10-4 Hyperbolas
A hyperbola contains two symmetrical parts called
branches.
A hyperbola 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.
Holt Algebra 2
10-4 Hyperbolas
Holt Algebra 2
10-4 Hyperbolas
Holt Algebra 2
10-4 Hyperbolas
The focus of a hyperbola is c units away from the
2
2
center on the transverse axis where c2 = a + b .
a is the length from the center to a vertex on the
transverse axis.
b is the length from the center to a co-vertex on the
conjugate axis.
Holt Algebra 2
10-4 Hyperbolas
The asymptotes for a hyperbola can be written from
the equation:
Horizontal Hyperbola
Vertical Hyperbola
𝑏
𝑦 − 𝑘 = ± (𝑥 − ℎ)
𝑎
𝑎
𝑦 − 𝑘 = ± (𝑥 − ℎ)
𝑏
Holt Algebra 2
10-4 Hyperbolas
Find the vertices, co-vertices, foci, and
asymptotes of each hyperbola, and then graph.
y2
=1
9
x2 –
49
Step 1 The equation is in the
form
x2
a2
–
y2
2 = 1 so the
b
transverse axis is horizontal
with center (0, 0).
Holt Algebra 2
10-4 Hyperbolas
Example 3A Continued
Step 2 Because a = 7 and b = 3, the vertices are
(–7, 0) and (7, 0) and the co-vertices are
(0, –3) and (0, 3).
Step 3 The equations of the asymptotes are
y=
Holt Algebra 2
3
7
x
and y = –
3
7
x.
10-4 Hyperbolas
Example 3A Continued
Step 4 Draw a box by using
the vertices and covertices. Draw the
asymptotes through
the corners of the box.
Step 5 Draw the hyperbola
by using the vertices
and the asymptotes.
Holt Algebra 2
10-4 Hyperbolas
Find the vertices, co-vertices, foci, and
asymptotes of each hyperbola, and then graph.
(x – 3)2
(y + 5)2
–
=1
9
49
The equation is in the form
(x – h)2
(y – k)2
–
= 1 , so the
2
2
a
b
transverse axis is horizontal
with center (3, –5).
Holt Algebra 2
10-4 Hyperbolas
Example 3B Continued
Step 2 Because a = 3 and b =7, the
vertices are (0, –5) and (6, –5)
and the co-vertices are (3, –12)
and (3, 2) .
Step 3 The equations of the asymptotes are
y+5=
Holt Algebra 2
7
3
(x – 3) and y = –
7
3
(x – 3).
10-4 Hyperbolas
Example 3B Continued
Step 4 Draw a box by using
the vertices and covertices. Draw the
asymptotes through
the corners of the box.
Step 5 Draw the hyperbola
by using the vertices
and the asymptotes.
Holt Algebra 2
10-4 Hyperbolas
Write an equation in standard form for the
hyperbola.
Step 1 Identify the form of
the equation.
The graph opens horizontally, so the equation
will be in the form of
Holt Algebra 2
x2
a2
–
y2
.
2 = 1
b
10-4 Hyperbolas
Example 2A Continued
Step 2 Identify the center and the vertices.
The center of the graph is (0, 0), and the
vertices are (–6, 0) and (6, 0), and the covertices are (0, –6), and (0, 6). So a = 6, and
b = 6.
Step 3 Write the equation.
Because a = 6 and b = 6, the equation of the
graph is
x2
62
Holt Algebra 2
–
2
y2
x
= 1, or
2
6
36
y2
–
= 1.
36
10-4 Hyperbolas
Write an equation in standard form for the
hyperbola.
(𝑦 − 𝑘)2
𝑥−ℎ
−
2
𝑎
𝑏2
2
=1
Center: (1, -5)
Transverse axis: a = 1
Conjugate axis: b = 3
(𝑦 + 5)2
𝑥−1
−
2
1
32
Holt Algebra 2
2
=1
10-4 Hyperbolas
Write an equation in standard form for a
hyperbola with center at the origin, vertex (4, 0),
and focus (10, 0).
Step 1 Because the vertex and the focus are on the
horizontal axis, the transverse axis is
horizontal and the equation is in the form
2
x2 – y = 1 .
2
a2
Holt Algebra 2
b
10-4 Hyperbolas
Example 2B Continued
Step 2 Use a = 4 and c = 10; Use c2 = a2 + b2 to
solve for b2.
102 = 42 + b2 Substitute 10 for c, and 4 for a.
84 = b2
Step 3 The equation of the hyperbola is
Holt Algebra 2
x2
16
y2
–
= 1.
84
10-4 Hyperbolas
HW pg. 748
#’s 8 – 15, 18 - 29
Holt Algebra 2