Hyberbola Conic Sections. Hyperbola The plane can intersect two nappes of the cone resulting in a hyperbola.

Hyberbola

Conic Sections

Hyperbola The plane can intersect

two nappes of the cone resulting in a hyperbola.

Hyperbola - DefinitionA hyperbola is the set of all points in a plane such that the difference in the distances from two points (foci) is constant.

| d1 – d2 | is a constant value.

Finding An Equation

Hyperbola

Hyperbola - DefinitionWhat is the constant value for the difference in the distance from the two foci? Let the two foci be (c, 0) and (-c, 0). The vertices are (a, 0) and (-a, 0).

| d1 – d2 | is the constant.

If the length of d2 is subtracted from the left side of d1, what is the length which remains?

| d1 – d2 | = 2a

Hyperbola - EquationFind the equation by setting the difference in the distance from the two foci equal to 2a.

| d1 – d2 | = 2a

2 2

22

1

2

( )

( )

x c y

d c

d

x y

Hyperbola - EquationSimplify:

2 2 2 2( ) ( ) 2x c y x c y a

Remove the absolute value by using + or -.

2 2 2 2( ) ( ) 2x c y x c y a

Get one square root by itself and square both sides.

2 2 2 2( ) ( ) 2x c y x c y a

2 22 2 2 2( ) ( ) 2x c y x c y a

2 2 2 2 2 2 2( ) ( ) 4 ( ) 4x c y x c y a x c y a

Hyperbola - Equation

Subtract y2 and square the binomials.

2 2 2 2 2 2 2( ) ( ) 4 ( ) 4x c y x c y a x c y a

2 2 2 2 2 2 22 2 4 ( ) 4x xc c x xc c a x c y a

Solve for the square root and square both sides.

2 2 24 4 4 ( )xc a a x c y

222 2 2( )xc a a x c y

2 2 2( )xc a a x c y

Hyperbola - Equation

Square the binomials and simplify.

222 2 2( )xc a a x c y

2 2 2 4 2 2 22 ( )x c xca a a x c y

2 2 2 4 2 2 2 22 2x c xca a a x xc c y 2 2 2 4 2 2 2 2 2 2 22 2x c xca a a x xca a c a y 2 2 4 2 2 2 2 2 2x c a a x a c a y

Get x’s and y’s together on one side. 2 2 2 2 2 2 2 2 4x c a x a y a c a

Hyperbola - Equation

Factor.

2 2 2 2 2 2 2 2 4x c a x a y a c a

2 2 2 2 2 2 2 2x c a a y a c a

Divide both sides by a2(c2 – a2)

2 2 2 2 2 22 2

2 2 2 2 2 2 2 2 2

x c a a c aa y

a c a a c a a c a

2 2

2 2 21

x y

a c a

Hyperbola - Equation

Let b2 = c2 – a2

2 2

2 2 21

x y

a c a

2 2

2 21

x y

a b where c2 = a2 + b2

If the graph is shifted over h units and up k units, the equation of the hyperbola is:

Hyperbola - Equation

2 2

2 21

x h y k

a b

where c2 = a2 + b2

Recognition:How do you tell a hyperbola from an ellipse?

Answer:A hyperbola has a minus (-) between the terms while an ellipse has a plus (+).

Graph - Example #1

Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Center: (-3, -2)

The hyperbola opens in the “x” direction because “x” is positive.

Transverse Axis: y = -2

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Vertices (2, -2) (-4, -2)

Construct a rectangle by moving 4 units up and down from the vertices.

Construct the diagonals of the rectangle.

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Draw the hyperbola touching the vertices and approaching the asymptotes.

Where are the foci?

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

The foci are 5 units from the center on the transverse axis.

2 2 2c a b 2

2

9 16

25

5

c

c

c

Foci: (-6, -2) (4, -2)

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Find the equation of the asymptote lines.

Slope =

Use point-slope formy – y1 = m(x – x1) since the center is on both lines.

3

4

-4

4

3

Asymptote Equations

Graph - Example #2

Hyperbola

Hyperbola - Graph

2 210 5 40 10 15 0y x y x

Sketch the graph without a grapher:

Recognition:How do you determine the type of conic section?

Answer:The squared terms have opposite signs.

Write the equation in hyperbolic form.

Hyperbola - Graph

2 210 5 40 10 15 0y x y x

Sketch the graph without a grapher:

2 210 40 5 10 15y y x x

2 210 4 ?? 5 2 ?? 15y y x x

2 210 4 5 2 54 1 401 5y y x x

2 210 2 5 1 50y x

2 210 2 5 1 50

50 50 50

y x

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Center: (-1, 2)

Transverse Axis Direction:Up/Down

Equation: x=-1

Vertices: Up/Down from the center or 5 1, 2 5

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Plot the rectangular points and draw the asymptotes.

Sketch the hyperbola.

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Plot the foci.

Foci:

2 2 2

2

2

10 5

15

15

c a b

c

c

c

1, 2 15

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Equation of the asymptotes:

52 1

10y x

22 1

2y x

Finding an EquationA problem for CSI!

Hyperbola

Hyperbola – Find an EquationThe sound of a gunshot was recorded at one microphone 0.54 seconds before being recorded at a second microphone. If the two microphones are 2,000 ft apart. Provide a model for the possible locations of the gunshot. (The speed of sound is 1100 ft/sec.)

The time between the shots can be used to calculate the difference in the distance from the two microphones.

1100 ft/sec * 0.54 sec = 594 ft. The constant difference in distance from the microphones is 594 ft.

Since the difference is constant, the equation must be a hyperbola. The points on the hyperbola are possible positions for the gunshot.

Hyperbola – Find an EquationTwo microphones are stationed 2,000 ft apart. The difference in distance between the microphones is 594 ft.

Let the center be at (0,0). The foci must be 2,000 ft apart.

The vertices are a possible position for the gunshot. The difference in the distance must be 594 feet between the vertices.

V

Let the vertices be at (+z, 0). Assuming z>0, then(z-(-1000)) – (1000-z) = 594z+1000-1000+z = 5942z = 594 or z = 297.

Hyperbola – Find an Equation

V

Start finding the model of the hyperbola.

V(297, 0)

V

V(-2970, 0)

2 2

21

88209

x y

b

The distance from the center to the foci (c) is 1000 ft. Find b.

Oops! We could have remembered the constant difference in distance is 2a! 2a = 594, a = 297.

2972 = 88209

Hyperbola – Find an Equation

V

V(294, 0)

V

V(294, 0)

The model is:

2 2 2

2 2 2

2

1000 297

911791

c a b

b

b

2 2 2 2

2 21 1

72900 911791 270 954.9

x y x yor

Hyperbola – Find an EquationThe gunshot was calculated to be at some point along the hyperbola.

Conic Section Recogition

Recognizing a Conic SectionParabola -

One squared term. Solve for the term which is not squared. Complete the square on the squared term.

Ellipse - Two squared terms. Both terms are the same “sign”.

Circle - Two squared terms with the same coefficient.

Hyperbola -

Two squared terms with opposite “signs”.