Hyperbolas, Exams of Pre-Calculus

Download Hyperbolas and more Exams Pre-Calculus in PDF only on Docsity!

Hyperbolas

Lesson 10-

A

hyperbola

is a set of points in a plane the difference of whose

distances from two fixed points, called

foci

, is a constant. F

1

F

2

d

1

d

2

P

For any point P that is on thehyperbola, d

2

• d

1

is always the

same. In this example, the origin is thecenter of the hyperbola. It ismidway between the foci.

F

F

V

V

C

The figure at the left is anexample of a hyperbola whosebranches open up and downinstead of right and left.

Since the transverse axis isvertical, this type ofhyperbola is often referredto as a vertical hyperbola. When the transverse axis ishorizontal, the hyperbola isreferred to as a horizontalhyperbola.

(x – h)

2

(y – k)

2

a

2

b

2

HorizontalHyperbola

(y – k)

2

(x – h)

2

b

2

a

2

Vertical Hyperbola

The center of a hyperbola is at the point (h, k) in either form For either hyperbola, a

2

• b

2

= c

2

Where c is the distance from the center to a focus point. The equations of the asymptotes for a HORIZONTAL HYPERBOLA are^ y =

(x – h) + k

and

y =

(x – h) + k

b a

b - a

Graph:

(x + 2)

2

(y – 1)

2

c

2

= 9 + 25 = 34

c =

√

34 = 5.

Foci: (-7.83, 1) and (3.83, 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

(y – 1)

2

(x – 3)

2

c

2

= 9 + 4 = 13

c =

√

13 = 3.

Foci: (3, 4.61) and (3, -2.61)

Center: (3, 1)

The hyperbola is vertical Graph: 9y

2

x

2

• 18y + 24x – 63 = 0

9(y

2

• 2y + ___) – 4(x

2

• 6x + ___) = 63 + ___ – ___

9

1

9

36

9(y – 1)

2

• 4(x – 3

(^2) )

= 36

Asymptotes:

y =

(x – 3) + 1 2 3

y =

(x – 3) + 1 2 3

Find the standard form equation of thehyperbola that is graphed at the right Vertical hyperbola (y – k) Center: (-1, -2)

2

(x – h)

2

b

2

a

2

a = 3 and

b = 5

(y + 2)

2

(x + 1)

2

M

2

M

1

An explosion is recorded by two microphones that are twomiles apart. M1 received the sound 4 seconds before M2.assuming that sound travels at 1100 ft/sec, determine thepossible locations of the explosion relative to the locationsof the microphones. (5280, 0)

(-5280, 0)

E(x,y)

Let us begin by establishing a coordinate systemwith 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

x

2

y

2

a

2

b

2

c

2

= a

2

• b

2

5280

2

= 2200

2

• b

2

b

2

= 23,038,

The explosion must by on the

hyperbola