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10-5 Day 1 Parabolas

Battle of the CST’s

Lesson Presentation

Lesson Quiz

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Write the standard equation of aparabola and its axis of symmetry.

Graph a parabola and identify its focus,directrix, and axis of symmetry.

Objectives

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In Chapter 5, you learnedthat the graph of aquadratic function is aparabola. Because aparabola is a conic section,it can also be defined interms of distance.

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A parabola is the set of all pointsP ( x , y ) in a plane that are anequal distance from both a fixedpoint, the focus, and a fixed

line, the directrix. A parabolahas a axis of symmetryperpendicular to its directrix andthat passes through its vertex.

The vertex of a parabola is themidpoint of the perpendicularsegment connecting the focusand the directrix.

The distance from a point to a line is defined asthe length of the line segment from the pointperpendicular to the line.

Remember!

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Use the Distance Formula to find the equation of aparabola with focus F (2, 4) and directrix y = –4.

Example 1: Using the Distance Formula to Write the

Equation of a Parabola

Definition of a parabola. PF = PD

Substitute (2, 4)

for (x 1, y 1 ) and (x, –4) for (x 2 , y 2 ).

Distance

Formula.

( x –

2)

2

+ (y –

4)

2

= (y + 4)

2

Square both sides.

Simplify.

Expand.( x – 2)2+ y

2 – 8y + 16 = y

2+ 8y + 16

Subtract y 2 and 16 from both sides.( x – 2)

2– 8y = 8y

( x – 2)2= 16y Add 8y to both sides.

Solve for y.

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Use the Distance Formula to find the equation of aparabola with focus F (0, 4) and directrix y = –4.

Definition of a parabola. PF = PD

Substitute (0, 4)

for (x 1, y 1 ) and

(x, –4) for (x 2 , y 2 ).

Distance Formula

Check for Understanding

x 2+ (y – 4)

2= (y + 4)

2 Square both sides.

Expand. x 2+ y

2 – 8y + 16 = y

2+ 8y +16

Simplify.

Subtract y 2 and 16 from both sides. x

2– 8y = 8y

x 2= 16y Add 8y to both sides.

Solve for y.

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Previously, you have graphedparabolas with vertical axes of

symmetry that open upward ordownward. Parabolas may also havehorizontal axes of symmetry and

may open to the left or right.The equations of parabolas use theparameter p. The | p| gives the

distance from the vertex to both thefocus and the directrix.

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Write the equation in standard form for theparabola.

Example 2A: Writing Equations of Parabolas

Step 1 Because the axis

of symmetry is verticaland the parabola opensdownward, the equationis in the form

y = x 2 with p < 0.1

4 p

Step 2 The distance from the focus (0, –5) tothe vertex (0, 0), is 5, so p = –5 and 4 p = –20.

Step 3 The equation of the parabola is .y = – x 2 1

20

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Example 2B: Writing Equations of Parabolas

vertex (0, 0), directrix x = –6

Write the equation in standard form for theparabola. Step 1 Because the directrix is a vertical

line, the equation is in the form . The

vertex is to the right of the directrix, so the

graph will open to the right.

Step 2 Because the directrix is x = –6, p = 6and 4 p = 24.

Step 3 The equation of the parabola is . x = y 2 1

24

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vertex (0, 0), directrix x = 1.25

Check for Understanding Write the equation in standard form for theparabola. Step 1 Because the directrix is a vertical line,

the equation is in the form of . The

vertex is to the left of the directrix, so the

graph will open to the left.

Step 2 Because the directrix is x = 1.25, p = –1.25 and 4 p = –5.

Step 3 The equation of the parabola is

END of Day 1

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10-5 Day 2 Parabolas

Battle of the CST’s

Lesson Presentation

Lesson Quiz

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Write the standard equation of aparabola and its axis of symmetry.

Graph a parabola and identify its focus,directrix, and axis of symmetry.

Objectives

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The vertex of a parabola may not always be theorigin. Adding or subtracting a value from x or y translates the graph of a parabola. Also notice thatthe values of p stretch or compress the graph.

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Example 3: Graphing Parabolas

Step 1 The vertex is (2, –3).

Find the vertex, value of p, axis of

symmetry, focus, and directrix of the

parabola Then graph. y + 3 = ( x – 2)2.1

8

Step 2 , so 4 p = 8 and p = 2.1

4 p

1

8 =

Step 4 The focus is (2, –3 + 2),

or (2,–1).

Step 3 The graph has a vertical axis of symmetry,with equation x = 2, and opens upward.

Step 5 The directrix is ahorizontal liney = –3 – 2, ory = –5.

C f

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Step 1 The vertex is (1, 3).

Find the vertex, value of p, axis of symmetry,focus, and directrix of the parabola. Then graph.

Step 2 , so 4 p = 12 and p = 3.1

4p

1

12 =

Check for understanding

Step 4 The focus is (1 + 3, 3),or (4, 3).

Step 5 The directrix is avertical line x = 1 – 3,

or x = –2.

Step 3 The graph has a horizontal axis of symmetry with equation y = 3, and opensright.

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Light or sound waves collectedby a parabola will be reflectedby the curve through the focus

of the parabola, as shown inthe figure. Waves emittedfrom the focus will be reflectedout parallel to the axis of symmetry of a parabola. Thisproperty is used incommunications technology.

E l 4 U i th E ti f P b l

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The cross section of a larger parabolic

microphone can be modeled by the

equation What is the length of

the feedhorn?

Example 4: Using the Equation of a Parabola

x = y 2.1

132

The equation for the cross section is in the form

x = y 2,

1

4 p so 4 p = 132 and p = 33. The focus

should be 33 inches from the vertex of the crosssection. Therefore, the feedhorn should be 33

inches long.

Ch k F U d t di

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Check For Understanding Find the length of the feedhorn for a microphone

with a cross section equation x = y 2

.1

44

The equation for the cross section is in the form

x = y 2,

1

4 p

so 4 p = 44 and p = 11. The focus

should be 11 inches from the vertex of the crosssection. Therefore, the feedhorn should be 11inches long.

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