OHHS

Pre-Calculus

Mr. J. Focht

8.1 Conic Section: the Parabola

• Geometry of a Parabola

• Translations of Parabolas

• Reflective Property

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Parabola

• Set of all points that are equidistant from a line and a point.

Directrix FocusVertex

Axis of Symmetry

Any point (x,y) is as far from the line as it is from the focus

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A Parabola is a Conic Section

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Parabola

All parallel beams reflect through the focus

Why is this important?

Think satellite dish, flashlight, headlight

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Parabola EquationVertex (0,0)

A(x,y)p

p

F(0, p)

D(x,-p)

p = Focal Length =

distance from focus to vertex8.1

Parabola EquationVertex (0,0)

A(x,y)p

p

F(0, p)

D(x,-p)

22 p)(yxAF 2 2AD (x-x) (y p)

y p

AF=AD8.1

Parabola EquationVertex (0,0)

2 2x (y p) y p

22 2( )x y p y p

AF=AD

2 2 22x y py p 2 22y py p

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Parabola EquationVertex (0,0)

2 2 2x + y - 2py + p = 2 2y + 2py + p2x - 2py = 2py

2x = 4py

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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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Parabola EquationsSummary

• x2 = 4py p > 0

• x2 = 4py p < 0

• y2 = 4px p > 0

• y2 = 4px p < 0

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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

(0,0)

F(0,-½)

D(0,½)

Y = ½

FW = 2

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Now You Try

• P. 641, #1 : Find the focus, the directrix, and the focal width of the parabola

x2 = 6y

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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 = -8xx=2

(-2,0) (0,0)

p = -2 p = -2

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Now You Try

• P. 641 #15: Find the standard form of a parabola with focus (0, 5), directrix y=-5

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Standard Form of the Equation of Vertex (h, k)

(y-k)2 = 4p(x-h)

(h, k)p

Dir

ectr

ix

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Standard Form of the Equation of Vertex (h, k)

(x-h)2 = 4p(y-k)

(h, k)

p

Directrix8.1

Example

• Find the equation of this parabola.

(-2,3)(-4,3)

p = distance from vertex to focus

(y-k)2 = 4p(x-h)

p = 2 h = -4 k = 3

(y-3)2 = 8(x+4)

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Example

• Find the equation of this parabola.

(3, -1)(-2, -1)

p = distance from vertex to focus

(y-k)2 = 4p(x-h)

p = -5 h = 3 k = -1

(y+1)2 = -10(x-3)

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Now You Try

• P. 641, #21

Find the equation of this parabola.

Focus (-2, -4), vertex (-4, -4)

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What’s the sign of p?

p > 0

p < 0

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Graphing a Parabola

• Graph (y-4)2 = 8(x-3)

• Change to 3)8(x4y

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Example

• 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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Example

(x-3)2 = 12(y+2)

h = 3 k = -2 p = 3

(3, -2)The focus is 3 above the vertex.

(3, 1)

The directrix is a horizontal line 3 below the vertex.

(3, -5)

y = -5

The line of symmetry passes through the vertex and focus

x = 3

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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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Sketch a Graph Example

• Graph (y-2)2 = 8(x-1)

• The vertex is (1,2)

• 4p = 8

• p = 2

• Focus = (3,2)

• Focal Width = 8

• 4 above & 4 below

the focus.

(1,2) (3,2)

(3,6)

(3,-2)

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Now Your Turn

• P. 641, #33: Sketch the parabola by hand: (x+4)2 = -12(y+1)

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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?

8.1

Application Example

x2 = 4py(1.5, 1) is on the

parabola.(1.5)2 = 4p(1)2.25 = 4pp = 2.25/4p = 0.5625 ftp = 6.75 inThe microphone should be

placed inside the reflector along its axis and 6.75 inches from its vertex.

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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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Home Work

• P. 641-642

• #2, 4, 12, 16, 22, 28, 34, 42, 50, 52, 56, 60, 65-70

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