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A Presentation for Analytic Geometry Topic
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Hyperbola
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.F2 F1
F2 F1
d1
d2
P
d2 – d1 is always the same.
F F
V V
F F
V V
C
F
F
V
V
C
The center is at the point (0, 0)
c2 = a2 + b2
c is the distance from the center to a focus point.
The foci are at (c, 0) and (-c, 0)
12
2
2
2
b
y
a
x
12
2
2
2
b
y
a
xThe conjugate points are at (0, b) and (0, -b)
The vertices are at (a, 0) and (-a, 0)
Length of the latus rectum is 2b2
a
12
2
2
2
b
y
a
x
Ends of the latus rectum:
a
bcL
2
1 ,
a
bcL
2
2 ,
a
bcL
2
3 ,
a
bcL
2
4 ,
Horizontal Hyperbola
Equation of the directrix
12
2
2
2
b
y
a
x
a
bxy
The center is at the point (0, 0)
c2 = a2 + b2
c is the distance from the center to a focus point.
The foci are at (0, c) and (0, -c)
12
2
2
2
b
x
a
y
12
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 2b2
a
12
2
2
2
b
x
a
y
Ends of the latus rectum:
c
a
bL ,
2
1
ca
bL ,
2
2
c
a
bL ,
2
3
c
a
bL ,
2
4
Vertical Hyperbola
Equation of the directrix
12
2
2
2
b
x
a
y
b
axy
1169
22
yx
Example 1.
12
2
2
2
b
y
a
x
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
Example 1.
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
V1(3, 0), V2 (-3, 0)
F1(5, 0), F2 (-5, 0)
Center at (0, 0)
Center at (0, 0)V1(a, 0), V2(-a, 0)
F1(c, 0), F2(-c, 0)
Example 1.
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
LR = 2b2 = 2(4)2 =32
a
Conjugate points(0, 4), (0, -4)
3 3
Conjugate Points(0, b), (0, -b)
LR = 2b2
a
Example 1.LR = 2b2 = 32
a 3
a
bcL
2
1 ,
a
bcL
2
2 ,
a
bcL
2
3 ,
a
bcL
2
4 ,
c2 = 25; c = 5
Endpoints of LR
3
16,51L
3
16,52L
3
16,53L
3
16,54L
Example 1.
Asymptotes
c2 = 25; c = 5b2 = 16; b = 4a2 = 9; a = 3
1169
22
yx
034 yx
034 yx
Asymptotes
1169
22
yx
Example 1.
V1(3, 0), V2 (-3, 0)
F1(5, 0), F2 (-5, 0)
Center at (0, 0)
3
16,51L
3
16,52L
3
16,53L
3
16,54L
Conjugate points(0, 4), (0, -4)
Endpoints of LR
Asymptotes
034 yx034 yx
Symmetric at x-axis
0y
1169
22
yx
Example 1.
4x + 3y = 0
4x – 3y = 0
1169
22
xy
Example 2.
12
2
2
2
b
x
a
y
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
Example 2.
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
V1(0, 3), V2 (0, -3)
F1(0, 5), F2 (0, -5)
Center at (0, 0)
Center at (0, 0)V1(0, a), V2(0, -a)
F1(0, c), F2(0, -c)
Example 2.
a2 = 9; a = 3
b2 = 16; b = 4
c2 = 25; c = 5
LR = 2b2 = 2(4)2 =32
a
Conjugate points(4, 0), (-4, 0)
3 3
Conjugate Points(b, 0), (-b, 0)
LR = 2b2
a
Example 2.LR = 2b2 = 32
a 3
c
a
bL ,
2
1
ca
bL ,
2
2
c
a
bL ,
2
3
c
a
bL ,
2
4
c2 = 25; c = 5
Endpoints of LR
5,3
161L
5,
3
162L
5,
3
163L
5,
3
164L
Example 2.
Asymptotes
c2 = 25; c = 5b2 = 16; b = 4a2 = 9; a = 3
1169
22
xy
043 yx
043 yx
Asymptotes
1169
22
xy
Example 2.
V1(0, 3), V2 (0, -3)
F1(0, 5), F2 (0, -5)
Center at (0, 0)
5,3
161L
5,
3
162L
5,
3
163L
5,
3
164L
Conjugate points(4, 0), (-4, 0)
Endpoints of LR
Asymptotes
043 yx043 yx
Symmetric at y-axis
0x
1169
22
xy
Example 2.
3x + 4y = 0
3x – 4y = 0
1)()(
2
2
2
2
b
ky
a
hx
1)()(
2
2
2
2
b
hx
a
ky
C (h, k)
)( hxa
bky
)( hxa
bky
Center (h, k)
F1(h+c, k)F2(h-c, k)
V1(h+a, k)
V2(h-a, k)
Example 1.
125
)1(
9
)2( 22
yx
125
)1(
9
)2( 22
yx
Graph: (x + 2)2 (y – 1)2 9 25
c2 = 9 + 25 = 34c = 34 = 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 53
y = (x + 2) + 153
-
Example 2.
116
)1(
9
)2( 22
xy
116
)1(
9
)2( 22
xy
Properties of this Hyperbola
Center ((1,2)
525
16943
4;16
3;9
22
222
2
2
c
c
bac
bb
aa
Foci: (1,7), (1, -3)
Vertices: (1,5), (1, -1)
The hyperbola is verticalTransverse Axis: parallel to y-axis
Properties of this Hyperbola
Asymptotes: 14
32 xy
01143
3384
0543
3384
)1(384
yx
xy
yx
xy
xy
)3,3
19();52,
3
161('
)3,3
13();52,
3
161('
)7,3
19();52,
3
161(
)7,3
13();52,
3
161(
R
L
R
L
Latera Recta
Ax2 + By2 + 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.
Example 1.
9x2 – 4y2 – 18x – 16y + 29 = 0
(y – 1)2 (x – 3)2 4 9
c2 = 9 + 4 = 13c = 13 = 3.61
Foci: (3, 4.61) and (3, -2.61)
– = 1
Center: (3, 1)
The hyperbola is vertical
Graph: 9y2 – 4x2 – 18y + 24x – 63 = 0
9(y2 – 2y + ___) – 4(x2 – 6x + ___) = 63 + ___ – ___ 91 9 36
9(y – 1)2 – 4(x – 3)2 = 36
Asymptotes: y = (x – 3) + 1 23
y = (x – 3) + 123
-
Find the standard form of the equation of a hyperbola given:
49 = 25 + b2
b2 = 24
Horizontal hyperbola
Foci: (-7, 0) and (7, 0)Vertices: (-5, 0) and (5, 0)
10
8
F FV V
Center: (0, 0)
c2 = a2 + b2
(x – h)2 (y – k)2
a2 b2– = 1
x2 y2
25 24– = 1
a2 = 25 and c2 = 49 C
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
b2 a2– = 1
a = 3 and b = 5
(y + 2)2 (x + 1)2
25 9– = 1
M2 M1
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 M2 4 seconds after it reached M1, the difference in the distances from the explosion to the two microphones must be
d2 d1
1100(4) = 4400 ft wherever E is
This fits the definition of an hyperbola with foci at M1 and M2
Since d2 – d1 = transverse axis, a = 2200
x2 y2
4,840,000 23,038,400– = 1
x2 y2
a2 b2
– = 1
c2 = a2 + b2
52802 = 22002 + b2
b2 = 23,038,400The explosion must be on the
hyperbola