1. OHHS Pre-Calculus Mr. J. Focht
Analytic Geometry in Two
and Three Dimensions
Chapter 8
OHHS
Pre-Calculus
Mr. J. Focht

2. 8.1 Conic Section: the Parabola
What
You'll
Learn
Geometry of a Parabola
Translations of Parabolas
Reflective Property
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3. Parabola
Set of all points that are equidistant from a line and a point.
Directrix
Focus
Vertex
Axis of Symmetry
Any point (x,y) is as far from the line as it is from the focus
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4. A Parabola is a Conic Section
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5. Parabola All parallel beams reflect through the focus
Why is this important?
Think satellite dish, flashlight, headlight
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6. Parabola Equation Vertex (0,0)
F(0, p) p
A(x,y) p
D(x,-p)
p = Focal Length =
distance from focus to vertex
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7. Parabola Equation Vertex (0,0)
A(x,y) p
F(0, p)
D(x,-p)
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8. Parabola Equation Vertex (0,0)
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9. Parabola Equation Vertex (0,0)
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10. Definition: Latus Rectum
Segment passing through the focus parallel to the directrix
Focal Width is the length of the Latus Rectum. This length is |4p| 4p
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11. Parabola Equations Summary
x2 = 4py p > 0
x2 = 4py p < 0
y2 = 4px p > 0
y2 = 4px p < 0
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12. Example
Find the focus, the directrix, and the focal width of the parabola y = -½x2.
First put into standard form
x2 = -2y
4p = -2
p = -½
D(0,½)
Y = ½
(0,0)
FW = 2
F(0,-½)
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13. Now You Try
P. 641, #1 : Find the focus, the directrix, and the focal width of the parabola
x2 = 6y
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14. Example
Find an equation in standard form for the parabola whose directrix is the line x=2 and whose focus is the point (-2,0).
y2 = 4px
y2 = -8x
x=2
(-2,0)
(0,0)
p = -2
p = -2
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15. Now You Try
P. 641 #15: Find the standard form of a parabola with focus (0, 5), directrix y=-5
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16. Standard Form of the Equation of Vertex (h, k)
(y-k)2 = 4p(x-h) p
Directrix
(h, k)
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17. Standard Form of the Equation of Vertex (h, k)
(x-h)2 = 4p(y-k) p
(h, k)
Directrix
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18. Example (y-3)2 = 8(x+4) Find the equation of this parabola.
(y-k)2 = 4p(x-h)
p = distance from vertex to focus
(-4,3)
(-2,3)
p = 2 h = k = 3
(y-3)2 = 8(x+4)
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19. Example (y+1)2 = -10(x-3) Find the equation of this parabola.
(y-k)2 = 4p(x-h)
p = distance from vertex to focus
(-2, -1)
(3, -1)
p = -5 h = k = -1
(y+1)2 = -10(x-3)
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20. Now You Try P. 641, #21 Find the equation of this parabola.
Focus (-2, -4), vertex (-4, -4)
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21. What’s the sign of p?
p > 0
p < 0
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22. Graphing a Parabola
Graph (y-4)2 = 8(x-3)
Change to
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23. Example x2-6x-12y-15=0 x2 -6x = 12y + 15 h = 3 k = -2 p = 3
Find the coordinates of the vertex and focus, and the equations of the directrix and axis of symmetry.
x2-6x-12y-15=0
Put the equation into standard form.
x2 -6x = 12y + 15
+ 9
+ 9
(x-3)2 = 12y+24
(x-3)2 = 12(y+2)
h = 3 k = -2 p = 3
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24. Example (x-3)2 = 12(y+2) h = 3 k = -2 p = 3 x = 3
(3, 1)
(3, -2)
The focus is 3 above the vertex.
y = -5
The directrix is a horizontal line 3 below the vertex.
(3, -5)
The line of symmetry passes through the vertex and focus
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25. Now Your Turn
P. 651 #49: Prove that the graph of the equation x2 + 2x – y + 3 = 0 is a parabola, and find its vertex, focus, and directrix.
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26. Sketch a Graph Example Graph (y-2)2 = 8(x-1) The vertex is (1,2)
Focus = (3,2)
Focal Width = 8
4 above & 4 below
the focus.
(3,6)
(1,2)
(3,2)
(3,-2)
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27. Now Your Turn
P. 641, #33: Sketch the parabola by hand: (x+4)2 = -12(y+1)
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28. Application Example
On the sidelines of each of its televised football games, the FBTV network uses a parabolic reflector with a microphone at the reflector’s focus to capture the conversations among players on the field. If the parabolic reflector is 3 ft across and 1 ft deep, where should the microphone be placed?
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29. Application Example x2 = 4py (1.5, 1) is on the parabola.
p = ft
p = 6.75 in
The microphone should be placed inside the reflector along its axis and 6.75 inches from its vertex.
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30. Your Turn Now
P. 652, #59 The mirror of a flashlight is a paraboloid of revolution. Its diameter is 6 cm and its depth is 2 cm. How far from the vertex should the filament of the light bulb be placed for the flashlight to have its beam run parallel to the axis of its mirror?
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31. Home Work P
#2, 4, 12, 16, 22, 28, 34, 42, 50, 52, 56, 60, 65-70
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32. The End