ohhs pre-calculus mr. j. focht. 8.1 conic section: the parabola geometry of a parabola translations...

OHHS

Pre-Calculus

Mr. J. Focht

8.1 Conic Section: the Parabola

• Geometry of a Parabola

• Translations of Parabolas

• Reflective Property

8.1

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

8.1

A Parabola is a Conic Section

8.1

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

8.1

Parabola EquationVertex (0,0)

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

2x = 4py

8.1

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

8.1

Parabola EquationsSummary

• x2 = 4py p > 0

• x2 = 4py p < 0

• y2 = 4px p > 0

• y2 = 4px p < 0

8.1

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

8.1

Now You Try

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

x2 = 6y

8.1

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

8.1

Now You Try

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

8.1

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

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

(h, k)p

Dir

ectr

ix

8.1

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)

8.1

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

8.1

Graphing a Parabola

• Graph (y-4)2 = 8(x-3)

• Change to 3)8(x4y

8.1

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

8.1

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

8.1

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.

8.1

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)

8.1

Now Your Turn

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

8.1

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.

8.1

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?

8.1

Home Work

• P. 641-642

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

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