Section 10.3: Hyperbolas
Objectives: Upon completion of this lesson, you will be able to:
 given the graph of a hyperbola, find the vertices, the foci, and the equations of the asymptotes, and write
its equation
 write the equation of a hyperbola given specific conditions and sketch its graph
 write the equation of a hyperbola in standard form, find the vertices, the foci, and the equations of the
asymptotes, and sketch its graph
 identify a given equation of a conic and its orientation
 find the points of intersection given a system of equations.
Required Reading, Video, Tutorial
 Read Swokowski/Cole: Section 10.3
o Disregard Examples 4 and 6
 Watch the Section 10.3 Video
 Complete the Section 10.3 Before Class WA assignment
Discussion
Definition: A hyperbola is a set of points in a plane for which the difference of the distances from two fixed
points, the foci, is a positive constant.
Please read and try to follow the derivation of the equation of a hyperbola, but you are not responsible for
reproducing this derivation.
The general forms of a hyperbola are given below.
Horizontal Hyperbola with center at  h, k  :
Vertical Hyperbola with center at  h, k  :
 x  h 2
a2
y  k
a2

2

y  k
2
b2
 x  h 2
b2
1
1
where c2  a 2  b2 .
For our purposes, these are the only equations necessary for working with hyperbolas, because if the center is at
the origin, these formulas reduce to those given in the text on page 740.
Notice the similarities and differences between the equations for ellipses and hyperbolas. With hyperbolas, it is
the minus sign that determines whether the hyperbola is horizontal or vertical, not the relative size of a to b.
Important reminders
 The relationship between the variables a, b, and c is c2  a 2  b2 .
 If the x 2 coefficient is positive, then the hyperbola is horizontal.
 If the y 2 coefficient is positive, then the hyperbola is vertical.
 The value of a is the distance from the center of the hyperbola to a vertex. This makes the length of the
transverse axis 2a.
Math 130: Section 10.3 - Hyperbolas
 The value of b is the distance from the center of the hyperbola to the midpoint of the rectangular box,
through which pass the asymptotes. This makes the length of the conjugate axis 2b.
 The value of c is the distance from the center of the hyperbola to a focus.
 When writing the equations of the asymptotes, use the point-slope form of a linear equation:
( y  k )  m( x  h) where  h, k  is the center of the hyperbola and m is the ratio of rise/run from the
center to the corner of the rectangular box drawn for the asymptotes. You may leave the equations in this
form.
When trying to write an equation for a hyperbola, it is beneficial to sketch the given information to determine if
the hyperbola is horizontal or vertical. Once this is known, the correct version of the equation can be selected,
and you are on your way.
Carefully study Examples 1, 2, 3, and 5 from the text. The author does a great job of working with hyperbolas
centered at the origin in Examples 1, 2, and 3.
Example 1: Write the equation of a hyperbola with foci at  3, 2  and  5, 2  and vertex at  2, 2  .
Solution: Begin by plotting the points. Notice that this is a horizontal hyperbola with a center at  1, 2  .
The distance from the center to a vertex is 3, so a  3.
The distance from the center to a focus is 4, so c  4.
Since a 2  b2  c 2 , then b  7.
Choose the formula for a horizontal hyperbola,
and substitute the determined values.
 x  h
a2
2
y k

b2
 x  1 2
9

2
 1,
 y  2
7
2
1
Example 2: Find the vertices, foci, and the equations of the asymptotes for the given hyperbola.
9 y 2  x 2  36 y  6 x  18  0
Solution: We will begin by converting the given equation to standard form.
©2014;Dr. Brenda Shryock
Page 2 of 4
Math 130: Section 10.3 - Hyperbolas
9 y 2  36 y  x 2  6 x  18
9( y  4 y  ___)  1( x  6 x  ___)  18
2
2
9( y  4 y  4)  1( x  6 x  9)  18  36  9
2
2
9( y  2)2  1( x  3) 2  9
group like variables
factor out leading coefficients
b
complete the square using  
2
factor
2
( y  2)2 ( x  3) 2

1
1
9
 3,  2 
Center:
a  1, b  3, c  10
Vertices:
 3,  2  1  3, 1 and 3,  3
3,  2 
Foci:
10

Asymptotes: y  2  
1
 x  3
3
 x 2  y 2  4
Example 3: Find the solution for the system of equations.  2
 x  y 2  1
Solution The first equation is a circle centered at the origin with a radius of 2. The second equation is a
horizontal hyperbola centered at the origin with vertices at (1, 0).
x2  y2  4
+
x2  y2  1
5
2 x2
x2 
5
5
so x  
2
2
2
 5
2
 
  y  1
 2
5
1  y2
2
y2 
Plug the value of x into either of the given equations.
3
3
so y  
2
2
©2014;Dr. Brenda Shryock
Page 3 of 4
Math 130: Section 10.3 - Hyperbolas
 5
 5
3
3
The four solutions (intersection points) are 
,
 .
 and   , 
2
2
2
2




Practice Problems
Answers to odd-numbered problems can be found at the end of your text.
Section Exercises on WA
5, 7, 12, 14, 17, 19, 22, 25, 27, 29, 37, 38, 39, 40, 41, 44, 45,47
10.3
©2014;Dr. Brenda Shryock
Page 4 of 4