1. DEFINITION OF A PARABOLA
Standard 4, 9, 17
PARABOLAS
DEFINITION OF A PARABOLA
PARTS OF A PARABOLA
SUMMARY OF FORMULAS
PROBLEM 1
PROBLEM 2
PROBLEM 3
PROBLEM 4
PROBLEM 5
PROBLEM 6
PROBLEM 7
PROBLEM 8
END SHOW
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2. ALGEBRA II STANDARDS THIS LESSON AIMS:
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
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3. DEFINITION OF A PARABOLA:
Standard 4, 9, 17
DEFINITION 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
focus
directrix
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4. Standard 4, 9, 17 PARTS OF A PARABOLA: y axis of symmetry latus rectum
focus
vertex
directrix
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5. PARABOLAS: SUMMARY Form of equation y = a(x-h) + k x = a(y-k) + h
Standard 4, 9, 17
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)
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 2 2 1 4a 1 4a 1 4a 1 4a 1 a 1 a
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6. Standard 4, 9, 17 (h, k + ) y = a(x-h) + k y = k -
Find the vertex, axis of symmetry, focus, directrix and Latus rectum from and graph it.
y= (x+5) + 2 2 1 6
(h, k ) 1 4a
=( -5, ) 1 4
Focus: 1 6
y= (x+5) + 2 2 1 6 1 4 6
=( -5, )
Rewriting the equation: 4 6 1 = 2 3 1 =
y= (x- -5) + (+2) 2 1 6 1 4 6 = 3 2 3 2
=( -5, )
y = a(x-h) + k 2 3 1 2
=( -5, ) 2+ 3 2 = 4 2 3 + 2
y = k - 1 4a
h= -5
Directrix: = 7 2
k= 2
y = 2 - 1
4( ) a= 1 6 3 1 6 7 2 3 1 2 6
Vertex: (h,k) = (-5, 2) 3 2 1
y = 2 -
Axis of symmetry:
x= -5
y = 1 2
2 - 3 2 = 4 2 3 - 2 1 6 = = 1 2 1 1 a =6
Latus rectum: 1 6
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7. Summarizing the information about the parabola and graphing it:
Standard 4, 9, 17
Summarizing the information about the parabola and graphing it:
x= -5 4 2 6 -2 -4 -6 8 10 -8
-10 x y
Vertex: (-5, 2)
Axis of symmetry:
x= -5
Focus:
( -5, ) 3 1 2
y = 1 2
Directrix:
y = 1 2
Latus rectum
= 6
a>0 so it opens upward.
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8. Standard 4, 9, 17 (h, k + ) y = a(x-h) + k y = k -
Find the vertex, axis of symmetry, focus, directrix and Latus rectum from and graph it.
y= (x+6) + 3 2 1 8
(h, k ) 1 4a
=( -6, ) 1 4
Focus: 1 8 -
y= (x+6) + 3 2 1 8 1 4 8 -
=( -6, ) 4 8 1 = - 1 2 = -
Rewriting the equation: 1 4 8 - = 2 1 - +
y= (x- -6) + (+3) 2 1 8
=( -6, ) -2
y = a(x-h) + k 2
= -2
=( -6, ) 2
y = k - 1 4a
=( -6, ) 1
h= -6
Directrix:
k= 3
y = 3 - 1
4( )
a= - 1 8 1 8 -
Vertex: (h,k) = (-6, 3)
y = 3 -
(-2)
Axis of symmetry:
x= -6
y = 3 + 2
y = 5 1 8 = - 1 1 a =8
Latus rectum: 1 8 -
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9. Summarizing the information about the parabola and graphing it:
Standard 4, 9, 17
Summarizing the information about the parabola and graphing it: y 8 4 12 -4 -8
-12 16 20
-16
-20 x
Vertex: (-6, 3)
x=- 6
Axis of symmetry:
x= -6
y= 5
Focus:
( -6, 1)
Directrix:
y = 5
Latus rectum =8
a<0 so it opens downward
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10. Standard 4, 9, 17 x = a(y-k) + h (h + , k) x = h - x = a(y-k) + h
Write 4x= y -4y + 8 in the form and graph it.
x = a(y-k) + h 2
Changing the form of the equation:
Vertex: (h,k) = (1, 2)
4x = y -4y + 8 2
Axis of symmetry:
y= 2 4 2
(h , k) 1 4a
=( , 2) 1 4
Focus:
4x = (y -4y )+ 8 - 2 1 4
(2) 2
4x = (y - 4y )+ 8 - 2 1 4
= ( , 2)
4x =(y -4y )+ 8 - 2 4
(4)
4x = (y -2) + 4 2
= ( , 2) 1
=(2,2)
x = h - 1 4a
Directrix:
x= (y-2) + 2 1 4
x = 1 - 1 4 1 4
x= (y-2) + 1 2 1 4
x = 1 - 1
Rewriting the equation:
h= 1 1 4 =
x= (y- +2) + (+1) 2 1 4
x= 0
k= 2 1
x = a(y-k) + h 2 a= 1 4 =4 1 a 1 4
Latus rectum:
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11. Summarizing the information about the parabola and graphing it:
Standard 4, 9, 17
Summarizing the information about the parabola and graphing it:
directrix 2 1 3 -1 -2 -3 4 5 -4 -5 x y
Vertex: (1,2)
latus rectum
Axis of symmetry:
y= 2
axis of symmetry
focus
Focus:
(2,2)
vertex
Directrix:
x =0
Latus rectum
= 4
a>0 so it opens rightward.
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12. Standard 4, 9, 17 x = a(y-k) + h (h + , k) x = h - x = a(y-k) + h
Write 4x= y -6y + 3 in the form and graph it.
x = a(y-k) + h 2
Vertex: (h,k) = ( , 3) 3 2 -
Changing the form of the equation:
4x = y -6y + 3 2
Axis of symmetry:
y= 3 6 2
(h , k) 1 4a
=( , 3) 1 4 3 2 -
Focus:
4x = (y -6y )+ 3 - 2 1 4
(3) 2
= ( , 3) 3 2 - 1 4
4x = (y - 6y )+ 3 - 2 3 2 - 1 + 2 3 2 - = +
4x =(y -6y )+ 3 - 2 9
(9)
= - 1 2
=( , 3) 3 2 - 1
4x = (y -3) - 6 2 1 2 -
x = h - 1 4a
=( ,3)
x= (y-3) - 2 1 4 6
Directrix:
x = 1 4 3 2 -
x= (y-3) - 2 1 4 3 1 4
x = 1 3 2 - 2
Rewriting the equation: h= 3 2 - 1 4 =
x= (y- +3) + ( ) 2 1 4 3 - x= 3 2 -
= - 5 2
k= 3 1
x = a(y-k) + h 2 a= 1 4 =4 1 a 1 4
Latus rectum:
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13. Summarizing the information about the parabola and graphing it: x = -
Standard 4, 9, 17
Summarizing the information about the parabola and graphing it: x
= - 5 2
Vertex: (h,k) = ( , 3) 3 2 - 2 1 3 -1 -2 -3 4 5 -4 -5 x y
Axis of symmetry:
y= 3
y= 3
Focus:
( ,3) 1 2 -
Directrix: x
= - 5 2
Latus rectum
= 4
a>0 so it opens rightward. 1 2
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14. Standard 4, 9, 17 y = a(x-h) + k (h, k + ) y = k - y = a(x-h) + k
Write y= 4x + 24x + 16 in the form and graph it.
y = a(x-h) + k 2
Changing the form of the equation:
Vertex: (h,k) = (-3,-20)
y = 4x + 24x + 16 2
Axis of symmetry:
x= -3 6 2
y = 4(x + 6x ) 2
(h, k ) 1 4a
=(-3, ) 1
4( )
Focus: 4
(3) 2
y = 4(x + 6x ) 2
=(-3, ) 1 16
y = 4(x + 6x ) 2 9
(9)
=(-3, )
-19 15 16
y = 4(x + 3) 2
y = k - 1 4a
Directrix:
y = 4(x + 3) - 20 2
y = -20- 1
4( )
Rewriting the equation: 4
y = -20 - 1 16
y = 4(x - -3) + (-20) 2
y = a(x-h) + k 2
y = -20 1 16
h= -3
a>0 so it opens upward.
k= -20 1 a 1 4
Latus rectum:
=.25
a= 4
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15. Summarizing the information about the parabola and graphing it:
Standard 4, 9, 17
Summarizing the information about the parabola and graphing it: 2 1 3 -1 -2 -3
-21
-18
-15 4 5 -4 -5
-12 -9 x y -6
-24
-27
Vertex: (-3,-20)
axis of symmetry
Axis of symmetry:
x= -3
Focus:
(-3, )
-19 15 16
Directrix:
y = -20 1 16
Latus rectum
= .25
a>0 so it opens upward.
latus rectum
focus
-20
vertex
directrix
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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
Since the directrix is a horizontal line the parabola is a vertical parabola. And the following applies:
If focus formula is
(h, k ) 1 4a
then
If directrix formula is
y = k - 1 4a
then
h = -3
and
k + 1 4a
= -2
k - 1 4a
= -6
Equation 1 1 4a +
Solving both equations by substitution:
= -2 1 4a
-6 + 1 4a
k = - 6 + 1 4a
Equation 2
= -2 2 4a
Substituting a in equation 2:
2 = 16a 4 8 1
k = -6 + 1 4
= -6 + 1 4 8
= -6 + 1 8 . 2
= 4 2 4a
a = 2 16
(4a)
k = -6 + 8 4 =
2 = 16a
a = 1 8
k = -4
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17. Summarizing information about the parabola:
Standard 4, 9, 17
Summarizing information about the parabola:
h= -3
So the equation is:
k= -4 a= 1 8
y= (x- -3) + (-4) 2 1 8
Vertex: (-3,-4)
y = (x+3) - 4 2 1 8
Axis of symmetry is x = -3
Focus: (-3, -2) 4 2 6 -2 -4 -6 8 10 -8
-10 x y 1 8 =
Directrix: y = -6
x=- 3 1 =8 1 a 1 8
Latus rectum:
y= -6
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18. Standard 4, 9, 17
Find the equation of the parabola shown in the graph. 2 1 3 -1 -2 -3 4 5 -4 -5 x y
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:
x= (y-1) -3 2 a
( )= (( )-1) -3 2 a -2 3
-2 = a(3-1) - 3 2
-2 = a(2) - 3 2
-2 = a(4) - 3
-2 = 4a - 3
From the graph:
Vertex = (-3,1)
1 = 4a
This is a horizontal parabola so:
4 4
x = a(y-k) + h 2
Now we substitute in the parabola’s equation:
a = 1 4
x= (y- ) + ( ) 2 a +1 -3
x= (y-1) - 3 2 1 4
x= (y-1) -3 2 a
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19. Standard 4, 9, 17
Find the equation of the parabola shown in the graph. 2 1 3 -1 -2 -3 4 5 -4 -5 x y
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:
y= (x-2) -4 2 a
( )= (( )-2) -4 2 a -3
-3 = a(0-2) - 4 2
-3 = a(-2) - 4 2
-3 = a(4) - 4
-3 = 4a - 4
From the graph:
Vertex = (2,-4)
1 = 4a
This is a vertical parabola so:
4 4
y = a(x-h) + k 2
Now we substitute in the parabola’s equation:
a = 1 4
y= (x- ) + ( ) 2 a +2 -4
y= (x-2) - 4 2 1 4
y= (x-2) -4 2 a
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