Advanced Mathematical Concepts
Chapter 10
I. Definitions, Vocab
A hyperbola is the set of all points in the plane in which the difference of the
distances form two distinct fixed points, called foci, is constant. Like Ellipses, they
can have a vertical or horizontal orientation.
As the graphs on the previous slide show, a hyperbola contains two
symmetrical parts called branches.
A hyperbola also has two axes of symmetry. The transverse axis of
symmetry contains the vertices and, if it were extended, the foci of the
hyperbola. The vertices of a hyperbola are the endpoints of the transverse
axis.
The conjugate axis of symmetry separates the two branches of the
hyperbola. The co-vertices of a hyperbola are the endpoints of the
conjugate axis. The transverse axis is not always longer than the conjugate
axis.
The standard form of the equation of a hyperbola depends on whether the
hyperbola’s transverse axis is horizontal or vertical.
The relationship 𝑎2 + 𝑏 2 = 𝑐 2 exists, where a is the length of the
semitranverse axis, b is the length of the conjugate axis, and c is the
distance between the center and the foci. For hyperbolas, the transverse
axis is not always larger than the conjugate axis.
Advanced Mathematical Concepts
Chapter 10
II. Standard Equations for a Hyperbola
Ex 1: Find the equation of the hyperbola with foci (5, 2) and (-1, 2) whose transverse axis is 4 units
long.
The location of the foci tells us that the transverse axis is parallel to the x-axis. Therefore, we will use the
formula
(𝑥−ℎ)2
𝑎2
−
(𝑦−𝑘)2
𝑏2
= 1. Now, to fill in the blanks.
Lets start with the center, which is the midpoint of the two foci.
You can sketch it out or use the midpoint formula.
5 + (-1) 2 + 2
or (2, 2)
2 , 2
Thus, h = 2 and k = 2 since (2, 2) is the center.
The transverse axis is 4 units long. Thus, 2a = 4 or a = 2. So, a2 = 4.
Use the equation b2 = c2 – a2 to find b2. Recall that c is the distance from the center to a focus. Here c = 3
(as shown on the graph).
b2 = 32 – 22 or 5
Advanced Mathematical Concepts
Chapter 10
(x - 2)2 (y - 2)2
=1
4
5
III. Graphing Hyperbolas
Ex 2: Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of
(y - 2)2 (x + 1)2
the graph
= 1. Then graph the equation.
64
36
Since the y terms are in the first expression, the hyperbola has a vertical transverse axis.
From the equation, h = -1, k = 2, a = 8, and b = 6. The center is
at (-1, 2).
4
The equations of the asymptotes are y – 2 = 3(x + 1).
The vertices are at (h, k  a) or (-1, 10) and (-1, -6).
Since c2 = a2 + b2, c = 10. Thus, the foci are located at (-1, 12)
and (-1, -8).
Graph the center, vertices, and the rectangular guide. Next
graph the asymptotes. Then sketch the hyperbola.
IV. General Form of a Hyperbola
Ex 3: Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of
the graph of 25x2 – 16y2 + 250x + 32y + 109 = 0. Then graph the equation.
Write the equation in standard form. Use the same process you used with ellipses.
25x2 – 16y2 + 250x + 32y + 109 = 0
25(x2 + 10x + ?) – 16(y2 – 2y + ?) = -109 + ? + ?
25(x2 + 10x + 25) – 16(y2 – 2y + 1) = -109 + 25(25) - 16(1)
25(x + 5)2 – 16(y – 1)2 = 500
(𝑥 + 5)2 (𝑦 − 1)2
−
=1
20
31.25
The center is at (-5, 1). Since the x terms are in the first expression, the transverse axis is horizontal.
Advanced Mathematical Concepts
Chapter 10
5 5
205
a = 2 5, b = 2 , c = 2
The foci are at -5 
205
2 , 1.
The vertices are at (-5  2 5, 1).
5
The asymptotes have equations y – 1 = 4(x + 5).
Graph the vertices and the rectangular guide. Next graph the
asymptotes. Then sketch the hyperbola.
V. Rectangular Hyperbolas
Rectangular Hyperbolas are a special case where they coordinate axes are the
asymptotes. The general form is xy = c, where c is a non-zero constant. The sign of
c determines the location of the branches of the hyperbola. If c is positive, the
branches are located in Q1 and Q3. If c is negative, the branches are located in Q2
and Q4.
Ex 4: Graph xy = 9.
Since c is positive, the hyperbola lies in the first and third
quadrants. The transverse axis is along the graph of y = x.
The coordinates of the vertices must satisfy the equation of
the hyperbola and also their graph must be points on the
transverse axis. Thus, the vertices are at (3, 3) and (-3, -3).
Advanced Mathematical Concepts
Chapter 10
VI. Eccentricity
𝒄
The shape of a hyperbola is determined by its eccentricity, which is defined as 𝒆 = . e is
𝒂
always between 0 and 1.
Write the equation of the hyperbola with center at (2, 1), a focus of (2, -3),
4
and eccentricity of .
3
Sketch the graph using the points given. Since the center and
focus have the same x-coordinate, the transverse axis is vertical.
Use the appropriate standard form.
The focus is 4 units below the center, so c = 4. Now use the
eccentricity to find the values of a2 and b2.
c
e =a
4 4
3 =a
3 =a
9 = a2
The equation is
b2 = a2(e2 - 1)
16
b2 = 9 9 - 1
b2 = 16 - 9 or 7
(y - 1)2 (x - 2)2
9 - 7 = 1.