Unit 8.1

Copyright © 2011 Pearson, Inc.

8.1

Conic Sections

and Parabolas

Copyright © 2011 Pearson, Inc. Slide 8.1 - 2

What you’ll learn about

Conic Sections

Geometry of a Parabola

Translations of Parabolas

Reflective Property of a Parabola

… and why

Conic sections are the paths of nature: Any free-moving

object in a gravitational field follows the path of a conic

section.


A Right Circular Cone (of two nappes)


Conic Sections and

Degenerate Conic Sections


Conic Sections and

Degenerate Conic Sections (cont’d)


Second-Degree (Quadratic) Equations

in Two Variables

The conic sections can defined algebraically as the

graphs of second - degree (quadratic) equations

in two variables, that is, equations of the form

Ax2 Bxy Cy2 Dx Ey F 0,

where A, B, and C, are not all zero.


Parabola

A parabola is the

set of all points in

a plane equidistant

from a particular

line (the directrix)

and a particular

point (the focus)

in the plane.


Graphs of x2 = 4py


Parabolas with Vertex (0,0)

Standard equation x2 = 4py y2 = 4px

Opens Upward or To the right or to the

downward left

Focus (0, p) (p, 0)

Directrix y = –p x = –p

Axis y-axis x-axis

Focal length p p

Focal width |4p| |4p|


Graphs of y2 = 4px


Example Finding an Equation of a

Parabola

Find an equation in standard form for the parabola

whose directrix is the line x 3 and whose focus is

the point ( 3,0).



Parabola

Because the directrix is x 3 and the focus is ( 3,0),

the focal length is 3 and the parabola opens to the left.

The equation of the parabola in standard from is:

y2 4 px

y2 12x

Find an equation in standard form for the parabola

whose directrix is the line x 3 and whose focus is

the point ( 3,0).


Parabolas with Vertex (h,k)

Standard equation (x– h)2 = 4p(y – k) (y – k)2 = 4p(x – h)

Opens Upward or To the right or to the left

downward

Focus (h, k + p) (h + p, k)

Directrix y = k-p x = h-p

Axis x = h y = k

Focal length p p

Focal width |4p| |4p|



Parabola

Find the standard form of the equation for the parabola

with vertex at (1,2) and focus at (1, 2).



Parabola

The parabola is opening downward so the equation

has the form

(x h)2 4 p( y k).

(h,k) (1,2) and the distance between the vertex and

the focus is p 4.

Thus, the equation is (x 1)2 16( y 2).

Find the standard form of the equation for the parabola

with vertex at (1,2) and focus at (1, 2).


Quick Review

1. Find the distance between ( 1,2) and (3, 4).

2. Solve for y in terms of x. 2y2 6x

3. Complete the square to rewrite the equation in vertex form.

y x2 2x 5

4. Find the vertex and axis of the graph of f (x) 2(x 1)2 3.

Describe how the graph of f can be obtained from the graph

of g(x) x2 .

5. Write an equation for the quadratic function whose graph

contains the vertex (2, 3) and the point (0,3).


Quick Review Solutions

2

2 2

1. Find the distance between ( 1,2) and (3, 4).

2. Solve for in terms of . 2 6

3. Complete the square to rewrite the equation in vertex form.

2 5

4. Find

52

3

( 1

the ver

4

tex

)

y x y x

y xy

y

x x

x

2

2

and axis of the graph of ( ) 2( 1) 3.

Describe how the graph of can be obtained from the graph

vertex:( 1,3); axis: 1; translation left 1 unit,

vertical stretch by a factor of

of ( ) .

2,

f x x

f

g xx x

2

5. Write an equation for the quadratic function whose graph

contains the vertex (2, 3) and

translation up 3 u

the point (0,3).

nits.

32 3

2y x

Date post:	15-Jul-2015
Category:	Education
Upload:	mark-ryder
View:	233 times
Download:	4 times

Unit 8.1

Education