10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
6.1 - 1
6.1
Parabolas
Conic Sections
Horizontal Parabolas
Geometric Definition and Equations of
Parabolas
An Application of Parabolas
6.1 - 2
Conic Sections
Parabolas, circles, ellipses, and hyperbolas form
a group of curves known as conic sections,
because they are the results of intersecting a
cone with a plane. The graphics shown on the
next slide illustrate these curves. We studied
circles and some parabolas (those that open up
or down, that is, vertical parabolas) in Chapters
2 and 3.
6.1 - 3
Conic Sections
6.1 - 4
Horizontal Parabolas
From Chapter 3, we know that the graph of
the equation
2
y  a( x  h )  k
is a parabola with vertex (h,k) and the vertical
line x = h as axis. If we subtract k from each
side, this equation becomes
y  k  a( x  h ) .
2
Subtract k.
(1)
6.1 - 5
Horizontal Parabolas
y  k  a( x  h )2
Subtract k.
(1)
The equation
x  h  a( y  k )2
also has a parabola as its graph. While the graph
of y – k = a(x – h)2 has a vertical axis, the graph
of x – h = a(y – k)2 has a horizontal axis. The
graph of the first equation is the graph of a
function (specifically a quadratic function), while
the graph of the second equation is not.
6.1 - 6
Parabola with Horizontal Axis
The parabola with vertex (h, k) and the
horizontal line y = k as axis has an equation
of the form
x  h  a( y  k )2 .
The parabola opens to the right if a > 0 and
to the left if a < 0.
6.1 - 7
Note
When the vertex (h, k) is (0, 0) and
a = 1 in
y  k  a( x  h )2
(1)
and
y  h  a( y  k ) ,
(2)
2
the equations y = x2 and
x = y2 result. Notice that
the graphs are mirror
images of each other with
respect to the line y = x.
6.1 - 8
Example 1
GRAPHING A PARABOLA WITH
A HORIZONTAL AXIS
Graph x + 3 = (y – 2)2. Give the domain and range.
Solution
The graph of x + 3 = (y – 2)2 or x – (–3) = (y – 2)2
has vertex (–3, 2) and opens to the right because
a = 1, and 1 > 0. Plotting a few additional
points gives the graph shown in the next slide.
6.1 - 9
GRAPHING A PARABOLA WITH
A HORIZONTAL AXIS
Example 1
Note that the graph is symmetric about its axis,
The domain is [–3, ), and the range is (–, ).
x
–3
–2
–2
1
1
y
2
3
1
4
0
6.1 - 10
GRAPHING A PARABOLA WITH A
HORIZONTAL AXIS
When an equation of a horizontal parabola is
given in the form
x  ay 2  by  c,
completing the square on y allows us to write
the equation in the form
x  h  a( y  k )2
and more easily find the vertex.
6.1 - 11
GRAPHING A PARABOLA WITH
A HORIZONTAL AXIS
Example 2
Graph. Give the domain and range.
x  2y  6 y  5
2
Solution
x  2y  6 y  5
2
x  2( y 2  3 y )  5
9 9
 2
x  2  y  3y     5
4 4

Factor out 2.
2
9
1 
Complete the square;  (3)  .
4
2 
6.1 - 12
GRAPHING A PARABOLA WITH
A HORIZONTAL AXIS
Example 2
9
 2
 9
x  2  y  3y    2     5
4

 4
Distributive property
2
3 1

x  2 y   
2 2

Factor; simplify.
6.1 - 13
Example 2
GRAPHING A PARABOLA WITH
A HORIZONTAL AXIS
1
3

x   2 y  
2
2

2
Subtract ½.
The vertex of the parabola is
1 3
 ,  .
2 2
The axis is the horizontal line
y = k, or y = –3/2. Using the
vertex and the axis and
plotting a few additional points
gives the graph. The domain
is [½, ) and the range is
(–, ).
6.1 - 14
Parabolas
A parabola is the set of points in a
plane equidistant from a fixed point
and a fixed line. The fixed point is
called the focus, and the fixed line is
called the directrix of the parabola.
6.1 - 15
Geometric Definition and Equations
of Parabolas
The axis of symmetry of a parabola passes
through the focus and is perpendicular to the
directrix. The vertex is the midpoint of the line
segment joining the focus and directrix on
the axis.
6.1 - 16
Geometric Definition and Equations
of Parabolas
We can find an equation
of a parabola from the
preceding definition. Let
p represent the directed
distance from the vertex
to the focus. Then the
directrix is the line
y = –p and the focus is
the point F(0, p).
6.1 - 17
Geometric Definition and Equations
of Parabolas
To find the equation of the
set of points that are the
same distance from the line
y = – p and the point (0, p),
choose one such point P
and give it coordinates (x, y).
Since d(P, F) and d(P, D)
must be equal, using the
distance formula gives the
following:
6.1 - 18
Geometric Definition and Equations
of Parabolas
d (P , F )  d (P , D )
( x  0)  ( y  p)  ( x  x )  ( y  (  p))
2
2
2
2
Distance formula
x  ( y  p)  ( y  p)
2
2
2
x  y  2yp  p  y  2yp  p
2
Remember
the middle
term.
2
2
2
Remember
the middle
term.
2
Square both sides and multiply.
x  4 py .
2
6.1 - 19
Parabola with Vertical Axis
and Vertex (0, 0)
The parabola with focus (0, p) and directrix y = – p
2
has equation x  4 py .
The parabola has vertical axis x = 0 and opens up if
p > 0 or down if p < 0.
6.1 - 20
Parabola with Horizontal Axis
and Vertex (0, 0)
The parabola with focus (p, 0) and directrix x = – p
has equation y 2  4 px.
The parabola has horizontal axis y = 0 and opens
to the right if p > 0 or to the left if p < 0.
6.1 - 21
Example 3
DETERMINING INFORMATION ABOUT
PARABOLAS FROM THEIR EQUATIONS
Find the focus, directrix, vertex, and axis of each
parabola. Then use this information to graph the
parabola.
a. x  8 y
Solution
2
The equation x2 = 8y has the form x2 = 4py, so
4p = 8, from which p = 2. Since the x-term is
squared, the parabola is vertical, with focus (0, p) =
(0, 2) and directrix y = – 2. The vertex is (0, 0) and
the axis of the parabola is the y-axis.
6.1 - 22
Example 3
DETERMINING INFORMATION ABOUT
PARABOLAS FROM THEIR EQUATIONS
6.1 - 23
Example 3
DETERMINING INFORMATION ABOUT
PARABOLAS FROM THEIR EQUATIONS
Find the focus, directrix, vertex, and axis of each
parabola. Then use this information to graph the
parabola.
b. y  28 x
Solution
2
The equation y2 = – 28x has the form y2 = 4px, with
4p = – 28, so p = – 7. The parabola is horizontal,
with focus (– 7, 0), directrix x = 7, vertex (0, 0) and
x-axis as axis of the parabola. Since p is negative,
the graph opens to the left.
6.1 - 24
Example 3
DETERMINING INFORMATION ABOUT
PARABOLAS FROM THEIR EQUATIONS
6.1 - 25
Example 4
WRITING EQUATIONS OF
PARABOLAS
Write an equation for each parabola.
a. focus (⅔, 0) and vertex at the origin
Solution
Since the focus (⅔, 0) and the vertex (0, 0) are
both on the x-axis, the parabola is horizontal. It
opens to the right because p = ⅔ is positive. The
equation, which will have the form y2 = 4px is
2 
y  4  x  , or
3 
2
8
y  x.
3
2
6.1 - 26
Example 4
WRITING EQUATIONS OF
PARABOLAS
2 
y  4  x  , or
3 
2
8
y  x.
3
2
6.1 - 27
WRITING EQUATIONS OF
PARABOLAS
Write an equation for each parabola.
Example 4
b. vertical axis, vertex at the origin, through
the point (–2, 12)
Solution
The parabola will have an equation of the form
x2 = 4py because the axis is vertical and the
vertex is (0, 0). Since the point (–2, 12) is on the
graph, it must satisfy the equation.
6.1 - 28
WRITING EQUATIONS OF
PARABOLAS
Example 4
x 2  4 py
( 2)  4 p(12)
2
4  48p
Let x = –2 and y = 12.
Multiply.
1
p
Solve for p.
12
Thus, x 2  4 py
1
 1
2
Let p = .
x  4   y,
12
 12 
1
2
which gives the equation x  y , or y  3 x 2 .
3
6.1 - 29
Equation Forms for
Translated Parabolas
A parabola with vertex (h, k) has an
equation of the form
or
( x  h )2  4 p( y  k )
Vertical axis
( y  k )2  4 p( x  h ),
Horizontal axis
where the focus is distance p from the
vertex.
6.1 - 30
WRITING AN EQUATION OF A
PARABOLA
Write an equation for the parabola with vertex (1, 3)
and focus (–1, 3) and graph it. Give the domain and
range.
Example 5
Solution
Since the focus is to the left of the vertex, the axis
is horizontal and the parabola opens to the left.
The distance between the vertex and the focus is
–1 – 1 or –2, so p = –2 (since the parabola opens
to the left). The equation of the parabola is shown
on the next slide.
6.1 - 31
WRITING AN EQUATION OF A
PARABOLA
Example 5
( y  k )  4 p( x  h )
Parabola with horizontal
axis
( y  3)  4( 2)( x  1)
Substitute for p, h, and k.
2
2
( y  3)  8( x  1).
2
Multiply.
The domain is
(– , 1], and the
range is (–, ).
6.1 - 32
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
The Parkes radio
telescope has a parabolic
dish shape with diameter
210 ft and depth 32 ft.
Because of this parabolic
shape, distant rays hitting
the dish will be reflected
directly toward the focus.
A cross section of the dish
is shown here.
6.1 - 33
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
a. Determine an
equation that models
this cross section by
placing the vertex
at the origin with the
parabola opening up.
6.1 - 34
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
Solution
Locate the vertex at the origin as shown below.
The form of the parabola is x2 = 4py. The parabola
 210
must pass through the point  2 ,32   (105,32).


6.1 - 35
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
Solution
x 2  4 py
Vertical parabola
(105)  4 p(32)
Let x = 105 and y = 32.
11,025  128p
Multiply.
2
11,025
p
.
128
Solve for p.
6.1 - 36
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
Example 6
Solution
The cross section can be modeled by the equation
x 2  4 py
 11,025 
x  4
y
 128 
2
Substitute for p.
11,025
x 
y.
32
2
6.1 - 37
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
b. The receiver must be
placed at the focus of
the parabola. How far
from the vertex of the
parabolic dish should
the receiver be located?
6.1 - 38
Example 6
MODELING THE REFLECTIVE
PROPERTY OF PARABOLAS
b. The distance between the vertex and
the focus is p. In part (a), we found
11,025
p
 86.1,
128
so the receiver should be located at
(0, 86.1) or 86.1 ft above the vertex.
6.1 - 39