1 PARTS OF A PARABOLA PARABOLAS PROBLEM 1 PROBLEM 2 PROBLEM 3 DEFINITION OF A PARABOLA Standard 4, 9, 17 SUMMARY OF FORMULAS PROBLEM 4 PROBLEM 5 PROBLEM

1

PARTS OF A PARABOLA

PARABOLAS

PROBLEM 1 PROBLEM 2

PROBLEM 3

DEFINITION OF A PARABOLA

Standard 4, 9, 17

SUMMARY OF FORMULAS

PROBLEM 4

PROBLEM 5 PROBLEM 6

PROBLEM 7 PROBLEM 8

END SHOW

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

2

STANDARD 4:

Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes

ESTÁNDAR 4:

Los estudiantes factorizan polinomios representando diferencia de cuadrados, trinomios cuadrados perfectos, y la suma de diferencia de cubos.

STANDARD 9:

Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a(x-b) + c.

ESTÁNDAR 9:

Los estudiantes demuestran y explican los efectos que tiene el cambiar coeficientes en la gráfica de funciones cuadráticas; esto es, los estudiantes determinan como la gráfica de una parabola cambia con a, b, y c variando en la ecuación y=a(x-b) + c

STANDARD 17:

Given a quadratic equation of the form ax + by + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.

Estándar 17:

Dada una equación cuadrática de la forma ax +by + cx + dy + e=0, los estudiantes pueden usar el método de completar al cuadrado para poner la ecuación en forma estándar y pueden reconocer si la gráfica es un círculo, elipse, parábola o hiperbola. Los estudiantes pueden graficar la ecuación

2

2

2 2

2 2

2

ALGEBRA II STANDARDS THIS LESSON AIMS:



3

Standard 4, 9, 17DEFINITION OF A PARABOLA:

A parabola is the set of all points in a plane that are the same distance from a given point called the focus and a given line called directrix.

x

y

directrix

focus



4

Standard 4, 9, 17PARTS OF A PARABOLA:

directrix

axis of symmetry

focusx

y

vertex

latus rectum



5

PARABOLAS: SUMMARY

Form of equation y = a(x-h) + k x = a(y-k) + h

Axis of symmetry x= h y = k

Vertex (h,k) (h,k)

Focus (h, k + ) (h + , k)

Directrix y = k - x = h -

Direction of openning

Upward if a>0

Downward if a<0

Right if a>0

Left if a<0

Length of latus rectum

units units

2 2

14a

14a

14a

14a

1a

1a

Standard 4, 9, 17



Standard 4, 9, 17

Rewriting the equation:

y = a(x-h) + k2

h= -5

k= 2

Vertex: (h,k) = (-5, 2)

Axis of symmetry: x= -5

Latus rectum:1a

1

y= (x+5) + 221

6

a=16

y= (x- -5) + (+2)21

6

16

=6

11

16

=

Find the vertex, axis of symmetry, focus, directrix and Latus rectum from and graph it.y= (x+5) + 2

216

Focus: (h, k + )1

4a

Directrix: y = k - 14a

y = 2 - 1

4( )

y = 2 -

16

=( -5, 2 + )1

4

=( -5, 2 + )1

46

23

11

= =32 =( -5, 2 + )

32

=( -5, )

16

32

1

46

46

11

=

2+ 32

22

= 42

32

+

=72

7 23

6 1

3 1 2

y = 122 - 3

222

= 42

32

-

=12

3 1 2



7

Standard 4, 9, 17

Summarizing the information about the parabola and graphing it:


Latus rectum = 6

a>0 so it opens upward.

Vertex: (-5, 2)

Focus: ( -5, )3 1 2

Directrix: y = 12

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

y = 12

x= -5



Standard 4, 9, 17


y = a(x-h) + k2

h= -6

k= 3

Vertex: (h,k) = (-6, 3)


Latus rectum:1a

1

a= - 18

y= - (x- -6) + (+3)218

=8

Focus: (h, k + )1

4a


y = 3 - 1

4( )

y = 3 -

=( -6, 3 + )1

4

=( -6, 3 + )

=( -6, 3 + )

=( -6, )

y = 3 + 2

1

-2

(-2)

= -2

y= - (x+6) + 321

8

18

-

1

48

-1

48

-48

11

=-

12

11

=-

=21-

+

=( -6, 3 - )2

18

-

y = 5

18

-

11

1 8

= -

Find the vertex, axis of symmetry, focus, directrix and Latus rectum from and graph it.y= - (x+6) + 3

218



9

Standard 4, 9, 17



Latus rectum =8

a<0 so it opens downward

Vertex: (-6, 3)

Focus: ( -6, 1)

Directrix: y = 5

y

84 12-4-8-12

4

8

12

-4

-8

-12

16 20-16-20

16

-16

20

x

x=- 6

y= 5



Standard 4, 9, 17

Write 4x= y -4y + 8 in the form and graph it.x = a(y-k) + h22

4x = y -4y + 82

4x = (y -4y + )+ 8 -2

42

242

2

(2)2(2)

2

4x = (y - 4y + )+ 8 - 2

4x =(y -4y + )+ 8 - 2 4 (4)

4x = (y -2) + 42

Changing the form of the equation:


x = a(y-k) + h2

h= 1

k= 2

Vertex: (h,k) = (1, 2)

Focus:

Directrix:

Axis of symmetry: y= 2

Latus rectum: 1a

1

4 4

x= (y-2) +21

444

x= (y-2) + 121

4

a=14

x= (y- +2) + (+1)21

4

(h + , k)14a

x = h - 14a

14

x = 1 - 14

x = 1 - 11

x= 0

14

=4

11

14

=

=(1 + , 2)1

4 14

= (1 + , 2)144

= ( 1 + , 2)11

=(2,2)



11

Standard 4, 9, 17


Vertex: (1,2)


Latus rectum = 4

a>0 so it opens rightward.

Focus: (2,2)

Directrix: x =021 3-1-2-3

1

2

3

-1

-2

-3

4 5-4-5

4

-4

5

x

y

axis of symmetry

directrix

vertex

latus rectum

focus



Standard 4, 9, 17

Write 4x= y -6y + 3 in the form and graph it.x = a(y-k) + h22

4x = y -6y + 32

4x = (y -6y + )+ 3 -2

62

262

2

(3)2(3)

2

4x = (y - 6y + )+ 3 - 2

4x =(y -6y + )+ 3 - 2 9 (9)

4x = (y -3) - 62



x = a(y-k) + h2

k= 3

Focus:

Directrix:


Latus rectum: 1a

1

4 4

x= (y-3) -21

464

a=14

(h + , k)14a

x = h - 14a

14

14

=4

11

14

=

14144

11

x= (y-3) -21

432

x= (y- +3) + ( )21

432

- h= 3

2-

Vertex: (h,k) = ( , 3)32

-

=( + , 3)1

4

32

-

= ( + , 3)32

-

=( + , 3)32

-

2 2

32

- 11+

32

-= +22

= - 12

=( ,3)12

-

x = - 14

32

-

x = - 11

32

- 2 2

x=32

- 22

- = - 52



13

Standard 4, 9, 17



Latus rectum = 4

a>0 so it opens rightward.

21 3-1-2-3

1

2

3

-1

-2

-3

4 5-4-5

4

-4

5

x

yVertex: (h,k) = ( , 3)3

2-

Focus: ( ,3)12

-

Directrix: x= - 52

12

y= 3

x= - 52



14

Standard 4, 9, 17

Write y= 4x + 24x + 16 in the form and graph it.y = a(x-h) + k22

y = 4x + 24x + 162

y = 4(x + 6x + )+ 16 -42

62

262

2

y = 4(x + 6x + )+ 16 -42 (3)2(3)

2

y = 4(x + 6x + )+ 16 -42 9 (9)

y = 4(x + 3) + 16 - 362

y = 4(x + 3) - 202



y = 4(x - -3) + (-20)2

y = a(x-h) + k2

h= -3k= -20a= 4

Vertex: (h,k) = (-3,-20)

Focus: (h, k + )1

4a =(-3, -20 + )1

4( )4

=(-3, -20 + )1

16

=(-3, )-191516


y = -20- 14( )4

y = -20 - 116

y = -201

16



Latus rectum: 1a

14 =.25


15

Standard 4, 9, 17


Vertex: (-3,-20)


Focus: (-3, )-191516

Directrix: y = -201

16


Latus rectum = .25

21 3-1-2-3

-21

-18

-15

4 5-4-5

-12

-9

x

y

-6

-3

-24

-27

directrix

axis of symmetry

vertex

latus rectum

focus-20


16

Standard 4, 9, 17

The coordinates of the focus and equation of the directrix of a parabola are as follows. Write an equation for the parabola and graph it.

(-3, -2); y = -6

If focus formula is (h, k + )1

4athen

k +1

4a = -2

If directrix formula is y = k - 14a

then

k -1

4a = -6

14a

14a++

k = - 6 +1

4a

Equation 1

Equation 2

Solving both equations by substitution:

+ = -21

4a-6 +1

4a

-6 + = -224a

+6 +6

= 42

4a (4a)(4a)

2 = 16a

2 = 16a16 16

a = 216

....

22

a = 18

Substituting a in equation 2:

k = -6 + 1

4 18

= -6 + 148

= -6 + 48

11

k = -6 + 84

= -6 + 2

k = -4

Since the directrix is a horizontal line the parabola is a vertical parabola. And the following applies:

h = -3 and


17

Standard 4, 9, 17

Summarizing information about the parabola:

h= -3k= -4

a=18

So the equation is:

y= (x- -3) + (-4)21

8

y = (x+3) - 421

8

Latus rectum: 1a

118

=8

11

18

=

Vertex: (-3,-4)

Focus: (-3, -2)

Directrix: y = -6

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

y= -6

x=- 3

Axis of symmetry is x = -3


18

Standard 4, 9, 17

21 3-1-2-3

1

2

3

-1

-2

-3

4 5-4-5

4

-4

5

x

y

Find the equation of the parabola shown in the graph.

From the graph:

Vertex = (-3,1)

This is a horizontal parabola so:

x= (y- ) + ( )2

a +1 -3

x= (y-1) -32

a

We know that the graph pass through point (x,y)=(-2,3). We use this point to substitute it on the equation to find a:

( )= (( )-1) -32

a-2 3

x= (y-1) -32

a

-2 = a(3-1) - 32

-2 = a(2) - 32

-2 = a(4) - 3

-2 = 4a - 3

+3 +3

1 = 4a

4 4

a = 14

Now we substitute in the parabola’s equation:

x= (y-1) - 321

4

x = a(y-k) + h2


19

Standard 4, 9, 17

21 3-1-2-3

1

2

3

-1

-2

-3

4 5-4-5

4

-4

5

x

y

Find the equation of the parabola shown in the graph.

From the graph:

Vertex = (2,-4)

This is a vertical parabola so:

y= (x- ) + ( )2

a +2 -4

y= (x-2) -42

a

We know that the graph pass through point (x,y)=(0,-3). We use this point to substitute it on the equation to find a:

( )= (( )-2) -42

a-3 0

y= (x-2) -42

a

-3 = a(0-2) - 42

-3 = a(-2) - 42

-3 = a(4) - 4

-3 = 4a - 4

+4 +4

1 = 4a

4 4

a = 14

Now we substitute in the parabola’s equation:

y= (x-2) - 421

4

y = a(x-h) + k2


Documents

1 PARTS OF A PARABOLA PARABOLAS PROBLEM 1 PROBLEM 2 PROBLEM 3 DEFINITION OF A PARABOLA Standard 4, 9, 17 SUMMARY OF FORMULAS PROBLEM 4 PROBLEM 5 PROBLEM