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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.
Copyright © 2011 Pearson, Inc. Slide 8.1 - 3
A Right Circular Cone (of two nappes)
Copyright © 2011 Pearson, Inc. Slide 8.1 - 4
Conic Sections and
Degenerate Conic Sections
Copyright © 2011 Pearson, Inc. Slide 8.1 - 5
Conic Sections and
Degenerate Conic Sections (cont’d)
Copyright © 2011 Pearson, Inc. Slide 8.1 - 6
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.
Copyright © 2011 Pearson, Inc. Slide 8.1 - 7
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.
Copyright © 2011 Pearson, Inc. Slide 8.1 - 8
Graphs of x2 = 4py
Copyright © 2011 Pearson, Inc. Slide 8.1 - 9
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|
Copyright © 2011 Pearson, Inc. Slide 8.1 - 10
Graphs of y2 = 4px
Copyright © 2011 Pearson, Inc. Slide 8.1 - 11
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).
Copyright © 2011 Pearson, Inc. Slide 8.1 - 12
Example Finding an Equation of a
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).
Copyright © 2011 Pearson, Inc. Slide 8.1 - 13
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|
Copyright © 2011 Pearson, Inc. Slide 8.1 - 14
Example Finding an Equation of a
Parabola
Find the standard form of the equation for the parabola
with vertex at (1,2) and focus at (1, 2).
Copyright © 2011 Pearson, Inc. Slide 8.1 - 15
Example Finding an Equation of a
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).
Copyright © 2011 Pearson, Inc. Slide 8.1 - 16
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).
Copyright © 2011 Pearson, Inc. Slide 8.1 - 17
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