analyticgeometry-hyperbola-111119062959-phpapp02.pdf

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1. Hyperbola

2. A hyperbola is the set of points in a plane, the
absolute value of the difference of whose
distances from two fixed points, called foci, is
a constant.
F
2 F
1

3. F
2 F
1
d
1
d
2
P
d
2
– d
1
is always the
same.

4. F F
V V

5. F F
V V
C

6. F
F
V
V
C

7. The center is at the point (0, 0)
c
2
= a
2
+ b
2
c is the distance from the
center to a focus point.
The foci are at (c, 0) and (-c, 0)
1
2
2
2
2
=-
b
y
a
x

8. 1
2
2
2
2
=-
b
y
a
x
The conjugate points are at (0, b)
and (0, -b)
The vertices are at (a, 0) and (-a, 0)
Length of the latus rectum is 2b
2
a

9. 1
2
2
2
2
=-
b
y
a
x
Ends of the latus rectum:
÷
÷
ø
ö
ç
ç
è
æ
a
b
cL
2
1
, ÷
÷
ø
ö
ç
ç
è
æ
-
a
b
cL
2
2
,
÷
÷
ø
ö
ç
ç
è
æ
--
a
b
cL
2
3
,
÷
÷
ø
ö
ç
ç
è
æ
-
a
b
cL
2
4,

10. Horizontal
Hyperbola
Equation of the directrix
1
2
2
2
2
=-
b
y
a
x
a
bx
y±=

11. The center is at the point (0, 0)
c
2
= a
2
+ b
2
c is the distance from the
center to a focus point.
The foci are at (0, c) and (0, -c)
1
2
2
2
2
=-
b
x
a
y

12. 1
2
2
2
2
=-
b
x
a
y
The conjugate points are at (b, 0)
and (-b, 0)
The vertices are at (0, a) and (0, -a)
Length of the latus rectum is 2b
2
a

13. 1
2
2
2
2
=-
b
x
a
y
Ends of the latus rectum:
÷
÷
ø
ö
ç
ç
è
æ
c
a
b
L ,
2
1 ÷
÷
ø
ö
ç
ç
è
æ
-c
a
b
L ,
2
2
÷
÷
ø
ö
ç
ç
è
æ
-- c
a
b
L ,
2
3 ÷
÷
ø
ö
ç
ç
è
æ
-c
a
b
L ,
2
4

14. Vertical
Hyperbola
Equation of the directrix
1
2
2
2
2
=-
b
x
a
y
b
ax
y±=

15. 1
169
22
=-
yx
Example 1.
1
2
2
2
2
=-
b
y
a
x
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5

16. Example 1.
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5
V
1
(3, 0), V
2
(-3, 0)
F
1
(5, 0), F
2
(-5, 0)
Center at (0, 0)
Center at (0, 0)
V
1
(a, 0), V
2
(-a, 0)
F
1
(c, 0), F
2
(-c, 0)

17. Example 1.
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5
LR = 2b
2
= 2(4)
2
=32

a
Conjugate points
(0, 4), (0, -4)
3 3
Conjugate Points
(0, b), (0, -b)
LR = 2b
2
a

18. Example 1.
LR = 2b
2
= 32
a3
÷
÷
ø
ö
ç
ç
è
æ
a
b
cL
2
1
,
÷
÷
ø
ö
ç
ç
è
æ
-
a
b
cL
2
2
,
÷
÷
ø
ö
ç
ç
è
æ
--
a
b
cL
2
3, ÷
÷
ø
ö
ç
ç
è
æ
-
a
b
cL
2
4
,
c
2
= 25; c = 5
Endpoints of LR
÷
ø
ö
ç
è
æ
3
16
,5
1
L ÷
ø
ö
ç
è
æ
-
3
16
,5
2
L
÷
ø
ö
ç
è
æ
--
3
16
,5
3
L ÷
ø
ö
ç
è
æ
-
3
16
,5
4L

19. Example 1.
Asymptotes
a
bx
y±=
c
2
= 25; c = 5
b
2
= 16; b = 4
a
2
= 9; a = 3
1
169
22
=-
yx
3
4x
y=
3
4x
y-=
034 =-yx
034 =+yx
Asymptotes

20. 1
169
22
=-
yx
Example 1.
V
1
(3, 0), V
2
(-3, 0)
F
1
(5, 0), F
2
(-5, 0)
Center at (0, 0)
÷
ø
ö
ç
è
æ
3
16
,5
1L ÷
ø
ö
ç
è
æ
-
3
16
,5
2L
÷
ø
ö
ç
è
æ
--
3
16
,5
3L ÷
ø
ö
ç
è
æ
-
3
16
,5
4L
Conjugate points
(0, 4), (0, -4)
Endpoints of LR
Asymptotes
034 =-yx
034 =+yx
Symmetric at x-axis
0=y

21. 1
169
22
=-
yx
Example 1.
4x + 3y = 0
4x – 3y = 0

22. 1
169
22
=-
xy
Example 2.
1
2
2
2
2
=-
b
x
a
y
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5

23. Example 2.
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5
V
1
(0, 3), V
2
(0, -3)
F
1
(0, 5), F
2
(0, -5)
Center at (0, 0)
Center at (0, 0)
V
1
(0, a), V
2
(0, -a)
F
1
(0, c), F
2
(0, -c)

24. Example 2.
a
2
= 9; a = 3
b
2
= 16; b = 4
c
2
= 25; c = 5
LR = 2b
2
= 2(4)
2
=32

a
Conjugate points
(4, 0), (-4, 0)
3 3
Conjugate Points
(b, 0), (-b, 0)
LR = 2b
2
a

25. Example 2.
LR = 2b
2
= 32
a3
÷
÷
ø
ö
ç
ç
è
æ
c
a
b
L ,
2
1 ÷
÷
ø
ö
ç
ç
è
æ
- c
a
b
L ,
2
2
÷
÷
ø
ö
ç
ç
è
æ
-- c
a
b
L ,
2
3 ÷
÷
ø
ö
ç
ç
è
æ
-c
a
b
L ,
2
4
c
2
= 25; c = 5
Endpoints of LR
÷
ø
ö
ç
è
æ
5,
3
16
1
L ÷
ø
ö
ç
è
æ
-5,
3
16
2
L
÷
ø
ö
ç
è
æ
-- 5,
3
16
3
L ÷
ø
ö
ç
è
æ
-5,
3
16
4L

26. Example 2.
Asymptotes
b
ax
y±=
c
2
= 25; c = 5
b
2
= 16; b = 4
a
2
= 9; a = 3
1
169
22
=-
xy
4
3x
y=
4
3x
y-=
043 =-yx
043 =+yx
Asymptotes

27. 1
169
22
=-
xy
Example 2.
V
1
(0, 3), V
2
(0, -3)
F
1
(0, 5), F
2
(0, -5)
Center at (0, 0)
÷
ø
ö
ç
è
æ
5,
3
16
1L ÷
ø
ö
ç
è
æ
-5,
3
16
2L
÷
ø
ö
ç
è
æ
-- 5,
3
16
3L ÷
ø
ö
ç
è
æ
-5,
3
16
4L
Conjugate points
(4, 0), (-4, 0)
Endpoints of LR
Asymptotes
043 =-yx
043 =+yx
Symmetric at y-axis
0=x

28. 1
169
22
=-
xy
Example 2.
3x + 4y = 0
3x – 4y = 0

30. 1
)()(
2
2
2
2
=
-
-
-
b
ky
a
hx
1
)()(
2
2
2
2
=
-
-
-
b
hx
a
ky
C (h, k)

31. )(hx
a
b
ky -=-
)(hx
a
b
ky --=-
÷
÷
ø
ö
ç
ç
è
æ
+-
a
b
kch
2
,
÷
÷
ø
ö
ç
ç
è
æ
--
a
b
kch
2
,
÷
÷
ø
ö
ç
ç
è
æ
++
a
b
kch
2
,
÷
÷
ø
ö
ç
ç
è
æ
-+
a
b
kch
2
,
Center (h, k)
F
1
(h+c, k)F
2
(h-c, k)
V
1
(h+a, k)
V
2
(h-a, k)

32. Example 1.
1
25
)1(
9
)2(
22
=
-
-
+ yx
1
25
)1(
9
)2(
22
=
-
-
+ yx

33. Graph: (x + 2)
2
(y – 1)
2

9 25
c
2
= 9 + 25 = 34
c = Ö34 = 5.83
Foci: (-7.83, 1) and (3.83, 1)
– = 1
Center: (-2, 1)
Horizontal hyperbola
Vertices: (-5, 1) and (1, 1)
Asymptotes: y = (x + 2) + 1
5
3
y = (x + 2) + 1
5
3
-

34. Example 2.
1
16
)1(
9
)2(
22
=
-
-
- xy
1
16
)1(
9
)2(
22
=
-
-
- xy

35. Properties of this Hyperbola
Center ((1,2)
525
16943
4;16
3;9
22
222
2
2
==
+=+=
+=
==
==
c
c
bac
bb
aa

36. Foci: (1,7), (1, -3)
Vertices: (1,5), (1, -1)

The hyperbola is vertical
Transverse
Axis: parallel to
y-axis

37. Properties of this Hyperbola
Asymptotes: ( )1
4
3
2 -±=- xy
01143
3384
0543
3384
)1(384
=-+
+-=-
=+-
-=-
-=-
yx
xy
yx
xy
xy

38. )3,
3
19
();52,
3
16
1('
)3,
3
13
();52,
3
16
1('
)7,
3
19
();52,
3
16
1(
)7,
3
13
();52,
3
16
1(
--+
----
++
-+-
R
L
R
L
Latera Recta

40. Ax
2
+ By
2
+ Cx + Dy + E = 0
1.Group the x terms together and y
terms together.
2.Complete the square.
3.Express in binomial form.
4.Divide by the constant term,
where the first term has a positive
sign.

41. Example 1.
9x
2
– 4y
2
– 18x – 16y + 29 = 0

42. (y – 1)
2
(x – 3)
2

4 9
c
2
= 9 + 4 = 13
c = Ö13 = 3.61
Foci: (3, 4.61) and (3, -2.61)
– = 1
Center: (3, 1)
The hyperbola is vertical
Graph: 9y
2
– 4x
2
– 18y + 24x – 63 = 0
9(y
2
– 2y + ___) – 4(x
2
– 6x + ___) = 63 + ___ – ___ 91 9 36
9(y – 1)
2
– 4(x – 3)
2
= 36
Asymptotes: y = (x – 3) + 1
2
3
y = (x – 3) + 1
2
3
-

43. Find the standard form of the
equation of a hyperbola given:
49 = 25 + b
2
b
2
= 24
Horizontal hyperbola
Foci: (-7, 0) and (7, 0)
Vertices: (-5, 0) and (5, 0)
10
8
F FV V
Center: (0, 0)
c
2
= a
2
+ b
2

(x – h)
2
(y – k)
2
a
2
b
2
– = 1
x
2
y
2
25 24
– = 1
a
2
= 25 and c
2
= 49 C

44. Center: (-1, -2)
Vertical hyperbola
Find the standard form equation of the
hyperbola that is graphed at the right
(y – k)
2
(x – h)
2
b
2
a
2
– = 1
a = 3 and b = 5
(y + 2)
2
(x + 1)
2
25 9
– = 1

45. M
2 M
1
An explosion is recorded by two microphones that are two
miles apart. M1 received the sound 4 seconds before M2.
assuming that sound travels at 1100 ft/sec, determine the
possible locations of the explosion relative to the locations
of the microphones.
(5280, 0)(-5280, 0)
E(x,y)
Let us begin by establishing a coordinate system
with the origin midway between the microphones
Since the sound reached M
2
4 seconds after it
reached M
1
, the difference in the distances from
the explosion to the two microphones must be
d
2
d
1
1100(4) = 4400 ft wherever E is
This fits the definition of an hyperbola with foci at M
1
and M
2
Since d
2
– d
1
= transverse axis, a = 2200
x
2
y
2
4,840,000 23,038,400
– = 1
x
2
y
2
a
2
b
2
– = 1
c
2
= a
2
+ b
2

5280
2
= 2200
2
+ b
2

b
2
= 23,038,400
The explosion must be on the
hyperbola