What type of conic is each?. Hyperbolas 5.4 (M3)

What type of conic is each?What type of conic is each?

2 2

2 2

2 2

2 2

2 2

1. 4 16

2. ( 2) ( 1) 9

3. 16( 3) 9( 1) 225

4. 4 18

5. 2 4 8 81 0

x y

x y

x y

x y y

x x y y

HyperbolasHyperbolas

5.4 (M3)5.4 (M3)

What is the equation?What is the equation?

2 2 1x y


2 2

2 2

1

1

x y

x y


22 1

4

yx


22

22

14

14

yx

yx


22 1

9

xy

What is the equation?What is the equation?2

2

22

19

19

xy

xy

What is the equation of each What is the equation of each graph?graph?

2 2 1x y

22 1

9

xy

22 1

4

yx

a.

b.

c.


22 1

16

xy


2

22

116

116

xy

xy


2 2

116 4

x y

What is the equation?What is the equation?2 2

2 2

116 4

14 16

x y

y x


22 1

25

xy


2

22

125

125

xy

xy

2 2 1x y

22 1

9

xy

22 1

4

yx

22 1

16

xy

2 2

14 16

y x

22 1

25

xy

Place the equation with the graph.

1.

6.5.4.

3.2.

What do you notice?What do you notice?

Standard Formula of a hyperbola is Standard Formula of a hyperbola is like an ellipse with subtraction.like an ellipse with subtraction.

Its appearance is a broken ellipse Its appearance is a broken ellipse turned in the opposite direction.turned in the opposite direction.

There are 2 asymptotes.There are 2 asymptotes. Positive term is the direction the Positive term is the direction the

“parabola” opens.“parabola” opens. The term subtracted becomes the The term subtracted becomes the

axis you “cut” from the ellipse.axis you “cut” from the ellipse.

Hyperbola ExamplesHyperbola Examples

Hyperbola NotesHyperbola NotesHorizontal Transverse AxisHorizontal Transverse Axis

Center Center (0,0)(0,0)

Vertices (a,0) Vertices (a,0) && (-a,0)(-a,0)

Foci (c,0) &Foci (c,0) & (-c, 0)(-c, 0)

AsymptotAsymptoteses


EquationEquation

x 2

a2 y2

b2 1

Foci : c2 a2 b2


To find To find asymptotesasymptotes

y b

ax

y b

ax

Hyperbola NotesHyperbola NotesVertical Transverse AxisVertical Transverse Axis

Center Center (0,0)(0,0)

Vertices (a,0) Vertices (a,0) && (-a,0)(-a,0)

Foci (c,0) &Foci (c,0) & (-c, 0)(-c, 0)

AsymptotAsymptoteses


EquationEquation

y 2

a2 x2

b2 1

Foci : c2 a2 b2


To find To find asymptotesasymptotes

y a

bx

y a

bx

Graphing a HyperbolaGraphing a Hyperbola1.1. Graph the center.Graph the center.2.2. Take the square root of the positive Take the square root of the positive

denominator to find the vertices.denominator to find the vertices.3.3. Graph the vertices with a point.Graph the vertices with a point.4.4. Take the square root of the negative Take the square root of the negative

denominator to find the co-vertices.denominator to find the co-vertices.5.5. Graph the co-vertices with a dash.Graph the co-vertices with a dash.6.6. Create the box.Create the box.7.7. Draw the 2 asymptotes as the 2 Draw the 2 asymptotes as the 2

diagonals.diagonals.8.8. Go back to the vertices & graph the Go back to the vertices & graph the

hyperbola going towards the asymptotes.hyperbola going towards the asymptotes.9.9. To find the foci, cTo find the foci, c22 =a =a22 + b + b22

Write an equation of the hyperbola Write an equation of the hyperbola with foci (-5,0) & (5,0) and vertices with foci (-5,0) & (5,0) and vertices

(-3,0) & (3,0)(-3,0) & (3,0)

c 2 a2 b2

52 32 b2

25 9 b2

b2 16

x 2

9

y2

161

a = 3 c = 5a = 3 c = 5

Write an equation of the hyperbola Write an equation of the hyperbola with foci (0,-6) & (0,6) and vertices with foci (0,-6) & (0,6) and vertices

(0,-4) & (0,4)(0,-4) & (0,4)

c 2 a2 b2

62 42 b2

36 16 b2

b2 20

y 2

16

x2

201

a = 4 c = 6a = 4 c = 6

EXAMPLE 1 Graph an equation of a hyperbola

Graph 25y2 – 4x2 = 100. Identify the vertices, foci, and asymptotes of the hyperbola.

SOLUTION

STEP 1Rewrite the equation in standard form.25y2 – 4x2 = 100 Write original equation.

25y2

100 – 4x2

100100 100= Divide each side by 100.

y2

4 –y2

25 = 1 Simplify.


STEP 2

Identify the vertices, foci, and asymptotes. Note that a2 = 4 and b2 = 25, so a = 2 and b = 5. The y2 - term is positive, so the transverse axis is vertical and the vertices are at (0, +2). Find the foci.

c2 = a2 – b2 = 22 – 52 = 29.

so c = 29.

The foci are at ( 0, + )29. (0, + 5.4).

The asymptotes are y = ab

+ x or 25

+ xy =


STEP 3

Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 4 units high and 2b = 10 units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

EXAMPLE 2 Write an equation of a hyperbola

Write an equation of the hyperbola with foci at (–4, 0) and (4, 0) and vertices at (–3, 0) and (3, 0).

SOLUTION

The foci and vertices lie on the x-axis equidistant from the origin, so the transverse axis is horizontal and the center is the origin. The foci are each 4 units from the center, so c = 4. The vertices are each 3 units from the center, so a = 3.

EXAMPLE 2 Write an equation of a hyperbola

Because c2 = a2 + b2, you have b2 = c2 – a2. Find b2.

b2 = c2 – a2 = 42 – 32 = 7

Because the transverse axis is horizontal, the standard form of the equation is as follows:

x2

32 –y2

7 = 1 Substitute 3 for a and 7 for b2.

x2

9 –y2

7 = 1 Simplify

GUIDED PRACTICE for Examples 1 and 2

Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola.

1. x2

16 –y2

49 = 1

SOLUTION

(+4, 0) , ( + ) ,65 , 0 74

+ xy =


2. y2

36 – x2 = 1

SOLUTION

(0, +6) , ( 0, + ) ,37 y = +6x


3. 4y2 – 9x2 = 36

SOLUTION

( 0, + ) ,13 (0, +3) ,32

+ xy =


Write an equation of the hyperbola with the given foci and vertices.

4. Foci: (–3, 0), (3, 0)Vertices: (–1, 0), (1, 0)

SOLUTION x2 –y2

8 = 1

5. Foci: (0, – 10), (0, 10)Vertices: (0, – 6), (0, 6)

SOLUTION = 1 y2

36 –x2

64

What type of conic is each?. Hyperbolas 5.4 (M3)

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