10-5 Hyperbolas
Quick Review
A hyperbola is the set of all points P such that the absolute value of the difference of the distances from P to two fixed points, the foci, is constant. There are two standard forms of hyperbolas centered at the origin. If
x
2
a
2
−
y
2
b
2
=
1
,
fraction x squared , over eh squared end fraction . minus . fraction y squared , over b squared end fraction . equals 1 comma the asymptotes are
y
=
±
b
a
x
,
y equals plus minus , b over eh , x comma the transverse axis is horizontal with vertices
(
±
a
,
0
)
,
open plus minus eh comma 0 close comma and the foci are
(
±
c
,
0
)
.
open plus minus c comma 0 close . If
y
2
a
2
−
x
2
b
2
=
1
,
fraction y squared , over eh squared end fraction . minus . fraction x squared , over b squared end fraction . equals 1 comma the asymptotes are
y
=
±
a
b
x
,
y equals plus minus , eh over b , x comma the transverse axis is vertical with vertices
(
0
,
±
a
)
,
open 0 comma plus minus eh close comma and the foci are
(
0
,
±
c
)
.
open 0 comma plus minus c close . In either case,
c
2
=
a
2
+
b
2
.
c squared , equals , eh squared , plus , b squared , .
Example
Find the foci of the graph of
x
2
25
−
y
2
9
=
1
.
fraction x squared , over 25 end fraction , minus , fraction y squared , over 9 end fraction , equals 1 .
The equation is in the form
x
2
a
2
−
y
2
b
2
=
1
,
fraction x squared , over eh squared end fraction . minus . fraction y squared , over b squared end fraction . equals 1 comma so the transverse axis is horizontal;
a
2
=
25
eh squared , equals 25 and
b
2
=
9
.
b squared , equals 9 .
Using the Pythagorean Theorem to find c,
c
=
25
+
9
=
34
≈
5.8
.
c equals , square root of 25 plus 9 end root , equals square root of 34 almost equal to 5.8 .
The foci,
(
±
c
,
0
)
,
open plus minus c comma 0 close comma are approximately (5.8, 0) and
(
−
5
.
8
,
0
)
.
open negative 5 . 8 comma 0 close .
Exercises
Find the foci of each hyperbola. Graph the hyperbola.
-
x
2
36
−
y
2
225
=
1
fraction x squared , over 36 end fraction , minus . fraction y squared , over 225 end fraction . equals 1
-
y
2
400
−
x
2
169
=
1
fraction y squared , over 400 end fraction . minus . fraction x squared , over 169 end fraction . equals 1
-
x
2
121
−
y
2
81
=
1
fraction x squared , over 121 end fraction . minus , fraction y squared , over 81 end fraction , equals 1
Write an equation of a hyperbola with the given foci and vertices.
- foci
(
±
17
,
0
)
,
open plus minus 17 comma 0 close comma vertices
(
±
8
,
0
)
open plus minus 8 comma 0 close
- foci
(
0
,
±
25
)
,
open 0 comma plus minus 25 close comma vertices
(
0
,
±
7
)
open 0 comma plus minus 7 close
- Find an equation that models the hyperbolic path of a spacecraft around a planet if
a
=
107
,
124
km
eh equals 107 comma 124 , km and
c
=
213
,
125
.
9
km
.
c equals 213 comma 125 . 9 , km , .
10-6 Translating Conic Sections
Quick Review
Substitute
(
x
−
h
)
open x minus h close for x and
(
y
−
k
)
open y minus k close for y to translate graphs of the conic sections.
Example
Identify and describe the conic section represented by the equation
2
x
2
+
3
y
2
+
4
x
+
12
y
−
22
=
0
.
2 bold italic x squared , plus 3 , bold italic y squared , plus 4 bold italic x plus 12 bold italic y minus 22 equals 0 .
By completing the square, the equation becomes
(
x
+
1
)
2
18
+
(
y
+
2
)
2
12
=
1
,
fraction open , x plus 1 , close squared , over 18 end fraction . plus . fraction open , y plus 2 , close squared , over 12 end fraction . equals 1 comma which is an ellipse.
The center is
(
−
1
,
−
2
)
open negative 1 comma negative 2 close and the major axis is horizontal.
Using the equation
c
2
=
a
2
−
b
2
,
c
=
6
,
c squared , equals , eh squared , minus , b squared , comma . c equals square root of 6 comma so the distance from the center of the ellipse to the foci is
6
.
square root of 6 .
Since the ellipse is centered at
(
−
1
,
−
2
)
open negative 1 comma negative 2 close and the major axis is horizontal, the foci are located
6
square root of 6 to the left and right of this center.
The foci are at
(
−
1
+
6
,
−
2
)
open . negative 1 plus square root of 6 comma negative 2 . close and
(
−
1
−
6
,
−
2
)
.
open . negative 1 minus square root of 6 comma negative 2 . close . .
Exercises
Write an equation of a conic section with the given characteristics.
- a circle with center (1, 1); radius 5
- an ellipse with center
(
3
,
−
2
)
;
open 3 comma negative 2 close semicolon vertical major axis of length 6; minor axis of length 4
- a hyperbola with vertices (3, 3) and (9, 3); foci (1, 3) and (11, 3)
Identify the conic section and sketch the graph. If it is a parabola, give the vertex. If it is a circle, give the center and radius. If it is an ellipse or a hyperbola, give the center and foci.
-
−
x
2
+
y
2
+
4
y
−
16
=
0
negative , x squared , plus , y squared , plus 4 y minus 16 equals 0
-
x
2
+
y
2
+
3
x
−
4
y
−
9
=
0
x squared , plus , y squared , plus 3 x minus 4 y minus 9 equals 0
-
x
2
+
x
−
y
−
42
=
0
x squared , plus x minus y minus 42 equals 0
