝑦)

By definition of a hyperbola, d2−d1𝑑2−𝑑1 is constant for any point (x,y)(𝑥,𝑦) on the hyperbola. We know that the difference of these distances is 2a2𝑎 for the vertex (a,0).(𝑎,0). It follows that d2−d1=2a𝑑2−𝑑1=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.

                                     d2−d1=(x−(−c))2+(y−0)2−−−−−−−−−−−−−−−−−−√−(x−c)2+(y−0)2−−−−−−−−−−−−−−−√=2a(x+c)2+y2−−−−−−−−−−√−(x−c)2+y2−−−−−−−−−−√=2a                          (x+c)2+y2−−−−−−−−−−√=2a+(x−c)2+y2−−−−−−−−−−√                            (x+c)2+y2=(2a+(x−c)2+y2−−−−−−−−−−√)2                   x2+2cx+c2+y2=4a2+4a(x−c)2+y2−−−−−−−−−−√+(x−c)2+y2                   x2+2cx+c2+y2=4a2+4a(x−c)2+y2−−−−−−−−−−√+x2−2cx+c2+y2                                            2cx=4a2+4a(x−c)2+y2−−−−−−−−−−√−2cx                                 4cx−4a2=4a(x−c)2+y2−−−−−−−−−−√                                     cx−a2=a(x−c)2+y2−−−−−−−−−−√                                 (cx−a2)2=a2((x−c)2+y2−−−−−−−−−−√)2                   c2x2−2a2cx+a4=a2(x2−2cx+c2+y2)                  c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2                                 a4+c2x2=a2x2+a2c2+a2y2                c2x2−a2x2−a2y2=a2c2−a4                  x2(c2−a2)−a2y2=a2(c2−a2)                            x2b2−a2y2=a2b2                           x2b2a2b2−a2y2a2b2=a2b2a2b2                                   x2a2−y2b2=1Distance FormulaSimplify expressions.Move radical to opposite side.Square both sides.Expand the squares.Expand remaining square.Combine like terms.Isolate the radical.Divide by 4.Square both sides.Expand the squares.Distribute a2.Combine like terms.Rearrange terms.Factor common terms.Set b2=c2−a2.Divide both sides by a2b2                                     𝑑2−𝑑1=(𝑥−(−𝑐))2+(𝑦−0)2−(𝑥−𝑐)2+(𝑦−0)2=2𝑎Distance Formula(𝑥+𝑐)2+𝑦2−(𝑥−𝑐)2+𝑦2=2𝑎Simplify expressions.                          (𝑥+𝑐)2+𝑦2=2𝑎+(𝑥−𝑐)2+𝑦2Move radical to opposite side.                            (𝑥+𝑐)2+𝑦2=(2𝑎+(𝑥−𝑐)2+𝑦2)2Square both sides.                   𝑥2+2𝑐𝑥+𝑐2+𝑦2=4𝑎2+4𝑎(𝑥−𝑐)2+𝑦2+(𝑥−𝑐)2+𝑦2Expand the squares.                   𝑥2+2𝑐𝑥+𝑐2+𝑦2=4𝑎2+4𝑎(𝑥−𝑐)2+𝑦2+𝑥2−2𝑐𝑥+𝑐2+𝑦2Expand remaining square.                                            2𝑐𝑥=4𝑎2+4𝑎(𝑥−𝑐)2+𝑦2−2𝑐𝑥Combine like terms.                                 4𝑐𝑥−4𝑎2=4𝑎(𝑥−𝑐)2+𝑦2Isolate the radical.                                     𝑐𝑥−𝑎2=𝑎(𝑥−𝑐)2+𝑦2Divide by 4.                                 (𝑐𝑥−𝑎2)2=𝑎2((𝑥−𝑐)2+𝑦2)2Square both sides.                   𝑐2𝑥2−2𝑎2𝑐𝑥+𝑎4=𝑎2(𝑥2−2𝑐𝑥+𝑐2+𝑦2)Expand the squares.                  𝑐2𝑥2−2𝑎2𝑐𝑥+𝑎4=𝑎2𝑥2−2𝑎2𝑐𝑥+𝑎2𝑐2+𝑎2𝑦2Distribute 𝑎2.                                 𝑎4+𝑐2𝑥2=𝑎2𝑥2+𝑎2𝑐2+𝑎2𝑦2Combine like terms.                𝑐2𝑥2−𝑎2𝑥2−𝑎2𝑦2=𝑎2𝑐2−𝑎4Rearrange terms.                  𝑥2(𝑐2−𝑎2)−𝑎2𝑦2=𝑎2(𝑐2−𝑎2)Factor common terms.                            𝑥2𝑏2−𝑎2𝑦2=𝑎2𝑏2Set 𝑏2=𝑐2−𝑎2.                           𝑥2𝑏2𝑎2𝑏2−𝑎2𝑦2𝑎2𝑏2=𝑎2𝑏2𝑎2𝑏2Divide both sides by 𝑎2𝑏2                                   𝑥2𝑎2−𝑦2𝑏2=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

x2a2−y2b2=1𝑥2𝑎2−𝑦2𝑏2=1

where

• the length of the transverse axis is 2a2𝑎
• the coordinates of the vertices are (±a,0)(±𝑎,0)
• the length of the conjugate axis is 2b2𝑏
• the coordinates of the co-vertices are (0,±b)(0,±𝑏)
• the distance between the foci is 2c,2𝑐, where c2=a2+b2𝑐2=𝑎2+𝑏2
• the coordinates of the foci are (±c,0)(±𝑐,0)
• the equations of the asymptotes are y=±bax𝑦=±𝑏𝑎𝑥

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

y2a2−x2b2=1𝑦2𝑎2−𝑥2𝑏2=1

where

• the length of the transverse axis is 2a2𝑎
• the coordinates of the vertices are (0,±a)(0,±𝑎)
• the length of the conjugate axis is 2b2𝑏
• the coordinates of the co-vertices are (±b,0)(±𝑏,0)
• the distance between the foci is 2c,2𝑐, where c2=a2+b2𝑐2=𝑎2+𝑏2
• the coordinates of the foci are (0,±c)(0,±𝑐)
• the equations of the asymptotes are y=±abx𝑦=±𝑎𝑏𝑥

See Figure 5b.

Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2.𝑐2=𝑎2+𝑏2. 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 a2𝑎2 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 x2a2−y2b2=1,𝑥2𝑎2−𝑦2𝑏2=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 y2a2−x2b2=1,𝑦2𝑎2−𝑥2𝑏2=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 a𝑎 using the equation a=a2−−√.𝑎=𝑎2.
3. Solve for c𝑐 using the equation c=a2+b2−−−−−−√.𝑐=𝑎2+𝑏2.

EXAMPLE 1

Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation y249−x232=1.𝑦249−𝑥232=1.

Solution

The equation has the form y2a2−x2b2=1,𝑦2𝑎2−𝑥2𝑏2=1, so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set x=0,𝑥=0, and solve for y.𝑦.

1=y249−x2321=y249−02321=y249y2=49y=±49−−√=±71=𝑦249−𝑥2321=𝑦249−02321=𝑦249𝑦2=49𝑦=±49=±7

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

c=a2+b2−−−−−−√=49+32−−−−−−√=81−−√=9𝑐=𝑎2+𝑏2=49+32=81=9

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

TRY IT #1

Identify the vertices and foci of the hyperbola with equation x29−y225=1.𝑥29−𝑦225=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 c2=a2+b2.𝑐2=𝑎2+𝑏2. Note that this equation can also be rewritten as b2=c2−a2.𝑏2=𝑐2−𝑎2. 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 x2a2−y2b2=1.𝑥2𝑎2−𝑦2𝑏2=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 y2a2−x2b2=1.𝑦2𝑎2−𝑥2𝑏2=1.
2. Find b2𝑏2 using the equation b2=c2−a2.𝑏2=𝑐2−𝑎2.
3. Substitute the values for a2𝑎2 and b2𝑏2 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 (±210−−√,0)?(±210,0)?

Solution

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form x2a2−y2b2=1.𝑥2𝑎2−𝑦2𝑏2=1.

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

The foci are (±210−−√,0),(±210,0), so c=210−−√𝑐=210 and c2=40.𝑐2=40.

Solving for b2,𝑏2, we have

b2=c2−a2b2=40−36b2=4Substitute for c2and a2.Subtract.𝑏2=𝑐2−𝑎2𝑏2=40−36Substitute for 𝑐2and 𝑎2.𝑏2=4Subtract.

Finally, we substitute a2=36𝑎2=36 and b2=4𝑏2=4 into the standard form of the equation, x2a2−y2b2=1.𝑥2𝑎2−𝑦2𝑏2=1. The equation of the hyperbola is x236−y24=1,𝑥236−𝑦24=1, as shown in Figure 6.

Figure 6

TRY IT #2

What is the standard form equation of the hyperbola that has vertices (0,±2)(0,±2) and foci (0,±25–√)?(0,±25)?

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 hℎ units horizontally and k𝑘 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 x𝑥 replaced by (x−h)(𝑥−ℎ) and y𝑦 replaced by (y−k).(𝑦−𝑘).

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

(x−h)2a2−(y−k)2b2=1(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1

where

• the length of the transverse axis is 2a2𝑎
• the coordinates of the vertices are (h±a,k)(ℎ±𝑎,𝑘)
• the length of the conjugate axis is 2b2𝑏
• the coordinates of the co-vertices are (h,k±b)(ℎ,𝑘±𝑏)
• the distance between the foci is 2c,2𝑐, where c2=a2+b2𝑐2=𝑎2+𝑏2
• 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 2a2𝑎 and its width is 2b.2𝑏. The slopes of the diagonals are ±ba,±𝑏𝑎, 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=±ba(x−h)+k.𝑦=±𝑏𝑎(𝑥−ℎ)+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

(y−k)2a2−(x−h)2b2=1(𝑦−𝑘)2𝑎2−(𝑥−ℎ)2𝑏2=1

where

• the length of the transverse axis is 2a2𝑎
• the coordinates of the vertices are (h,k±a)(ℎ,𝑘±𝑎)
• the length of the conjugate axis is 2b2𝑏
• the coordinates of the co-vertices are (h±b,k)(ℎ±𝑏,𝑘)
• the distance between the foci is 2c,2𝑐, where c2=a2+b2𝑐2=𝑎2+𝑏2
• the coordinates of the foci are (h,k±c)(ℎ,𝑘±𝑐)

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

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 c2=a2+b2.𝑐2=𝑎2+𝑏2. 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 (x−h)2a2−(y−k)2b2=1.(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=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 (y−k)2a2−(x−h)2b2=1.(𝑦−𝑘)2𝑎2−(𝑥−ℎ)2𝑏2=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 a2𝑎2 by solving for the length of the transverse axis, 2a2𝑎 , which is the distance between the given vertices.
4. Find c2𝑐2 using hℎ and k𝑘 found in Step 2 along with the given coordinates for the foci.
5. Solve for b2𝑏2 using the equation b2=c2−a2.𝑏2=𝑐2−𝑎2.
6. Substitute the values for h,k,a2,ℎ,𝑘,𝑎2, and b2𝑏2 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)?

Solution

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

(x−h)2a2−(y−k)2b2=1(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1

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

(h,k)=(0+62,−2+(−2)2)=(3,−2)(ℎ,𝑘)=(0+62,−2+(−2)2)=(3,−2)

Next, we find a2.𝑎2. The length of the transverse axis, 2a,2𝑎, is bounded by the vertices. So, we can find a2𝑎2 by finding the distance between the x-coordinates of the vertices.

2a=|0−6|2a=6 a=3a2=92𝑎=|0−6|2𝑎=6 𝑎=3𝑎2=9

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

h+c=83+c=8     c=5   c2=25ℎ+𝑐=83+𝑐=8     𝑐=5   𝑐2=25

Next, solve for b2𝑏2 using the equation b2=c2−a2:𝑏2=𝑐2−𝑎2:

b2=c2−a2   =25−9   =16𝑏2=𝑐2−𝑎2   =25−9   =16

Finally, substitute the values found for h,k,a2,ℎ,𝑘,𝑎2, and b2𝑏2 into the standard form of the equation.

(x−3)29−(y+2)216=1(𝑥−3)29−(𝑦+2)216=1

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 x2a2−y2b2=1𝑥2𝑎2−𝑦2𝑏2=1 for horizontal hyperbolas and the standard form y2a2−x2b2=1𝑦2𝑎2−𝑥2𝑏2=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 x2a2−y2b2=1,𝑥2𝑎2−𝑦2𝑏2=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=±bax𝑦=±𝑏𝑎𝑥
   2. If the equation is in the form y2a2−x2b2=1,𝑦2𝑎2−𝑥2𝑏2=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=±abx𝑦=±𝑎𝑏𝑥
3. Solve for the coordinates of the foci using the equation c=±a2+b2−−−−−−√.𝑐=±𝑎2+𝑏2.
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 y264−x236=1.𝑦264−𝑥236=1. Identify and label the vertices, co-vertices, foci, and asymptotes.

Solution

The standard form that applies to the given equation is y2a2−x2b2=1.𝑦2𝑎2−𝑥2𝑏2=1. Thus, the transverse axis is on the y-axis

The coordinates of the vertices are (0,±a)=(0,±64−−√)=(0,±8)(0,±𝑎)=(0,±64)=(0,±8)

The coordinates of the co-vertices are (±b,0)=(±36−−√,0)=(±6,0)(±𝑏,0)=(±36,0)=(±6,0)

The coordinates of the foci are (0,±c),(0,±𝑐), where c=±a2+b2−−−−−−√.𝑐=±𝑎2+𝑏2. Solving for c,𝑐, we have

c=±a2+b2−−−−−−√=±64+36−−−−−−√=±100−−−√=±10𝑐=±𝑎2+𝑏2=±64+36=±100=±10

Therefore, the coordinates of the foci are (0,±10)(0,±10)

The equations of the asymptotes are y=±abx=±86x=±43x𝑦=±𝑎𝑏𝑥=±86𝑥=±43𝑥

Plot 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 extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 8.

Figure 8

TRY IT #4

Graph the hyperbola given by the equation x2144−y281=1.𝑥2144−𝑦281=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 (x−h)2a2−(y−k)2b2=1(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1 for horizontal hyperbolas, and (y−k)2a2−(x−h)2b2=1(𝑦−𝑘)2𝑎2−(𝑥−ℎ)2𝑏2=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 (x−h)2a2−(y−k)2b2=1,(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=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=±ba(x−h)+k𝑦=±𝑏𝑎(𝑥−ℎ)+𝑘
   2. If the equation is in the form (y−k)2a2−(x−h)2b2=1,(𝑦−𝑘)2𝑎2−(𝑥−ℎ)2𝑏2=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=±ab(x−h)+k𝑦=±𝑎𝑏(𝑥−ℎ)+𝑘
3. Solve for the coordinates of the foci using the equation c=±a2+b2−−−−−−√.𝑐=±𝑎2+𝑏2.
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 9×2−4y2−36x−40y−388=0.9𝑥2−4𝑦2−36𝑥−40𝑦−388=0. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Solution

Start 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.

(9×2−36x)−(4y2+40y)=388(9𝑥2−36𝑥)−(4𝑦2+40𝑦)=388

Factor the leading coefficient of each expression.

9(x2−4x)−4(y2+10y)=3889(𝑥2−4𝑥)−4(𝑦2+10𝑦)=388

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

9(x2−4x+4)−4(y2+10y+25)=388+36−1009(𝑥2−4𝑥+4)−4(𝑦2+10𝑦+25)=388+36−100

Rewrite as perfect squares.

9(x−2)2−4(y+5)2=3249(𝑥−2)2−4(𝑦+5)2=324

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

(x−2)236−(y+5)281=1(𝑥−2)236−(𝑦+5)281=1

The standard form that applies to the given equation is (x−h)2a2−(y−k)2b2=1,(𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1, where a2=36𝑎2=36 and b2=81,𝑏2=81, or a=6𝑎=6 and b=9.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)(ℎ,𝑘)=(2,−5)
• the coordinates of the vertices are (h±a,k)=(2±6,−5),(ℎ±𝑎,𝑘)=(2±6,−5), or (−4,−5)(−4,−5) and (8,−5)(8,−5)
• the coordinates of the co-vertices are (h,k±b)=(2,−5±9),(ℎ,𝑘±𝑏)=(2,−5±9), or (2,−14)(2,−14) and (2,4)(2,4)
• the coordinates of the foci are (h±c,k),(ℎ±𝑐,𝑘), where c=±a2+b2−−−−−−√.𝑐=±𝑎2+𝑏2. Solving for c,𝑐, we have

c=±36+81−−−−−−√=±117−−−√=±313−−√𝑐=±36+81=±117=±313

Therefore, the coordinates of the foci are (2−313−−√,−5)(2−313,−5) and (2+313−−√,−5).(2+313,−5).

The equations of the asymptotes are y=±ba(x−h)+k=±32(x−2)−5.𝑦=±𝑏𝑎(𝑥−ℎ)+𝑘=±32(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 #5

Graph the hyperbola given by the standard form of an equation (y+4)2100−(x−3)264=1.(𝑦+4)2100−(𝑥−3)264=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.

Solution

We 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: x2a2−y2b2=1,𝑥2𝑎2−𝑦2𝑏2=1, where the branches of the hyperbola form the sides of the cooling tower. We must find the values of a2𝑎2 and b2𝑏2 to complete the model.

First, we find a2.𝑎2. Recall that the length of the transverse axis of a hyperbola is 2a.2𝑎. This length is represented by the distance where the sides are closest, which is given as 6060 meters. So, 2a=60.2𝑎=60. Therefore, a=30𝑎=30 and a2=900.𝑎2=900.

To solve for b2,𝑏2, we need to substitute for x𝑥 and y𝑦 in our equation using a known point. To do this, we can use the dimensions of the tower to find some point (x,y)(𝑥,𝑦) that lies on the hyperbola. We will 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 represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,

x2a2−y2b2=1         b2=y2x2a2−1            =(79.6)2(36)2900−1            ≈14400.3636Standard form of horizontal hyperbola.Isolate b2Substitute for a2,x,and yRound to four decimal places𝑥2𝑎2−𝑦2𝑏2=1Standard form of horizontal hyperbola.         𝑏2=𝑦2𝑥2𝑎2−1Isolate 𝑏2            =(79.6)2(36)2900−1Substitute for 𝑎2,𝑥,and 𝑦            ≈14400.3636Round to four decimal places

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

x2900−y214400.3636 =1,orx2302−y2120.00152 =1𝑥2900−𝑦214400.3636 =1,or𝑥2302−𝑦2120.00152 =1

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.

Figure 12