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