8.5 Graph and Write Equations of Hyperbolas

p.518

What are the parts of a hyperbola?

What are the standard form equations of a hyperbola?

How do you know which way it opens?

Given a & b, how do you find the value of c?

How do you graph a hyperbola?

Why does drawing a box make graphing easier?

How do you write the equation from a graph?

Hyperbolas• Like an ellipse but instead of the sum of

distances it is the difference• A hyperbola is the set of all points P such that

the differences from P to two fixed points, called foci, is constant

• The line thru the foci intersects the hyperbola @ two points (the vertices)

• The line segment joining the vertices is the transverse axis, and it’s midpoint is the center of the hyperbola.

• Has 2 branches and 2 asymptotes• The asymptotes contain the diagonals of a

rectangle centered at the hyperbolas center

(0,b)

(0,-b)

Vertex (a,0)Vertex (-a,0)

Asymptotes

FocusFocus

Vertical transverse axis

12

2

2

2

b

x

a

y

Standard Form of Hyperbola w/ center @ origin

12

2

2

2

b

y

a

x

EquationTransverse

AxisAsymptotes Vertices

Horizontal y=+/- (b/a)x (+/-a,0)

Vertical y=+/- (a/b)x (0,+/-a)12

2

2

2

b

x

a

y

Foci lie on transverse axis, c units from the center c2 = a2+b2

Graph 4x2 – 9y2 = 36

• Write in standard form (divide through by 36)

• a=3 b=2 – because x2 term is ‘+’ transverse axis is horizontal & vertices are (-3,0) & (3,0)

• Draw a rectangle centered at the origin.

• Draw asymptotes.

• Draw hyperbola.

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

SOLUTION

STEP 1 Rewrite the equation in standard form.

25y2 – 4x2 = 100 Write original equation.

25y2

100 – 4x2

100100 100= Divide each side by 100.

y2

4 –x2

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.

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

1. x2

16 –y2

49 = 1

SOLUTION

STEP 1 The equation is in standard form. x2

16 –y2

49 = 1STEP 2

Identify the vertices, foci, and asymptotes. Note that a2 = 16 and b2 = 49, so a = 4 and b = 7. The x2 - term is positive, so the transverse axis is horizontal and the vertices are at (+4, 0). Find the foci.

c2 = a2 + b2 = 42 + 72 = 65.

so c = + 65.

The foci are at ( + )65 , 0

STEP 3

Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 8 units high and 2b = 14 units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

Write the equation of a hyperbola with foci (0,-3) & (0,3) and vertices

(0,-2) & (0,2).

• Vertical because foci & vertices lie on the y-axis• Center @ origin because focus & vertices are

equidistant from the origin• Since c=3 & a=2, c2 = b2 + a2

• 9 = b2 + 4• 5 = b2

• +/-√5 = b

154

22

xy

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.

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

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

4. Foci: (– 3, 0), (3, 0)Vertices: (– 1, 0), (1, 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 3 units from the center, so c = 3. The vertices are each 1 units from the center, so a = 1.

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

b2 = c2 – a2 = 32 – 12 = 8Because the transverse axis is horizontal, the standard form of the equation is as follows:

x2

12 –y2

8 = 1 Substitute 1 for a and 8 for b2.

x2 –y2

8 = 1 Simplify

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

SOLUTION The foci and vertices lie on the y-axis equidistant from the origin, so the transverse axis is vertical and the center is the origin. The foci are each 10 units from the center, so c = 10. The vertices are each 6 units from the center, so a = 6.

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

b2 = c2 – a2 = 102 – 62 = 64

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

y2

362 –x2

64 = 1 Substitute 6 for a and 64 for b2.

Simplify= 1 y2

36 –x2

64

Photography

You can take panoramic photographs using a hyperbolic mirror. Light rays heading toward the focus behind the mirror are reflected to a camera positioned at the other focus as shown. After a photograph is taken, computers can “unwrap” the distorted image into a 360° view.

• Write an equation for the cross section of the mirror.

• The mirror is 6 centimeters wide. How tall is it?

SOLUTION

STEP 1 From the diagram, a = 2.81 and c = 3.66.

To write an equation, find b2. b2 = c2 – a2 = 3.662 – 2.812 5.50

Because the transverse axis is vertical, the standard form of the equation for the cross section of the mirror is as follows: y2

2.812– x2

5.50 = 1 or y2 7.90

– x2 5.50 = 1

STEP 2

Find the y-coordinate at the mirror’s bottom edge. Because the mirror is 6 centimeters wide, substitute x = 3 into the equation and solve.

Substitute 3 for x.

y2 20.83

= 1

y2 7.90

32 5.50

–

Solve for y2.

y – 4.56 Solve for y.So, the mirror has a height of – 2.81 – (– 4.56) = 1.75 cm.

• What are the parts of a hyperbola?

Vertices, foci, center, transverse axis & asymptotes• What are the standard form equations of a hyperbola?

• How do you know which way it opens?

Transverse axis is always over a• Given a & b, how do you find the value of c?

c2 = a2 + b2 • How do you graph a hyperbola?

Plot a and b, draw a box with diagonals. Draw the hyperbola following the diagonals through the vertices.

12

2

2

2

b

y

a

x1

2

2

2

2

b

x

a

y

• Why does drawing a box make graphing easier?

The diagonals of the box are the asymptotes of the hyperbola.

• How do you write the equation from a graph?

Identify the transverse axis, find the value of a and b (may have to use c2 = a2 + b2) and substitute into the equation.

8.5 Assignment, day 1

Page 521, 3-13 odd, 19-25 odd

8.5 Hyperbolas, day 2

What are the standard form equations of a hyperbola if the center has been translated?

How do you graph a translated hyperbola?

How do you write the equation of a translated hyperbola?

Translated Hyperbolas

In the following equations the point (h,k) is the center of the hyperbola.Horizontal axis Vertical axis

1

2

2

2

2

b

ky

a

hx 1

2

2

2

2

b

hx

a

ky

Remember c2 = a2 + b2

Graphing the Equation of a Translated Hyperbola

Graph (y + 1) 2

– = 1.(x + 1) 2

4

SOLUTION

The y 2-term is positive, so thetransverse axis is vertical. Sincea 2 = 1 and b 2 = 4, you know thata = 1 and b = 2.

Plot the center at (h, k) = (–1, –1). Plot the vertices 1 unit above and below the center at (–1, 0) and (–1, –2).

Draw a rectangle that is centered at (–1, –1) and is 2a = 2 units high and 2b = 4 units wide.

(–1, –2)

(–1, 0)

(–1, –1)

Graphing the Equation of a Translated Hyperbola

SOLUTION

Draw the asymptotes through the corners of the rectangle.

Draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

(–1, –2)

(–1, 0)

(–1, –1)

Graph (y + 1) 2

– = 1.(x + 1) 2

4

The y 2-term is positive, so thetransverse axis is vertical. Sincea 2 = 1 and b 2 = 4, you know thata = 1 and b = 2.

Graph (y – 3)2

4–

(x + 1)2

9= 1

SOLUTION

STEP 1 Compare the given equation to the standard forms of equations of hyperbolas. The equation’s form tells you that the graph is a hyperbola with a vertical transverse axis. The center is at (h, k) = (– 1, 3). Because a2 = 4 and b2 = 9, you know that a = 2 and b = 3.

STEP 2Plot the center, vertices, and foci. The vertices lie a = 2 units above and below the center, at (21, 5) and (21, 1). Because c2 = a2 + b2 = 13, the foci lie c = 13 3.6 units above and below the center, at (– 1, 6.6) and (– 1, – 0.6).

STEP 3Draw the hyperbola. Draw a rectangle centered at (−1, 3) that is 2a = 4 units high and 2b = 6 units wide. Draw the asymptotes through the opposite corners of the rectangle. Then draw the hyperbola passing through the vertices and approaching the asymptotes.

Graph (x + 3)2 –(y – 4)2

9= 1

SOLUTION

STEP 1 Compare the given equation to the standard forms of equations of hyperbolas. The equation’s form tells you that the graph is a hyperbola with a vertical transverse axis. The center is at (h, k) = (– 3, 4). Because a2 = 1 and b2 = 4, you know that a = 1 and b = 2.

3.

STEP 2 Plot the center, vertices, and foci. The vertices lie a = 1 units above and below the center, at (–2, 4) and (–4, 4). Because c2 = a2 + b2 = 5, the foci lie c = 5 units above and below the center, at (– 3 + ,4 ) and (– 3, – , 4).5

5

STEP 3

Draw the hyperbola. Draw a rectangle centered at ( –3, 4) that is 2a = 2 units high and 2b = 4 units wide. Draw the asymptotes through the opposite corners of the rectangle.

Write the equation of the hyperbola in standard form.

16y2 −36x2 + 9 = 0

16y2 −36x2 = −9

19

36

9

16 22

xy

1

369

169

22

xy

1

169

41

22

yx

161

161

361361

Write an equation for the hyperbola.

Vertices at (5, −4) and (5,4) and foci at (5,−6) and (5,6).

Draw a quick graph.

Equation will be:

1

2

2

2

2

b

hx

a

ky

Center (5,0) (h,k)

a = 4, c = 6

c2 = a2 + b2

62 = 42 + b2

36 = 16 + b2

20 = b2

1

2

2

2

2

b

hx

a

ky

1

2

2

2

2

b

hx

a

ky

1

20

5

16

22

xy

Center?

(5,4)

(5,−4)

(5,6)

(5,−6)

• What are the standard form equations of a hyperbola if the center has been translated?

1)()(

2

2

2

2

b

ky

a

hx1

)()(2

2

2

2

b

hx

a

ky

8.5 Assignment, day 2

Page 521, 4-10 even, 18-24 even

Page 531, 5, 6, 11, 19-20