### Ch 6 Analytic Geometry

```Chapter 6: Analytic Geometry
6.1 Circles and Parabolas
6.2 Ellipses and Hyperbolas
6.3 Summary of the Conic Sections
6.4 Parametric Equations
Slide 6-2
6.1 Circles and Parabolas
• Conic Sections
– Parabolas, circles, ellipses, hyperbolas
Slide 6-3
6.1 Circles
A circle is a set of points in a plane that are equidistant
from a fixed point. The distance is called the radius of
the circle, and the fixed point is called the center.
•
•
A circle with center (h, k) and radius r has length
2
2
r  ( x  h )  ( y  k ) to some point (x, y) on
the circle.
Squaring both sides yields the center-radius
form of the equation of a circle.
r  ( x  h)  ( y  k )
2
2
2
Slide 6-4
6.1 Finding the Equation of a Circle
Example Find the center-radius form of the equation
of a circle with radius 6 and center (–3, 4). Graph the
circle and give the domain and range of the relation.
Solution
Substitute h = –3, k = 4, and r = 6 into the
equation of a circle.
6  ( x  (  3 ))  ( y  4 )
2
2
36  ( x  3 )  ( y  4 )
2
2
2
Slide 6-5
6.1 Graphing Circles with the Graphing
Calculator
Example Use the graphing calculator to graph the
circle x 2  y 2  9 in a square viewing window.
Solution
x  y 9
2
2
y 9x
2
2
y   9x
Let y1 
2
9  x and y 2   9  x .
2
2
Slide 6-6
6.1 Graphing Circles with the Graphing
Calculator
• TECHNOLOGY NOTES:
– Graphs in a nondecimalwindow may not be connected
– Graphs in a rectangular (non-square) window look like
an ellipse
Slide 6-7
6.1 Finding the Center and Radius of a
Circle
Example Find the center and radius of the circle
with equation x 2  6 x  y 2  10 y  25  0.
Solution Our goal is to obtain an equivalent
equation of the form r 2  ( x  h ) 2  ( y  k ) 2 .
We complete the square in both x and y.
x  6 x  y  10 y   25
2
2
( x  6 x  9)  ( y  10 y  25)   25  9  25
2
2
( x  3)  ( y  5)  9
2
2
( x  3)  ( y  2)  3
2
2
2
The circle has center (3, –2) with radius 3.
Slide 6-8
6.1 Equations and Graphs of Parabolas
A parabola is a 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, the directrix.
• For example, let the directrix be the line y = –c and
the focus be the point F with coordinates (0, c).
Slide 6-9
6.1 Equations and Graphs of Parabolas
• To get the equation of the set of points that are the
same distance from the line y = –c and the point
(0, c), choose a point P(x, y) on the parabola. The
distance from the focus, F, to P, and the point on
the directrix, D, to P, must have the same length.
d (P, F )  d (P, D)
( x  0)  ( y  c) 
( x  x )  ( y  (  c ))
x  y  2 yc  c 
y  2 yc  c
2
2
2
2
2
2
2
x  y  2 yc  c  y  2 yc  c
2
2
2
2
2
2
2
x  4 cy
2
Slide 6-10
6.1 Parabola with a Vertical Axis
The parabola with focus (0, c) and directrix y = –c has
equation x2 = 4cy. The parabola has vertical axis x = 0,
opens upward if c > 0, and opens downward if c < 0.
• The focal chord through the focus and perpendicular to the
axis of symmetry of a parabola has length |4c|.
– Let y = c and solve for x.
x  4 cy
2
x  4c
2
2

x   2 c or 2 c
The endpoints of the chord are ( x, c), so the length is |4c|.
Slide 6-11
6.1 Parabola with a Horizontal Axis
The parabola with focus (c, 0) and directrix x = –c
has equation y2 = 4cx. The parabola has horizontal
axis y = 0, opens to the right if c > 0, and to the left
if c < 0.
• Note: a parabola with a horizontal axis is not a function.
• The graph can be obtained using a graphing calculator by
solving y2 = 4cx for y: y   2 cx .
Let y1  2 cx and y 2   2 cx and graph each half of the
parabola.
Slide 6-12
6.1 Determining Information about
Parabolas from Equations
Example Find the focus, directrix, vertex, and axis
of each parabola.
2
(a) x 2  8 y
(b) y   28 x
Solution
(a) 4 c  8
c  2
Since the x-term is squared, the
parabola is vertical, with focus
at (0, c) = (0, 2) and directrix
y = –2. The vertex is (0, 0), and
the axis is the y-axis.
Slide 6-13
6.1 Determining Information about
Parabolas from Equations
(b) 4 c   28
c  7
The parabola is horizontal,
with focus (–7, 0), directrix
x = 7, vertex (0, 0), and
x-axis as axis of the parabola.
Since c is negative, the graph
opens to the left.
Slide 6-14
6.1 Writing Equations of Parabolas
Example Write an equation for the parabola with
vertex (1, 3) and focus (–1, 3).
Solution
Focus lies left of the vertex implies the
parabola has
- a horizontal axis, and
- opens to the left.
Distance between vertex and
focus is 1–(–1) = 2, so c = –2.
( y  3 )  4 (  2 )( x  1)
2
( y  3 )   8 ( x  1)
2
Slide 6-15
6.1 An Application of Parabolas
Example Signals coming in
parallel to the axis of a parabolic
reflector are reflected to the focus,
thus concentrating the signal.
The Parkes radio telescope has a
parabolic dish shape with diameter 210 feet and depth 32 feet.
Slide 6-16
6.1 An Application of Parabolas
(a) Determine the equation describing the cross section.
(b) The receiver must be placed at the focus of the parabola.
How far from the vertex of the parabolic dish should the
Solution
(a) The parabola will have the form y = ax2 (vertex at the
origin) and pass through the point  2102 , 32   (105 , 32 ).
32  a (105 )
a 
32
105
2
32

32
y 
2
,
so the cross section can be described
by
11 , 025
2
x .
11 , 025
Slide 6-17
6.1 An Application of Parabolas
(b) Since
y 
32
2
x ,
11 , 025
4c 
1
a
4c 
11 , 025
32
c
11 , 025
 86 . 1 .
128
The receiver should be placed at (0, 86.1), or
86.1 feet above the vertex.