Transcript 11.2 Hyperbolas - Martin's Classroom Website

Objectives:
Define a hyperbola
2. Write the equation of a hyperbola
3. Identify important characteristics of hyperbolas
4. Graph hyperbolas
1.
Hyperbola
 The set of all points for which the difference of the
distances from two points is constant.
Equation of a Hyperbola Centered
on the Origin
Characteristics of a Hyperbola
Important Facts:






The hyperbola bends toward
the foci
The positive term
determines which way the
hyperbola opens
The distance between the
foci is 2c
The distance between the
vertices is 2a
The center is the midpoint
between the foci and the
midpoint between the
vertices
c2 = a2 + b2
Example #1
 Show that the graph of the equation is a hyperbola.
Graph it and its asymptotes. Find the equations of the
asymptotes, and label the foci and the vertices.
2
2
4 y  16x  64
4 y 2 16 x 2 64


64
64
64
2
2
y
x

1
16 4
y 2 x2
 2 1
2
4
2
a  4, b  2
c  42  22
c  16  4
c  20  2 5
Asymptotes :
a
y x
b
4
y x
2
y  2 x
Example #1
 Show that the graph of the equation is a hyperbola.
Graph it and its asymptotes. Find the equations of the
asymptotes, and label the foci and the vertices.
2
2
4 y  16x  64
Equation :
2
2
y
x

1
2
2
4
2
Foci :
0,2 5 
Asymptotes :
y  2 x
Vertices :
0,4 
y
10
9
8
7
6
5
4
3
2
1
–10–9 –8 –7 –6 –5 –4 –3 –2 –1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
1 2 3 4 5 6 7 8 9 10
x
Example #2
 Graph the following hyperbola using a graphing
calculator.
2
2
5 x  4 y  20
 4 y 2  20  5 x 2
2
20

5
x
y2 
4
20  5 x 2
y
4
Example #3A
A. Find the equation of the hyperbola that has vertices
at (2, 0) and (-2, 0) and passes through 4, 3


Then sketch its graph by using the asymptotes, and
label the foci.
With the vertices on the
2
x
2
 y 1
4
c  2 1
2
 4 1
 5
2
b
y x
a
1
y x
2
x-axis, this implies a = 2.
x2 y 2
 2 1
2
2
b
4
2
2
 3
2

2
2
b
16 3
 2 1
4 b
1
3
4  2 1
b
3
 2  3
b
3  3b 2
b 1
Example #3A
A. Find the equation of the hyperbola that has vertices
at (2, 0) and (-2, 0) and passes through 4, 3


Then sketch its graph by using the asymptotes, and
label the foci.
y
6
Equation :
Foci :

x2
2
 y 1
4
5 ,0

Asymptotes :
1
y x
2
5
4
3
2
1
–6
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
–6
1
2
3
4
5
6
x
Example #3B
B. Find the equation of a hyperbola with y-intercepts at
1
±7 and an asymptote at y  x
3
With it intersecting the y-axis,
this implies that a = 7. From
the equation of the asymptote
we get:
a
7 1
 
b
b 3
b  21
y
2
7
2

x
2
2
21
1
Example #3C
C. Find the equation of a hyperbola with foci at (±8, 0)


and a vertex at 5 2,0
a  5 2, c  8
8
5 2 
2
64  50  b
14  b
2
b  14
2
b
2
x
2
8
2

y
2
5 2 
2
x2 y2

1
64 50
1
Example #4
 An airplane crashed and was heard by a park ranger
and by a family camping in a park. The park ranger
and the family are ¼ mile apart and the ranger heard
the sound 1 second before the family. The speed of
sound in air is approximately 1100 feet per second.
Describe the possible locations of the plane crash.
The family and the ranger are placed at opposite foci of a hyperbolic
curve. The crash occurred closer to the ranger than the family so the
crash occurred on the branch of the hyperbola closest to the ranger.
Since sound travels at 1100 ft/sec, after 1 sec it will have traveled 1100 ft.
This implies the crash was 1100 ft closer to the ranger than the family,
which also means the vertices are 1100 ft apart. Since 1 mile has 5280 ft,
¼ a mile is 5280 ÷ 4 = 1320 ft, which is the distance between the foci.
Example #4
 Describe the possible locations of the plane crash.
Distance between foci:
Distance between vertices:
1320 ft
1100 ft
2a  1100
660  550 2  b 2
a  550
2c  1320
435,600  302,500  b
133,100  b
c  660
b  365
x2
550
2

1200
1000
2
800
600
2
400
200
Ranger
Family
–1200–1000–800 –600 –400 –200
–200
y2
365
y
2
1
The crash occurred somewhere on the left
branch of the hyperbola.
–400
–600
–800
–1000
–1200
200 400 600 800 10001200
x