### Learning Outcomes

Derive an equation for a hyperbola centered at the originWrite an equation for a hyperbola centered at the originSolve an applied problem involving hyperbolas

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.

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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

ight)

ight)

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.

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.

## Standard Form of the Equation of a Hyperbola Centered at the Origin

Let

ight)

ight)**foci** of a hyperbola centered at the origin. The hyperbola is the set of all points

ight)

ight)

If

ight)

ight)

ight)

ight)=a+c

ight)

ight)

ight)-left(c-a

ight)=2a

If

ight)

ight) ext{ to }left(x,y

ight)\ &{d}_{1}= ext{the distance from }left(c,0

ight) ext{ to }left(x,y

ight)end{align}

By definition of a hyperbola,

vert

ight)

ight)

vert=2a**distance formula**, but is again beyond the scope of this text. The standard form of an equation of a hyperbola centered at the origin with vertices

ight)

ight)

### A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center

ight)*x*-axis is

where

the length of the transverse axis is

ight)

ight)

ight)

The standard form of the equation of a hyperbola with center

ight)*y*-axis is

where

the length of the transverse axis is

ight)

ight)

ight)

Note that the vertices, co-vertices, and foci are related by the equation

(a) Horizontal hyperbola with center

ight)

ight)

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

Determine whether the transverse axis lies on the *x*– or *y*-axis. Notice that

a. If the equation has the form *x*-axis. The vertices are located at

ight)

ight)

b. If the equation has the form *y*-axis. The vertices are located at

ight)

ight)

Solve for

### Example: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the **hyperbola** with equation

Show Solution

The equation has the form *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

The foci are located at

ight)

Therefore, the vertices are located at

ight)

ight)

## 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

ight)

### How To: Given the vertices and foci of a hyperbola centered at left(0, ext{0}

ight) , write its equation in standard form.

ight)

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Determine whether the transverse axis lies on the *x*– or *y*-axis.If the given coordinates of the vertices and foci have the form

ight)

ight)*x*-axis. Use the standard form

ight)

ight)*y*-axis. Use the standard form

### Example: 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

ight)

ight)?

Show Solution

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

The vertices are

ight)

The foci are

ight)

Solving for

Finally, we substitute

## 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 **hyperbola** will be

ight)

ight)

ight)

### A General Note: Standard Forms of the Equation of a Hyperbola with Center (*h*, *k*)

The standard form of the equation of a hyperbola with center

ight)*x*-axis is

ight)}^{2}}{{a}^{2}}-dfrac{{left(y-k

ight)}^{2}}{{b}^{2}}=1

the length of the transverse axis is

ight)

ight)

ight)

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is

ight)**point-slope formula**, it is simple to show that the equations of the asymptotes are

ight)+k

The standard form of the equation of a hyperbola with center

ight)*y*-axis is

ight)}^{2}}{{a}^{2}}-dfrac{{left(x-h

ight)}^{2}}{{b}^{2}}=1

the length of the transverse axis is

ight)

ight)

ight)

Using the reasoning above, the equations of the asymptotes are

ight)+k

(a) Horizontal hyperbola with center

ight)

ight)

Like hyperbolas centered at the origin, hyperbolas centered at a point

ight)

### How To: Given the vertices and foci of a hyperbola centered at left(h,k

ight) , write its equation in standard form.

ight)

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

ight)}^{2}}{{a}^{2}}-dfrac{{left(y-k

ight)}^{2}}{{b}^{2}}=1*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

ight)}^{2}}{{a}^{2}}-dfrac{{left(x-h

ight)}^{2}}{{b}^{2}}=1

ight)

### Example: 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

ight)

ight)

ight)

ight)?

Show 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

ight)}^{2}}{{a}^{2}}-dfrac{{left(y-k

ight)}^{2}}{{b}^{2}}=1

First, we identify the center,

ight)

ight)

ight)

ight)=left(dfrac{0+6}{2},dfrac{-2+left(-2

ight)}{2}

ight)=left(3,-2

ight)

Next, we find *x*-coordinates of the vertices.

Now we need to find

ight)

ight)=left(-2,-2

ight)

ight)=left(8,-2

ight)*x*-coordinate from either of these points to solve for

ight)

Next, solve for

Finally, substitute the values found for

ight)}^{2}}{9}-dfrac{{left(y+2

ight)}^{2}}{16}=1

## 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. For example a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

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: Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown below. 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.

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.

Show 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:

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First, we find

To solve for

ight)*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,

ight)}^{2}}{frac{{left(36

ight)}^{2}}{900}-1} && ext{Substitute for }{a}^{2},x, ext{ and }y \ &approx 14400.3636 && ext{Round to four decimal places} end{align}

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