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.
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.
Equations of 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.
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) in a plane such that the difference of the distances between (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.
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) and (c,0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x,y) such that the difference of the distances from (x,y) to the foci is constant.
If (a,0) is a vertex of the hyperbola, the distance from (−c,0) to (a,0) is a−(−c)=a+c. The distance from (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
If (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)
By definition of a hyperbola, d2−d1 is constant for any point (x,y) on the hyperbola. We know that the difference of these distances is 2a for the vertex (a,0). It follows that d2−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)2x2+2cx+c2+y2=4a2+4a(x−c)2+y2+(x−c)2+y2x2+2cx+c2+y2=4a2+4a(x−c)2+y2+x2−2cx+c2+y22cx=4a2+4a(x−c)2+y2−2cx4cx−4a2=4a(x−c)2+y2cx−a2=a(x−c)2+y2(cx−a2)2=a2[(x−c)2+y2]2c2x2−2a2cx+a4=a2(x2−2cx+c2+y2)c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2a4+c2x2=a2x2+a2c2+a2y2c2x2−a2x2−a2y2=a2c2−a4x2(c2−a2)−a2y2=a2(c2−a2)x2b2−a2y2=a2b2a2b2x2b2−a2b2a2y2=a2b2a2b2a2x2−b2y2=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
This equation defines a hyperbola centered at the origin with vertices (±a,0) and co-vertices (0±b).
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 (0,0) and transverse axis on the x-axis is
a2x2−b2y2=1
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=±abx
The standard form of the equation of a hyperbola with center (0,0) and transverse axis on the y-axis is
a2y2−b2x2=1
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=±bax
Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.
(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 and foci.
Determine whether the transverse axis lies on the x- or y-axis. Notice that a2 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.
a. If the equation has the form a2x2−b2y2=1, then the transverse axis lies on the x-axis. The vertices are located at (±a,0), and the foci are located at (±c,0).
b. If the equation has the form a2y2−b2x2=1, then the transverse axis lies on the y-axis. The vertices are located at (0,±a), and the foci are located at (0,±c).
Solve for a using the equation a=a2.
Solve for c using the equation c=a2+b2.
Example: Locating a Hyperbola’s Vertices and Foci
Identify the vertices and foci of the hyperbola with equation 49y2−32x2=1.
Answer:
The equation has the form a2y2−b2x2=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, and solve for y.
1=49y2−32x21=49y2−32021=49y2y2=49y=±49=±7
The foci are located at (0,±c). Solving for c,
c=a2+b2=49+32=81=9
Therefore, the vertices are located at (0,±7), and the foci are located at (0,9).
Try It
Identify the vertices and foci of the hyperbola with equation 9x2−25y2=1.
Answer: Vertices: (±3,0); Foci: (±34,0)
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 units horizontally and k units vertically, the center of the hyperbola will be (h,k). This translation results in the standard form of the equation we saw previously, with x replaced by (x−h) and y replaced by (y−k).
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 (h,k) and transverse axis parallel to the x-axis is
a2(x−h)2−b2(y−k)2=1
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 of the rectangle is 2a and its width is 2b. The slopes of the diagonals are ±ab, and each diagonal passes through the center (h,k). Using the point-slope formula, it is simple to show that the equations of the asymptotes are y=±ab(x−h)+k.
The standard form of the equation of a hyperbola with center (h,k) and transverse axis parallel to the y-axis is
a2(y−k)2−b2(x−h)2=1
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=±ba(x−h)+k.
(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, and foci that are related by the equation 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), write its equation in standard form.
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 a2(x−h)2−b2(y−k)2=1.
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 a2(y−k)2−b2(x−h)2=1.
Identify the center of the hyperbola, (h,k), using the midpoint formula and the given coordinates for the vertices.
Find a2 by solving for the length of the transverse axis, 2a , which is the distance between the given vertices.
Find c2 using h and k found in Step 2 along with the given coordinates for the foci.
Solve for b2 using the equation b2=c2−a2.
Substitute the values for h,k,a2, and b2 into the standard form of the equation determined in Step 1.
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 (0,−2) and (6,−2) and foci at (−2,−2) and (8,−2)?
Answer:
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
a2(x−h)2−b2(y−k)2=1
First, we identify the center, (h,k). The center is halfway between the vertices (0,−2) and (6,−2). Applying the midpoint formula, we have
(h,k)=(20+6,2−2+(−2))=(3,−2)
Next, we find a2. The length of the transverse axis, 2a, is bounded by the vertices. So, we can find a2 by finding the distance between the x-coordinates of the vertices.
2a=∣0−6∣2a=6a=3a2=9
Now we need to find 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), and substituting h=3,
h+c=83+c=8c=5c2=25
Next, solve for b2 using the equation b2=c2−a2:
b2=c2−a2=25−9=16
Finally, substitute the values found for h,k,a2, and b2 into the standard form of the equation.
9(x−3)2−16(y+2)2=1
Try It
What is the standard form equation of the hyperbola that has vertices (1,−2) and (1,8) and foci (1,−10) and (1,16)?
Answer: 25(y−3)2+144(x−1)2=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.
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.
Answer:
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: a2x2−b2y2=1, where the branches of the hyperbola form the sides of the cooling tower. We must find the values of a2 and b2 to complete the model.
First, we find a2. Recall that the length of the transverse axis of a hyperbola is 2a. This length is 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 for x and y in our equation using a known point. To do this, we can use the dimensions of the tower to find some point (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,
a2x2−b2y2=1b2=a2x2−1y2=900(36)2−1(79.6)2≈14400.3636Standard form of horizontal hyperbola.Isolate b2Substitute for a2,x, and yRound to four decimal places
The sides of the tower can be modeled by the hyperbolic equation
900x2−14400.3636y2=1,or302x2−120.00152y2=1
Try It
A design for a cooling tower project is shown below. 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.
Answer:
The sides of the tower can be modeled by the hyperbolic equation. 400x2−3600y2=1or 202x2−602y2=1.
Graph Hyperbolas
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 a2x2−b2y2=1 for horizontal hyperbolas and the standard form a2y2−b2x2=1 for vertical hyperbolas.
How To: Given a standard form equation for a hyperbola centered at (0,0), sketch the graph.
Determine which of the standard forms applies to the given equation.
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.
If the equation is in the form a2x2−b2y2=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=±abx
If the equation is in the form a2y2−b2x2=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=±bax
Solve for the coordinates of the foci using the equation c=±a2+b2.
Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.
Example: Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form
Graph the hyperbola given by the equation 64y2−36x2=1. Identify and label the vertices, co-vertices, foci, and asymptotes.
Answer:
The standard form that applies to the given equation is a2y2−b2x2=1. Thus, the transverse axis is on 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
Therefore, the coordinates of the foci are (0,±10)
The equations of the asymptotes are y=±bax=±68x=±34x
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.
Try It
Graph the hyperbola given by the equation 144x2−81y2=1. Identify and label the vertices, co-vertices, foci, and asymptotes.
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 a2(x−h)2−b2(y−k)2=1 for horizontal hyperbolas, and a2(y−k)2−b2(x−h)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), sketch the graph.
Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
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.
If the equation is in the form a2(x−h)2−b2(y−k)2=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=±ab(x−h)+k
If the equation is in the form a2(y−k)2−b2(x−h)2=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=±ba(x−h)+k
Solve for the coordinates of the foci using the equation c=±a2+b2.
Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.
Example: Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form
Graph the hyperbola given by the equation 9x2−4y2−36x−40y−388=0. Identify and label the center, vertices, co-vertices, foci, and asymptotes.
Answer:
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.
(9x2−36x)−(4y2+40y)=388
Factor the leading coefficient of each expression.
9(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+36−100
Rewrite as perfect squares.
9(x−2)2−4(y+5)2=324
Divide both sides by the constant term to place the equation in standard form.
36(x−2)2−81(y+5)2=1
The standard form that applies to the given equation is a2(x−h)2−b2(y−k)2=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=±313
Therefore, the coordinates of the foci are (2−313,−5) and (2+313,−5).
The equations of the asymptotes are y=±ab(x−h)+k=±23(x−2)−5.
Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola.
Try It
Graph the hyperbola given by the standard form of an equation 100(y+4)2−64(x−3)2=1. Identify and label the center, vertices, co-vertices, foci, and asymptotes.
Answer:
center: (3,−4); vertices: (3,−14) and (3,6); co-vertices: (−5,−4); and (11,−4); foci: (3,−4−241) and (3,−4+241); asymptotes: y=±45(x−3)−4
Key Equations
Hyperbola, center at origin, transverse axis on x-axis
a2x2−b2y2=1
Hyperbola, center at origin, transverse axis on y-axis
a2y2−b2x2=1
Hyperbola, center at (h,k), transverse axis parallel to x-axis
a2(x−h)2−b2(y−k)2=1
Hyperbola, center at (h,k), transverse axis parallel to y-axis
a2(y−k)2−b2(x−h)2=1
Key Concepts
A hyperbola is the set of all points (x,y) in a plane such that the difference of the distances between (x,y) and the foci is a positive constant.
The standard form of a hyperbola can be used to locate its vertices and foci.
When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form.
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.
Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides.
Glossary
center of a hyperbola the midpoint of both the transverse and conjugate axes of a hyperbola
conjugate axis the axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as its endpoints
hyperbola the set of all points (x,y) in a plane such that the difference of the distances between (x,y) and the foci is a positive constant
transverse axis the axis of a hyperbola that includes the foci and has the vertices as its endpoints
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