5.4 hyperbolas 1 please note the minus in the middle. a “+” in the middle makes the graph an...

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5.4 Hyperbolas 1 Please note the minus in the middle. A “+” in the middle makes the graph an ellipse. A minus in the middle will give us a hyperbola which looks like the following graphs below. In the last section, we graphed ellipses of the form: 2 2 2 2 x y 1. a b We will now “focus” on equations of the form: 2 2 2 2 x y 1 a b 2 2 2 2 y x 1. b a and Next Slide 2 2 2 2 x y 1 a b 2 2 2 2 y x 1 b a

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5.4 Hyperbolas

1

Please note the minus in the middle. A “+” in the middle makes the graph an ellipse. A minus in the middle will give us a hyperbola which looks like the following graphs below.

In the last section, we graphed ellipses of the form: 2 2

2 2

x y1.

a bWe will now “focus” on equations of the form:

2 2

2 2

x y1

a b

2 2

2 2

y x1.

b aand

Next Slide

●●

2 2

2 2

x y1

a b

2 2

2 2

y x1

b a

5.4 Hyperbolas

2Next Slide

Definition: A hyperbola is the collection of all points in the plane the difference of whose distance from two fixed points called the foci is a positive constant.

●●

●●

●(c,0)(-c,0) (a,0)(-a,0) (0,b)

(0,-b)

foci

foci

(0,c)

(0,-c)

vertices

vertices

Horizontal Transverse Axis Vertical Transverse Axis

A hyperbola which opens left and right has a horizontal transverse axis. A hyperbola which opens up and down has a vertical transverse axis. The midpoint of the transverse axis is the center.

5.4 Hyperbolas

3

Standard Equation: Hyperbola with Transverse Axis on the x Axis (Opens Left and Right)

The standard equation of a hyperbola with its center at (0,0) and its transverse axis on the x axis is

2 2

2 2

x y1

a b

where the vertices are (-a,0) and (a,0), and the foci are at (-c,0) and (c,0), and 2 2 2c a b .

The hyperbola’s shape will also be determined by two asymptotes. The equation of these asymptotes are the lines:

b

y xa

Some instructors will require you to state the equations of these asymptotes and other instructors may not. The equations of the asymptotes are simple to state when the center is at the origin. If the center is not at the origin, the equation of the asymptotes are:

Next slide

b

y x h ka

5.4 Hyperbolas

4

2 2 x yFind the vertices, the foci, and sketch the hypExample erbola 1.

9 1

4 .

2 2

2 2

x yCompare to the standard form 1 to determine a and b.

a b

x

yThe vertices are (-3,0) and (3,0).

2 2 2fociTo find the , use c a b . 2c 9 4

c 13 3.6

The foci lie on the x axis. Therefore the

focicoordinates

of the are 13,0 and 13,0 .

Next Slide

Plot all four of these points (-3,0), (3,0), (0,2) and (0,-2).

Now, create a dashed box using the four points.

Next, create the two dashed asymptotes drawing diagonals through the corners of the box.

The two branches of the hyperbola opening left and right at the vertices can be sketched using the diagonals as guidelines. The branches get closer and closer to the asymptotes but never cross it.

2 2a 9 and b 4 then a 3 and b 2

Center: (0,0)

5.4 Hyperbolas

5

Your Turn Problem #1

y

vertices: (-2,0), (2,0)Answer

foci: 2 5,0 and 2 5,0

x

2 2 x y

Find the vertices, the foci, and sketch the hyperbola 1. 4 16

5.4 Hyperbolas

6Next slide

2 2

2 2

y x1

b a

2 2

2 2

x y1

a bx is positive:opens left and right y is positive:

opens up and down

Standard Equation: Hyperbola with Transverse Axis on the y Axis (Opens up and down)The standard equation of a hyperbola with its center at (0,0) and its transverse axis on the y axis is

2 2

2 2

y x1

b a

where the vertices are (0,-b) and (0,b), and the foci are at (0,-c) and (0,c), and 2 2 2c a b .

The hyperbola’s shape will again be determined by two asymptotes. The equation of these asymptotes are still the lines:

by x.

a

It may seem confusing to determine if the branches open left and right or up and down. Once the hyperbola is in standard form, the vertices from which the branches open will be on the axis that comes first in the equation, which will have a positive coefficient.

5.4 Hyperbolas

7

2 2 y xFind the vertices, the foci, and sketch the hypeExampl rbola 1.

4

9e 2.

2 2

2 2

y xCompare to the standard form 1 to determine a and b.

b a

x

y

The vertices are (0,-2) and (0,2).

2 2 2fociTo find the , use c a b . 2c 9 4

c 13 3.6

The foci lie on the y axis. Therefore the

focicoordinates

of the are 0, 13 and 0, 13 .

Next Slide

Plot all four of these points (-3,0), (3,0), (0,2) and (0,-2).

Now, create a dashed box using the four points.Next, create the two dashed asymptotes by drawing diagonals through the corners of the box.The two branches of the hyperbola opening up and down at the vertices can be sketched using the diagonals as guidelines.

2 2b 4 and a 9 then b 2 and a 3

Center: (0,0)

5.4 Hyperbolas

8

Your Turn Problem #2

x

y

vertices: (0,-4), (0,4)

Answer:

foci: 0, 17 and 0, 17

2 2 y x

Find the vertices, the foci, and sketch the hyperbola 1. 16 1

5.4 Hyperbolas

9

2 2 x 4y 4 4 4 4

2 2Find the vertices, the foci, and sketch the hyperbola xExampl 4y .e 3. 4

Determine a and b. a 2 and b 1

x

yThe vertices are (-2,0) and (2,0).

2 2 2fociTo find the , use c a b . 2c 4 1

c 5 2.2

The foci lie on the x axis. Therefore the

foci coordinates

of the are 0, 5 and 0, 5 .

Next Slide

Now, create a dashed box using the four points.

Next, create the two dashed asymptotes by drawing diagonals through the corners of the box.The two branches of the hyperbola opening left and right at the vertices can be sketched using the diagonals as guidelines.

Plot all four of these points (0,-1), (0,1), (2,0) and (-2,0).

To obtain the standard form, divide both sides by 4 and simplify to obtain a “1” on the RHS.

2 2 x y

1 4 1

5.4 Hyperbolas

10

Your Turn Problem #3

x

y

vertices: (0,-5), (0,5)

Answer:

foci: 0, 34 and 0, 34

2 2Find the vertices, the foci, and sketch the hyperbola 9y 25x 225.

5.4 Hyperbolas

11

Hyperbolas whose center is not at the origin.

The standard form for a hyperbola where the center is at (h, k) is:

2 2

2 2

x h y k 1

a b

This hyperbola opens left and right.

Next Slide 2 2 2

Note: To find the foci which lie on the transverse axis: c a b .

2 2

2 2

y k x h 1

b aor

2x is positive.2y is positive.

This hyperbola opens up and down.

y

x

(h,k)

(h,k+b)

(h,k-b) ● ●

●(h-a,k)

(h+a,k)

(h,k+c)

(h,k-c)

●x(h-a,k)(h-c,k)

● ●(h+c,k)

●(h+a,k)

(h,k+b)

(h,k-b)●

y

●(h,k)

5.4 Hyperbolas

12

2 2y 3 x 2

116 4

523,2

523,2

2 2a 4 and b 16 then a 2 and b 4

x

y

The vertices are (2,1) and (2,-7).

2 2 2fociTo find the , use c a b .

2c 4 16

c 20 2 5 4.5

The center (2,is 3).

(2,-3)

(2,-7)

(2,1)

(0,-3)

(4,-3)

foThe foci lie on the transverse axis. Therefore the coordinatesof the are 2, 3 2 5 and 2, 3 2 5 or (2,1.5), (

i2,

c7.5)

Next Slide

Plot all four of these points (0,-3), (4,-3), (2,1) and (2-7).

Now, create the dashed box using the four points and draw the asymptotes through the corners. Then sketch the hyperbola.

Find the center, vertices, foci, and sketch theExample 4. hyper bola:

Compare to the standard form to determine the center, a, and b.

5.4 Hyperbolas

13

Your Turn Problem #4

(-1,-1)

(-1,2)

(-6,2)

x

● ●

center: (-1,2) vertices: (-6,2), (4,2)

Answer:

(4,2)

(-1,5)

● ●

y

foci: 1 34,2 , 1 34,2

1 34,2 1 34,2

2 2x 1 y 2

Find the center, vertices, the foci of the hyperbola 1,25 9

and sketch the hyperbola.

5.4 Hyperbolas

14

The hyperbola in the previous example was given in standard form.

If the hyperbola is given in general form,

we will need to convert it to standard form before graphing.

2 2 ay bx cx dy e 0,

Next Slide

2 2 ax by cx dy e 0 or

2 2

2 2

x h y k 1

a b

2 2

2 2

y k x h 1

b aor

Factoring and completing the square will be necessary to obtain the standard form:

5.4 Hyperbolas

15

Example 5.Write the given equation of the hyperbola in standard form:Group the x terms separately from the y terms and move the constant to the RHS.

2 24x 9y 8x 54y 113 0

2 24 x 2x 9 y y 6 113

2 24x 8x 9y 54y 113

Complete the square for both trinomials. The numbers added in the parentheses are 1 and 9. We need to add the same “value” to the RHS. The values are 4 and -81.

4 811 9

2 2

4 x 1 9 y 3 36

Write each perfect square trinomial as a binomial squared and add the constants on the RHS.

Factor out the ‘4” from the first pair and the ‘-9’ from the second pair. Leave a space at

the end of each set of parentheses to add the appropriate number when completing the square.

Finally, divide by 36 on both sides and simplify to obtain the hyperbola in standard form.

36 36 36

2 2x 1 y 3

Answer : 19 4

Answer: 2 2 x 5 y 3

14 5

Your Turn Problem #5

Write the given equation of the hyperbola in standard form: 2 25x 4y 50x 24y 69 0

5.4 Hyperbolas

16

a 3 and b 2 1.4

x

y

The vertices are (-1,1) and (5,1).

The center (2is ,1).

1st write in standard form using completing the square.

19 y2y9 x4x2 22

181y92x2 22

1

21y

92x 22

(5,1)(-1,1)

21,2

21,2

2 22x 9y 8x 18y 19 0Find the center, vertices, foci, and sketch theExample 6. hyper bola:

2 2 2fociTo find the , use c a b .

2c 9 2

c 11 3.3

The are 2 11, 1 and 2 11, 1 or ( 1.3,1), (5.

foci3, 1)

Next Slide

2 22x 8x 9y 18y 19

+8 - 94 1

1,112

1,112

(2,1)

5.4 Hyperbolas

17

(-3,-1)

(-1,1)(-5,1)

x

● ●

y center: (-3,1) vertices: (-3,3), (-3,-1)

Answer:

(-3,3)

(-3,1)

2213,,221 3, :foci 221 3,-

221 3,

The EndB.R.2-1-07

Your Turn Problem #6

2 2

Find the center, vertices, and foci of the hyperbolay x 2y 6x 12 0, and sketch the hyperbola.