2.7 more parabolas a& hyperbolas (optional) x

Hyperbolas and More Parabolas (optional)


Hyperbolas and More Parabolas (optional)Using similar methods to analyze Ax2 + By2 + Cx + Dy = E

we obtain the graphs of hyperbolas and parabolas.



In the special case B = 0 (or A = 0),



In the special case B = 0 (or A = 0), after dividing by A (by B),

the equation becomes 1x2 + #x + #y = # or 1y2 + #x + #y = #,









their graphs are parabolas. Assuming both variables x and y

remained in the equation(

(





their graphs are parabolas.

1x2 + #x + #y = # or

1y2 + #x + #y = #Parabolas:

Assuming both variables x and y


(






1x2 + #x + #y = # or

1y2 + #x + #y = #

If the equation Ax2 + By2 + Cx + Dy = E

has A and B of opposite signs,

Parabolas:



(






1x2 + #x + #y = # or

1y2 + #x + #y = #


has A and B of opposite signs, after dividing by A, we have 1x2

+ ry2 + #x + #y = #, with r < 0.

Parabolas:



(






1x2 + #x + #y = # or

1y2 + #x + #y = #



+ ry2 + #x + #y = #, with r < 0. These are hyperbolas.

Parabolas:



(


1x2 + #x + #y = # or

1y2 + #x + #y = #

(r < 0)

Hyperbolas: 1x2 + ry2 + #x + #y = #



+ ry2 + #x + #y = #, with r < 0. These are hyperbolas.

Parabolas:

Using similar methods to analyze Ax2 + By2 + Cx + Dy = E




their graphs are parabolas. Assuming both variables x and y


(

(in general)

Conic SectionsAfter dividing Ax2 + By2 + Cx + Dy = E by A, we obtain

1x2 + ry2 + #x + #y = #.


1x2 + ry2 + #x + #y = #. Let’s use the simpler cases 1x2 + ry2 = 1

to study the geometry.



to study the geometry. Starting with r = 1, a circle,

circle

r = 1




as r becomes smaller = ¼,

circle

r = 1




as r becomes smaller = ¼, 1/9, 1/16… ,

circle

r = 1





circle

r = 1





1x2 + y2 = 11

16

4

1

circle

r = 1





the circle is elongated to taller and taller ellipses.

circle ellipses

1x2 + y2 = 11

16

4

1

r = 1






When r = 0, we’ve 1x2 + 0y2 = 1,

circle ellipses

. … r → 0

1x2 + y2 = 11

16

r = 1/16

4

1

r = 1






When r = 0, we’ve 1x2 + 0y2 = 1, which is a pair of lines x = ±1,

circle ellipses

. … r → 0

1x2 + y2 = 11

16

r = 1/16

4

1

r = 1







circle ellipses

. … r → 0

1x2 + y2 = 11

16

r = 1/16

4

1

1

1x2 = 1or

x = ±1

r = 1 r = 0two lines







i.e. the ellipses are elongated

into two parallel lines.

circle ellipses

. … r → 0

1x2 + y2 = 11

16

r = 1/16

4

1

1

1x2 = 1or

x = ±1

r = 1 r = 0two lines

HyperbolasJust as all the other conic sections, hyperbolas are defined

by distance relations.

Hyperbolas

Given two fixed points, called foci, a hyperbola is the set

of points whose difference of the distances to the foci is

a constant.

Just as all the other conic sections, hyperbolas are defined


A

If A, B and C are points on a hyperbola as shown

Hyperbolas



a constant.

B

C



A

a2

a1

If A, B and C are points on a hyperbola as shown then

a1 – a2

Hyperbolas



a constant.

B

C



A

a2

a1

b2

b1


a1 – a2 = b1 – b2

Hyperbolas



a constant.

B

C



A

a2

a1

b2

b1


a1 – a2 = b1 – b2 = c2 – c1 = constant.

c1

c2

Hyperbolas



a constant.

B

C



HyperbolasA hyperbola has a “center”,

HyperbolasA hyperbola has a “center”, and two straight lines that

cradle the hyperbolas which are called asymptotes.



There are two vertices, one for each branch.



There are two vertices, one for each branch. The asymptotes

are the diagonals of a rectangle with the vertices of the

hyperbola touching the rectangle.



There are two vertices, one for each branch. The asymptotes

are the diagonals of a rectangle with the vertices of the

hyperbola touching the rectangle.

HyperbolasThe center-rectangle is defined by the x-radius a, and y-

radius b as shown.

a

b


radius b as shown. Hence, to graph a hyperbola, we find

the center and the center-rectangle first.

a

b



the center and the center-rectangle first.

a

b



the center and the center-rectangle first. Draw the

diagonals of the rectangle which are the asymptotes.

a

b




diagonals of the rectangle which are the asymptotes. Label

the vertices and trace the hyperbola along the asymptotes.

a

b

Hyperbolas

a

b

The location of the center, the x-radius a, and y-radius b may

be obtained from the equation.

The center-rectangle is defined by the x-radius a, and y-



diagonals of the rectangle which are the asymptotes. Label

the vertices and trace the hyperbola along the asymptotes.

HyperbolasThe equations of hyperbolas have the form

Ax2 + By2 + Cx + Dy = E

where A and B are opposite signs.


Ax2 + By2 + Cx + Dy = E

where A and B are opposite signs. By completing the square,

they may be transformed into the standard forms below.

(x – h)2 (y – k)2

a2 b2


Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2

(h, k) is the center.


Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2

x-rad = a, y-rad = b



Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2 – = 1

(x – h)2 (y – k)2

a2 b2




Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2

y-rad = b, x-rad = a

– = 1

(x – h)2 (y – k)2

a2 b2




Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2


– = 1

(h, k)

Open in the x direction

(x – h)2 (y – k)2

a2 b2




Ax2 + By2 + Cx + Dy = E



– = 1 (x – h)2(y – k)2

a2b2


– = 1

(h, k)

Open in the x direction

(h, k)

Open in the y direction

HyperbolasFollowing are the steps for graphing a hyperbola.


1. Put the equation into the standard form.



2. Read off the center, the x-radius a, the y-radius b, and

draw the center-rectangle.





3. Draw the diagonals of the rectangle, which are the

asymptotes.






asymptotes.

4. Determine the direction of the hyperbolas and label the

vertices of the hyperbola.






asymptotes.


vertices of the hyperbola. The vertices are the mid-points

of the edges of the center-rectangle.






asymptotes.


vertices of the hyperbola. The vertices are the mid-points

of the edges of the center-rectangle.

5. Trace the hyperbola along the asymptotes.

HyperbolasExample A. List the center, the x-radius, the y-radius.

Draw the rectangle, the asymptotes, and label the vertices.

Trace the hyperbola.

(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)




(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2




(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2

Hyperbolas

(3, -1)

42

Example A. List the center, the x-radius, the y-radius.



(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2

Hyperbolas

(3, -1)

42




(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2

The hyperbola opens

left-rt




(x – 3)2 (y + 1)2

42 22– = 1

(3, -1)

42

Center: (3, -1)

x-rad = 4

y-rad = 2

The hyperbola opens

left-rt and the vertices

are (7, -1), (-1, -1) .




(x – 3)2 (y + 1)2

42 22– = 1

(3, -1)

42

Center: (3, -1)

x-rad = 4

y-rad = 2

The hyperbola opens


are (7, -1), (-1, -1) .

Hyperbolas

(3, -1)(7, -1)(-1, -1) 4

2




(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2

The hyperbola opens


are (7, -1), (-1, -1) .

Hyperbolas

(3, -1)(7, -1)(-1, -1) 4

2




(x – 3)2 (y + 1)2

42 22– = 1

Center: (3, -1)

x-rad = 4

y-rad = 2

The hyperbola opens


are (7, -1), (-1, -1) .

Hyperbolas

(3, -1)(7, -1)(-1, -1) 4

2




(x – 3)2 (y + 1)2

42 22– = 1

When we use completing the square to get to the standard

form of the hyperbolas, depending on the signs,

we add a number or subtract a number from both sides.

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard

form. List the center, the x-radius, the y-radius.



Hyperbolas





Group the x’s and the y’s:

Hyperbolas






4y2 – 16y – 9x2 – 18x = 29

Hyperbolas






4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients

Hyperbolas







4(y2 – 4y ) – 9(x2 + 2x ) = 29

Hyperbolas







4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square

Hyperbolas








4(y2 – 4y + 4 ) – 9(x2 + 2x ) = 29

Hyperbolas








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29

Hyperbolas








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16

16

Hyperbolas








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36 divide by 36 to get 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

9(x + 1)24(y – 2)2

36 36


– = 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

9(x + 1)24(y – 2)2

36 36


– = 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

9

9(x + 1)24(y – 2)2

36 36


– = 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

9 4

9(x + 1)24(y – 2)2

36 36


– = 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

(y – 2)2 (x + 1)2

32 22– = 1

9 4

9(x + 1)24(y – 2)2

36 36


– = 1








4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9

16 –9

Hyperbolas

(y – 2)2 (x + 1)2

32 22– = 1

Center: (-1, 2), x-rad = 2, y-rad = 3

The hyperbola opens up and down.

9 4

(-1, 2)

Hyperbolas

Center: (-1, 2),

x-rad = 2,

y-rad = 3

(-1, 2)

(-1, 5)

(-1, -1)

Hyperbolas

Center: (-1, 2),

x-rad = 2,

y-rad = 3


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

(-1, 2)

(-1, 5)

(-1, -1)

Hyperbolas

Center: (-1, 2),

x-rad = 2,

y-rad = 3


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

Hyperbolas

More Graphs of Parabolas

The graphs of the equations of the form

y = ax2 + bx + c and x = ay2 + bx + c

are parabolas.


Each parabola has a vertex and the center line that contains

the vertex.



are parabolas.



the vertex.



are parabolas.



the vertex. Suppose we know another point on the parabola,



are parabolas.






are parabolas.




then the reflection of the point across the center line is also

on the parabola.



are parabolas.





on the parabola.



are parabolas.





on the parabola. There is exactly one parabola that goes

through these three points.



are parabolas.





on the parabola. There is exactly one parabola that goes

through these three points.



are parabolas.


When graphing parabolas, we must also give the x-intercepts

and the y-intercept.




The y-intercept is obtained by setting x = 0 and solve for y.





The x-intercept is obtained by setting y = 0 and solve for x.







The graphs of y = ax2 + bx = c are up-down parabolas.







If a > 0, the parabola opens up.

a > 0








If a < 0, the parabola opens down.

a > 0 a < 0









a > 0 a < 0

Vertex Formula (up-down parabolas) The x-coordinate of

the vertex of the parabola y = ax2 + bx + c is at x = .-b2a

More Graphs of ParabolasFollowing are the steps to graph the parabola y = ax2 + bx + c.


1. Set x = in the equation to find the vertex.-b2a


1. Set x = in the equation to find the vertex.

2. Find another point, use the y-intercept (0, c) if possible.

-b2a




3. Locate its reflection across the center line, these three

points form the tip of the parabola. Trace the parabola.

-b2a






4. Set y = 0 and solve to find the x intercept.

-b2a

Example A. Graph y = –x2 + 2x + 15







-b2a

The vertex is at x = = 1

Example A. Graph y = –x2 + 2x + 15 –(2)

2(–1)







-b2a

The vertex is at x = = 1y = 16.


2(–1)







-b2a


Example A. Graph y = –x2 + 2x + 15 (1, 16)–(2)

2(–1)







-b2a


The y-intercept is at (0, 15)


2(–1)







-b2a

(1, 16)




2(–1)







-b2a

(1, 16)

(0, 15)



Plot its reflection (2, 15).


2(–1)







-b2a

(1, 16)

(0, 15)



Plot its reflection (2, 15).


2(–1)







-b2a

(1, 16)

(0, 15) (2, 15)



Plot its reflection (2, 15)

Draw,

Example A. Graph y = –x2 + 2x + 15 (1, 16)

(0, 15) (2, 15)–(2)

2(–1)







-b2a




Draw, set y = 0 to get x-int:


(0, 15) (2, 15)–(2)

2(–1)







-b2a





–x2 + 2x + 15 = 0


(0, 15) (2, 15)–(2)

2(–1)







-b2a





–x2 + 2x + 15 = 0

x2 – 2x – 15 = 0


(0, 15) (2, 15)–(2)

2(–1)







-b2a





–x2 + 2x + 15 = 0

x2 – 2x – 15 = 0

(x + 3)(x – 5) = 0

x = –3, x = 5


(0, 15) (2, 15)–(2)

2(–1)







-b2a





–x2 + 2x + 15 = 0

x2 – 2x – 15 = 0

(x + 3)(x – 5) = 0

x = –3, x = 5


(0, 15) (2, 15)

(-3, 0) (5, 0)

–(2)

2(–1)







-b2a

The graphs of the equations

x = ay2 + by + c

are parabolas that open sideways.



x = ay2 + by + c



If a>0, the parabola opensto the right.


x = ay2 + by + c




If a<0, the parabola opensto the left.


x = ay2 + by + c


Each sideways parabola is symmetric to a horizontal center

line.





x = ay2 + by + c



line. The vertex of the parabola is on this line.





x = ay2 + by + c



line. The vertex of the parabola is on this line. If we know the

location of the vertex and another point on the parabola, the

parabola is completely determined.





x = ay2 + by + c



line. The vertex of the parabola is on this line. If we know the

location of the vertex and another point on the parabola, the

parabola is completely determined. The vertex formula is the

same as before except it's for the y coordinate.




More Graphs of ParabolasVertex Formula (sideways parabolas)

The y coordinate of the vertex of the parabola x = ay2 + by + c

is at y = .–b2a

Following are steps to graph the parabola x = ay2 + by + c.



is at y = .–b2a


1. Set y = in the equation to find the x coordinate of the

vertex.

–b2a



is at y = .–b2a



vertex.

2. Find another point; use the x intercept (c, 0) if it's not the

vertex.

–b2a



is at y = .–b2a



vertex.


vertex.

3. Locate the reflection of the point across the horizontal

center line, these three points form the tip of the parabola.

Trace the parabola.

–b2a



is at y = .–b2a



vertex.


vertex.

3. Locate the reflection of the point across the horizontal

center line, these three points form the tip of the parabola.

Trace the parabola.

4. Set x = 0 to find the y intercept.

–b2a



is at y = .–b2a

Example B. Graph x = –y2 + 2y + 15



Vertex: set y = =1 –(2)2(–1)



Vertex: set y = =1 then x = –(1)2 + 2(1) + 15 = 16 –(2)2(–1)




so v = (16, 1).




so v = (16, 1).

Another point:

Set y = 0 then x = 15

or (15, 0).




so v = (16, 1).

Another point:


or (15, 0).


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).

(15, 0)


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).

Plot its reflection.

(15, 0)


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2).

(15, 0)


(16, 1)

(15, 2)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2).

Draw. (15, 0)


(16, 1)

(15, 2)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2)

Draw. (15, 0)

(15, 2)


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2)

Draw. Get y-int:

–y2 + 2y + 15 = 0(15, 0)

(15, 2)


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2)

Draw. Get y-int:

–y2 + 2y + 15 = 0

y2 – 2y – 15 = 0

(y – 5) (y + 3) = 0

y = 5, -3

(15, 0)

(15, 2)


(16, 1)



so v = (16, 1).

Another point:


or (15, 0).


It's (15, 2)

Draw. Get y-int:

–y2 + 2y + 15 = 0

y2 – 2y – 15 = 0

(y – 5) (y + 3) = 0

y = 5, -3

(15, 0)

(15, 2)


(16, 1)

(0, -3)



The y-intercept is (o, c) obtained by setting x = 0.

The x-intercept is obtained by setting y = 0 and solve the

equation 0 = ax2 + bx + c which may or may not have real

number solutions. Hence there might not be any x-intercept.


Following are the steps to graph a parabola y = ax2 + bx + c.






-b2a

The graph of y = ax2 + bx = c are up-down parabolas.



More Graphs of ParabolasExercise A. Graph the following parabolas. Identify which

direction the parabolas face, determine the vertices using

the vertex method. Label the x and y intercepts, if any.

1. x = –y2 – 2y + 15 2. y = x2 – 2x – 15

3. y = x2 + 2x – 15 4. x = –y2 + 2y + 15

5. x = –y2 – 4y + 12 6. y = x2 – 4x – 21

7. y = x2 + 4x – 12 8. x = –y2 + 4y + 21

9. x = –y2 + 4y – 4 10. y = x2 – 4x + 4

11. x = –y2 + 4y – 4 12. y = x2 – 4x + 4

13. y = –x2 – 4x – 4 14. x = –y2 – 4y – 4

15. x = –y2 + 6y – 40 16. y = x2 – 6x – 40

17. y = –x2 – 8x + 48 18. x = y2 – 8y – 48

19. x = –y2 + 4y – 10 20. y = x2 – 4x – 2

21. y = –x2 – 4x – 8 22. x = –y2 – 4y – 5

EllipsesB. Complete the square of the following hyperbola-equations

if needed. Find the centers and the radii of the hyperbolas.

Draw and label the vertices.

1. x2 – 4y2 = 1 2. 9x2 – 4y2 = 1

3. 4x2 – y2/9 = 1 4. x2/4 – y2/9 = 1

5. 0.04x2 – 0.09y2 = 1 6. 2.25x2 – 0.25y2 = 1

7. x2 – 4y2 = 100 8. x2 – 49y2 = 36

9. 4x2 – y2/9 = 9 10. x2/4 – 9y2 = 100

13. (x – 4)2 – 9(y + 1)2 = 25

17. y2 – 8x – 4x2 + 24y = 29

15. x2 – 6x – 25y2 = 27

18. 9y2 – 18y – 25x2 + 100x = 116

11. x2 – 4y2 + 8y = 5 12. y2 – 8x – 4x2 + 24y = 29

14. 9(x + 2)2 – 4(y + 1)2 = 36

16. x2 – 25y2 – 100y = 200


1. (0, 3)

(0, -5)

(16, 1)

3.

(3, 0)(-5, 0)

(-1, -16)5.

(0, 2)

(0, -6)

(16, -2)

(2, 0)(-6, 0)

(-2, -16)

7.

(Answers to odd problems) Exercise A.

More Graphs of Parabolas9.

(-16, -2)

(-16, 6)

(0, 2)

11.

(-2, 0)

(0, -4)

13.

(-16, -2)

(-16, 6)

(0, 2)

15.

(-31, 3)

(-40, 0)

More Graphs of Parabolas17. 19.

21.

(-4, 64)

(-12, 0) (4, 0)(-10, 0)

(-6, 2)

(-2, -4)

(0, -8)

1. Center: (0, 0)

x radius: 1

y radius: 1/2

Exercise B.

(-1, 0) (1, 0)

3. Center: (0, 0)

x radius: 1/2

y radius: 3

(0.5, 0)(-0.5, 0)

Ellipses

5. Center: (0, 0)

x radius: 5

y radius: 3.33

(5, 0)(-5, 0)

7. Center: (0, 0)

x radius: 10

y radius: 5

(10, 0)(-10, 0)

9. Center: (0, 0)

x radius: 1.5

y radius: 9

11. Center: (0, 1)

x radius: 1

y radius: 1/2

Ellipses

13. Center: (4, -1)

x radius: 5

y radius: 9/3

15. Center: (3, 0)

x radius:

y radius:

(-1.5, 0) (1.5, 0) (1, 1)(-1, 1)

(5, 0)(-5, 0) (9, 0)(-3, 0)

17. Center: (-1,-12)

x radius: 6.5

y radius: 13

Ellipses

(-1, 1)

(-1, -25)

Education

2.7 more parabolas a& hyperbolas (optional) x