The quadratic functions of Chap­
ter 5 have general equations of
the form y = ax 2 + bx + c. If
the equation also has a y2- term
or an xy-term, the relation is still
quadratic, but may not be a
function. In this chapter you will
find out what the graphs of these
relations look like, and how they
are related to slices of a cone.
These "conic sections" are good
mathematical models of the paths
of planets, spacecraft, and other
objects traveling under the
influence of gravity.
'~
/' /'
~ Meteorite
~
"
'-...
'-...
'-...
'-..
'-..
-----­
'-..
'-..
x
9·1
9-1
I INTRODUCTION TO
Introduction to Quadratic Relations
QUADRc~TIC
RELATIONS
DEFINITION QUADRATIC RELATION
A quadratic relation is a relation specified by an equation or in­
equality of the form
Ax 2
+ Bxy + Cy2 +
Dx
+ By + F
=
0,
where A, B, C, D, E, and F stand for constants, and where the "="
sign may be replaced by an inequality sign.
It will be your objective in the next few sections to determine what the
graph of such a relation might look like and how the six constants affect
the graph.
To begin this objective, you will plot the graphs of several such relations
pointwise, and attempt to discover what geometrical figure each one is.
EXERCISE 9-1
Plot a graph of each relation. Select values of x and calculate the corre­
sponding values of y until you have enough points to draw a smooth curve.
Use a calculator or square root tables to approximate any radicals. When
you have finished, see if you can come to any conclusions about the shape,
size, and location of each graph.
461
J,
(ildPI( 'r 'i
i!?-'
II :1~~
1-1:. (iI
,i
I
L-__J ~ .j
QU<ldri:ltic Hdations and Systems
I
+ y" == 25}
{(x, y): x" + y2 + 6x = 16}
{(x, y): x" + y2 - 4y == 21} {(x, y): x 2 + )'" + 6x - 4y = 12} {(x, y): x 2
1.
2.
3.
4,
'"
!
I'
CIl<CLES
In Exercise 9-1 you discovered that the graphs of some quadratic relations
look like circles. In this section you will learn why this is true, and use the
results to sketch the graphs rapidly.
()IJj('t li\(':
the equation (lr inequali [y of a circle or circular region. be able to
Jr;J\\' the ~.:raph
(,; \ l'
To accomplish this objective, you will start with the geometric definition
of a circle, and use the definition to find out what the equation of a circle
looks like,
DEH'IlTIO'l
GEOMETRICAL DEFINITION OF CIRCLE
A circle is a set of points in a plane, each of which is equidistant
from afixed point called the center.
Suppose that a circle of radius r units has its center at the fixed point
(h, k) in a Cartesian coordinate system, as shown in Figure 9-2a. If (x. y)
y
x
Figure 9-2a _
I
9-2
Circles
is a point on the circle, then the distance between (x, y) and (h, k) is r
units, the radius of the circle. This distance can be expressed in terms of
the coordinates of the two points using the Distance Formula.
Background: The Distance Formula
Objective:
Express the distance between two points in a Cartesian coordinate system
in terms of the coordinates of the two points.
You recall from your work with slopes of linear functions that the rise and
run, ay and ax, for the line between the points (x" yd and (X2, Y2) are
ay
= Y2 - YI
ax = X2
XI.
Figure 9-2b shows that ay, ax, and the distance d between the two points
are the measures of the sides of a right triangle. By Pythagoras, then,
d 2 = (ax)2
+ (ay)2
CD.
By substitution,
d 2 = (X2
xlF +
®.
(Yz - YI}2
y
Ll.y
Y2- Y '
I
__ J
x
Ll.x~X2-Xl
The Distance Formula
Figure 9-2b _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
Taking the positive square root of both members gives
d = V(X2 -
XI)2
+ (Y2 - YI)2
@.
463
464
CilapWr~)
~!B
Quadratic Helations and S"SWl11S
Equations CD, ®, and ® are different forms of the distance formula. Each
form has certain advantages for different applications. The formula is, of
course, just a fancy form of the Pythagorean Theorem.
Returning to the problem of finding the equation of a circle, the distance r
between the points (x, y) and (h. k) in Figure 9-2a can be expressed using
form ® of the distance formula as:
! (x -
h)2 + (y - k)2 = r21
where the center is at (h, k), and the radius is r. This is called the general
equation of a circle.
If the center is at the origin, then (h, k)
reduces to
x2
+ y2
(0, 0). In this case the equation
=:
= r2
To show that an equation such as in Problem 4 of Exercise 9-1 is really
that of a circle, it is sufficient to transform the equation into the general
form, above. This transformation may be accomplished by completing the
square.
EXAMPLE
Graph the relation
x 2 + y2 + 6x
4y - 12
=:
0.
Solution:
Commuting and associating the terms containing x with each other, and
the terms containing y with each other, then adding the constant 12 to
both members gives
(x 2 + 6x
)
+ (y2
4y
)
=:
12.
The spaces are left inside the parentheses for completing the square. as
you did for equations of parabolas in Chapter 5. Adding 9 and 4 on the
left to complete the squares, and on the right to balance the equation,
gives
(x"
+
6x
+
9)
+
(y2 - 4y
+
4) = 12
+
9
+
4,
from which
(x + 3)2 + (y
2)2
25.
Therefore, the graph will be a circle of radius 5 units, centered at
(h, k) = (-3,2), as shown in Figure 9-2c. The graph you drew for Prob­
lem 4 of Exercise 9-1 should have looked something like this figure. The
9-2
Circles
y
x
Ix + 3)2 + Iy - 2)2
25
Figure 9-2c _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
prior knowledge that the graph is a circle of known center and radius,
however, will allow you to plot the graph much more quickly, in accor­
•
dance with your objective for this section_
If the open sentence is an inequality, then the graph will be the region in­
side the circle if the symbol is "<," as indicated in Figure 9-2d, or outside
the circle if the symbol is ">."
An objective stated in Section 9-1 is to determine the effects of the con­
stants A, B, C, D, E, and F on the graph of
Ax2
+
Bxy
+
Cy2
+
Dx
+
Ey
+
F
= O.
The above leads to a conclusion about the relative sizes of A and C.
y
x
Ix + 3)2 + (y
2)2
< 25
Figure 9-2d _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
The graph of a quadratic relation will be a circle if the coefficients of
the x 2 term and the y2 term are equal (and the xy term is zero).
465
(:lldptcrq
-lG()
()II(HiI,:!lf
Hcidlj'lll.'",
IIltl..."
,"'""
The following exercise is designed to give you practice drawing graphs of
circles and circular regions given the equation or inequality.
Do These Quickly
The following problems are intended to refresh your skills. You should be
able to do all 10 in less than 5 minutes.
Sketch the graph of:
Q I.
A linear function. Q2.
A quadratic function. Q3.
A decreasing exponential function. Q4.
An inconsistent linear system. Q5.
A linear inequality. Q6.
A function with a removable discontinuity at x
Q7.
A function with a vertical asymptote at x = 2. = 3. Write the general equation of:
Q8.
A linear function. Q9.
A quadratic function. Q10.
An exponen tial function. For Problems 1 through 12, complete the square (if necessary) to find the
center and radius. Then draw the graph.
1.
x"
2.
5
+
+
,
)'2 -
yl
3.
X' + y'
4.
Xl
+
5.
2
+
x
+
8y
+
6x - 4y
y2 - 8x
+
5
0
12x - 2y + 21 = 0
>
0
- II
<
56
0
12
+ lOy
+ 6y -
+ 16x
+ y2
6.
+
,
lOx
4x - 18}' + 69 ::;,. 0
~ 7.
X"
+ y" +
4x - 5
8.
X"
+
14y
+
0
48
=
0
0
9-2
Circles
9. x 2 + y2
10. x 2 + y2
11. x 2 + y2
= 49
=4
= 0 (Watch for a surprise!)
12. x 2 + y2 + 16 = 0 (Watch for a different surprise!)
For Problems 13 through 18, write an equation of the circle described.
l? (13);
Center at (7, 5), containing (3, - 2)
14. Center at (-4, 6), containing (-2, -3)
15. Center at (-9, -2), containing the origin
16. Center at (5, -4), containing (0, 3)
---'~
i"} 11-1) Cen ter at the origin, con taining (- 6, - 8)
18. Cen ter at the origin, containing (- 5, 1)
Iq i i)
Write a general equation for a circle
a. of radius r, centered at a point on the x-axis,
b. of radius r, centered at a point on the y-axis,
c. of radius r, centered at the origin.
20. What do you suppose is meant by a "point circle?" How can you tell
from the equation that a circle will be a point circle?
21. Sometimes the graph of a quadratic relation such as you have been
plotting turns out to have no points at all! How can you tell from the
equation whether or not this will happen?
22. Introduction to Ellipses In this section you have found that the
graph of a quadratic relation is a circle if x 2 and y 2 have equal
coefficients. In this problem you will find out what the graph looks
like if the coefficients are not equal, but have the same sign. Do the
following things for the relation whose equation is
9x 2 + 25y2 = 225.
a. Solve the equation for y in terms of x.
b. Explain why there are two values of y for each value of x be­
tween -5 and 5.
c. Explain why there are no real values of y for x > 5 and for
x < -5.
d. Calculate values of y for each integer value of x from - 5 to 5,
inclusive. Use a calculator or square root tables to approximate
any radicals which do not come out integers. Round off to one
decimal place.
e. Plot the graph. If you have been successful, you should have a
closed figure called an ellipse.
467
4(jH
q-3
Clldp!er
q
ig!lk
~.-.
(.,2u(\(lr'lTic Hc\cltions <H1<i Systems
I-t~
LLLlPSLS
In Section 9-2 you learned that the graph of a quadratic relation is a circle
if and y2 have the same coefficient. If x 2 and y2 have different
coefficients, but the same sign, then the graph is called an "ellipse." If you
worked Problem 22 of Exercise 9-2 you found by pointwise plotting that
the graph of
9x 2
+ 25y2
225
is as shown in Figure 9-3a. In this section you will learn properties of el­
lipses that will allow you to sketch their graphs quickly. You will also use
computer graphics to confirm that your sketches are correct.
v
3
I
o
-5
5
... x
-3
9x 2 + 25v 2
225
Figure 9-3a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
Various parts of the ellipse are given special names, as shown in Figure
9-3b. All ellipses look rather like flattened circles. Thus, if you know
where the center is, and how long the major and minor axes are, you can
sketch the graph quickly. The trick is to transform the equation so that
these lengths show up in it. Starting with
9x 2 + 25y2 = 225,
~
~I
Major
Vertex
-k
01
c:
__
__ _
AXIS
Vertex
Center
:i!
Figure 9·3b _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
9-3
Ellipses
you divide by 225 to make the right member equal 1.
9x 2
225
25y2
+ 225
= 1
x2
- +"- = 1
25
9
(~r + (~r = 1
The "5" under the x is the distance from the center to the graph in the x­
direction. The "3" under the y is the distance from the center to the graph
in the y-direction. In this text these distances will be called the x-radius
and the y-radius (see Figure 9-3c).
DEFINITION
X-RADIUS AND Y·RADIUS
If an ellipse has its major and minor axes parallel to the coordinate
axes, then:
the x-radius is the distance from the center to the ellipse in the .1'­
direction, and
the y.radius is the distance from the center to the ellipse in the y­
direction.
By comparing the graphs in Figures 9-3b and 9-3c, you can tell that the x­
and y-radii are each half of either the major or minor axis. Half the major
y
r:I
x-ri'dius )
I
x
Figure9.3c _____________________________________________
469
470
Chapter 9
[g] ~!
Quadratic Relations and Systems
axis is called the semi-major axis. Half the minor axis is called the semi­
minor axis. Since the ellipse can be oriented with the major axis vertical
as in Figure 9-3c, or horizontal as in Figure 9-3a, the semi-major axis is
either the y-radius or the x-radius, whichever is larger.
Letting r" and ry stand for the x- and y-radii, respectively, the general
equation of an ellipse centered at the origin is
+ (2::.)2 = 1
(~)2
r"
ry
If the center is at (h, k) instead of at the origin, (0, 0), then x and y in the
k), as they were for
above equation will be replaced by (x - h) and (y
the circle.
CONCLt:SION
The graph of
h)2 (y - k)2
X -+--=1
r"
ry
(
•
is an ellipse, centered at (h, k), having points r" units from the center
in the positive and negative x-direction. and ry units from the center
in the positive and negative y-direction.
EXAMPLE 1
Sketch the graph of
25x 2 + 9y2 - 200x + 18y + 184 = O.
Solution:
To sketch the graph quickly you must transform it to the form above by
completing the square. First you subtract 184 from both members and as­
sociate what remains on the left, getting
(25x 2
-
200x)
+ (9y2 + 18y) = -184.
Factoring out the coefficients of and y2 gives
25(x 2 - 8x
) + 9(y2 + 2y
-184.
Spaces are left in the parentheses for completing the square. Adding 16
and 1 completes the squares on the left, and requires that 25 . 16 and
9 . 1 be added on the right, giving
25(x 2 8x + 16) + 9(y2 + 2y + 1) = -184 + 25 . 16 + 9 . 1.
Writing the left member in terms of perfect squares, and doing the arith­
metic on the right gives
25(x - 4)2
+ 9(y + 1)2
=
225.
9·3
Ellipses
Dividing both members by 225 produces the desired form of the equation,
ex -
4)2
1
+
9
or
(x ~ 4
r (y ; r
+
1
= 1.
So the graph will be an ellipse; centered at (4, -1), with x-radius 3 units
and y-radius 5 units.
The graph may now be drawn quickly by plotting the four critical points,
then sketching the ellipse, as shown in Figure 9-3d.
J
I
I
1 2
T
I
51
I
I
I
3 4
I
-2
-- 3
4
I
5 6
----+­
3/1
3
Center I
-I
I
-­
7
x
-51
I
5
I
•
6
Plot the center and
x· and y- radii
y
4
1
3
7
x
-5
6
Figure 9-3d
Sketch the ellipse
•
471
472
Chapter D
[gJg
Quadratic RelCltions Clnd Systems
"<" or ">," then the
graph will be the region inside the ellipse or outside the ellipse, respec­
tively.
If the "=" sign is replaced by one of the inequalities
There is a pair of points associated with an ellipse that has an important
geometrical property. Suppose that you tie two pins to a piece of string in
such a way that there are 10 cm of string between them. Then you stick
the pins at the points F! = (4, 0) and F2 = (-4, 0) in a Cartesian coordi­
nate system that has l-cm squares (see Figure 9-3e). Placing a pencil as
shown in the figure, and keeping the string tight, you can draw a curve.
The curve turns out to be an ellipse, the same one as in Figure 9-3a~
This construction leads to a geometrical definition of the ellipse.
L
P'"
P'"
):
s",os
"'1
10 em
,..
F2
Fl = (-4,0)
(4,0)
x
v
_I
ee::=!~~-i~~:L~~~IL-~
5
x
-3
Figure 9-3e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
9·3
Ellipses
DEFINITION GEOMETRICAL DEFINITION OF ELLIPSE
An ellipse is a set of points in a plane. For each point, the sum of its
distances, d1 + d z • from two fixed points Fl and Fz• is constant.
Each point, FI and Fz , is called afocus. The name is chosen because
sound or light emitted from one focus will be "focused" toward the other
one by bouncing off the ellipse. (The plural of focus is "foci," the HC" be­
ing pronounced like an "s. ")
The distance from the center to a focus is called the focal radius. By plac­
ing the pencil at two strategic positions, you can find out how to calculate
the focal radius. Placing it on the x-axis as in Figure 9-3f, you can see
that the length of the major axis is equal to the length of the string. That
is, the constant sum of the distances, d l + d z , equals the length of the
major axis.
Figure 9-3f _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
By placing the pencil on the y-axis as in Figure 9-3g, a right triangle is
formed by the two axes and the string. The hypotenuse is equal to half the
length of the string.
Figure 9-3g
2
<3
b
2
2
+c
473
474
(:/mprcr
(~uddrdti('
<)
Helations dl1d Systems
If the x- and y-radii are known, then the semi-major axis is the larger one
and the semi-minor axis is the smaller, Letting a stand for the semi-major
axis, b stand for the semi-minor axis, and c stand for the focal radius,
then the following conclusion is true:
CO.'\CL LSIO.'\
In an ellipse,
If:
a is the length of semi-major axis, b is the length of semi-minor axis, c is the focal radius, and d l and d2 are the distances from a point (x, y) on the ellipse to the two foci, Then:
dl
+
a2
c2
= a2
2a
d2
= length of major axis,
= b 2 + c 2 , from which
-
b2•
OI)j(,CTi\'{':
Gi \'en the e4uation of an ellipse,
a. ~[..;etch the (!raph, b, calculate the focal radiu, and plot the foci. c. .:heck. the ,~etch hy computer graphic,.
EXA:\1PLE 2
Find the foci of
25x 2 + 9y2 - 200x + 18y + 184
0,
Solution:
This is the ellipse in Example I. Completing the square gives
(x ; 4
So a
r (y ; r
+
I
=
5 and b = 3. Using the fact that c 2 = a 2
c2
=
52 ­
= 16
:. c = 4.
32
l.
-
b2
9·3
Ellipses
475
Since the major axis is in the y-direction, the foci will be located verti­
cally, ±4 units from the center (Figure 9-3h).
y
II
, .. x
Figure 9-3h _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
•
The exercise that follows is designed to give you practice in sketching
graphs of ellipses and finding their foci.
EXERCISE 9·3
Do These Quickly
The following problems are intended to refresh your skills. You should be
able to do all 10 in less than 5 minutes.
+ (y - 2)2
Q2. Where is the center of the circle (x + 3)2 + (y
4)2
2
Q3. Complete the square: x + 20x + __ . Q4. Write as the square of a binomial: x 2 - 8x + 16 Q 1.
What is the radius of the circle (x - 5)2
Q5.
Find k if (3,4) is on the graph of y = kx 2 • Q6.
Solve:
Q7.
Vx + 3 = 7 Solve: Ix - 111 = 2 Q8.
Simplify: 7
Q9.
Find 90% of 90. + 3(x -
5) 49? 25? 476
Chapter n
QIO.
For
a. b. c. Quadratic Relations and Systems
~~
Solve: -3x < 12
Problems 1 through 16,
Sketch the graph of the relation,
Calculate the focal radius and plot the two foci,
Confirm your sketch by plotting the original equation, using PLOT
CONIC on the accompanying computer disk, or similar program.
1. 1.
5,
4x 2 + 9y2 - 16x + 90y + 205 = 0
2. 4x 2 + 36y2 + 40x - 288y + 532
0
3. 49x 2 + 16y2 + 98x - 64y - 671
0
4. 25x 2 + 4y2 - 150x + 32y + 189 = 0
5. x 2 + 4y2 + lOx + 24y + 45 = 0
6. 16x 1 + y2
1.
l28x - 20y + 292
0
2
7. 25x + 9y2 + 50x
36y - 164 < 0
8. 4x 2 + 36y2 + 48x + 216y + 324 :2: 0
9. 16x2 + 25y2
300y + 500 = 0
+ 9y2 216x = 0
II. 11. IOOx + 36y2 > 3600
12. 25x 2 + 49y2 ~ 1225
13. 12xl + y2 = 48
10. 36x 1
2
15. 5x 2 + 8y2
= 77
14.
16.
+ 6y2 = 25
l1x2 + 5y2
224
Xl
17. Stadium Problem Suppose that you are Chief Mathematician for
Ornery & Sly Construction Company. Your company has a contract
to build a football stadium in the form of two concentric ellipses,
with the field inside the inner ellipse, and the seats between the two
ellipses. The seats are in the intersection of the graphs of
x 2 + 4y2 :2: 100 and 25x 2 + 36y2 ~ 3600,
where each unit on the graph represents 10 meters.
a. Draw a graph of the seating area.
b. From a handbook, you find that the area of an elliptical region is
7Tab, where a and b are the semi-axes, and 7T = 3.14159. The
Engineering Department estimates that each seat occupies 0.8
square meter. What is the seating capacity of the stadium?
18. Show that the equation for an ellipse,
_
X- h)2 (y - k)2 -1,
(-rx- + -ry
reduces to the equation of a circle if rx = ry. 9-4
Hyperbolas
19. Introduction to Hyperbolas The ellipses you have plotted in this
section have equations in which the x 2 and y2 terms have the same
sign. In this problem you will find out what the graph looks like if
they have opposite signs. Do the following things for the relation
whose equation is
9x 2
-
16y2
144,
a. Solve the equation for y in terms of x.
b. Explain why there will be no real values of y for -4 < x < 4.
c. Explain why there will be two values of y for each value of x
when x > 4 or x < -4.
d. Make a table of values of y for each integer value of x from 4 to
8, inclusive, finding decimal approximations from square root ta­
bles or by calculator, where necessary.
e. Explain why you can use the same table of values of y for values
of x between -4 and -8.
f. Plot a graph of the relation from x = -4 to x = -8, and from
x
4 to x = 8.
g. On the same Cartesian coordinate system, plot graphs of the
lines
3
4
y = -x
and
y
3
4x .
If your graph in part f is correct, then it should have these two
lines as diagonal asymptotes. The graph is called a hyperbola,
and is a non-closed curve with two "branches."
9-4
I HYPERBOLAS
In the last section you learned that an ellipse has an equation in which the
and y2 coefficients are unequal, but have the same sign. If x 2 and y2
have opposite signs, such as
9x 2
-
16y2
=
144,
then the graph looks like that in Figure 9-4a, and is called a hyperbola.
You may have discovered this fact already if you worked Problem 19 in
Exercise 9-3.
Objective:
Given the equation of a hyperbola, be able to sketch its graph quickly.
477
478
Chapter 9
Quadratic Relations and Systems
v
,
~,
• ,'<iSy
,'>:1.0
4
,0,&
3
2
,
'"
x
1­I\4X)2
Figure 9-4a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
Hyperbolas have two disconnected branches. Each branch approaches diag­
onal asymptotes. You can see why by transforming the equation so that y is
by itself on the left side.
-16y 2
16y2
= -9x 2 +
= 9(x
3
2
16) -
.
2
.v = ± 4 Yx
144 Multiply by
factor out 9.
16
-
Subtract 9x 2 •
I and
Divide by 16, then take
the square root.
Two observations about the graph may be made from this equation:
1. There are no real values of y when x is between -4 and 4 because the
radicand, x 2 - 16, would be negative. This fact explains why the graph
is split into two branches.
2. As x becomes larger, the difference between
and v? gets
close to zero. For example, if x = 100, then
Vx
2
-
16
=
V1Ooo0 -
16
= Y9984
= 99.92,
while v? = V1Ooo0
100. Consequently, the larger x gets, the closer
y gets to ± iv?, or y = ± ~ x. The lines y = ± ~ x are thus asymptotes
of the graph. Their slopes are ~ and - ~ .
9·4
Hyperbolas
The 4 and 3 in the slopes of the asymptotes can be made to show up in
the equation by making the right member equal 1.
9x 2 - 16y2
144 Given equation
x2
Divide by 144.
9
16
(~r (~y
In this form, the 4 under the x is in the same position as the x-radius for
an ellipse. The 3 under the y is in the same position as the y-radius. As
shown in Figure 9-4b, the x-radius is, in this case, the distance from the
center to the vertex of the hyperbola in the x-direction. The y-radius does
not show up directly on the graph. It is the distance from the vertex to the
asymptote, either upward or downward.
y
""
./
./
x
~
v·radius
"
I
" ",."
""
"
" '"
Figure 9·4b _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
If the signs had been reversed, like this,
+ (y)2_
- (-X)2
4
3
1
'
then the graph would have two branches opening in the y-direction. The
asymptotes would still have slopes ~ and - ~. But the vertices would be 3
units in from the center in the y-direction since the y-radius is 3. The x­
radius would tell how far to go in the x-direction to get from the vertex to
the asymptotes. The graph of this hyperbola is shown in Figure 9-4c.
479
480
Chapter 9
[g]
Quadratic Helations and Systems
\"
""
/
""
/
/'
" "'­
4
/'
.3
/'
/'
" "'­
/'
/'
x
/,"",-
}y>"
x-radiUS>\/,/'
4
;<x-radiUS
radlw, "
/'
.,AI----
"
" "
--~
/'
/'
x -')
r
~(4)-+(3)2= I
Figure 9·4c _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
The equations for the two hyperbolas,
-"4
(X)2
+
(y)2
3"
= I and
(~r (~r =
are alike except that the signs are reversed on the left side. The two result­
ing hyperbolas are said to be conjugates of each other. As shown in Fig­
ure 9-4d, they have the same asymptotes and the same x- and y-radii. But
one opens in the y-direction and the other opens in the x-direction. When
the equation is written in the form above, with 1 on the right side, the
hyperbola opens in the direction of the squared term that has the positive
sign.
The axis that goes across the hyperbola, from vertex to vertex, is called
the transverse axis. The other one is called the conjugate axis because it is
the axis of the conjugate hyperbola.
There is a geometrical definition of hyperbola similar to that of ellipse. For
any given point on an ellipse, the sum of its distances from the two foci is
a constant. Hyperbolas also have two foci. In this case, the difference be­
tween the two distances is constant.
9-4
Hyperbolas
y
Transverse I Conjugate
x
r,
- ':.l ,
m:=
"
r,.=a
r,,=h
(~)L(~)2= I
y
"
~ Conjugate hyperbolas
/
x
(~)2+_(~)2:= I
Figure 9-4d _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
DEFINITION
GEOMETRICAL DEFINITION OF HYPERBOLA
A byperbola isa set of points in a plane. For each point (x, y) on
the hyperbola, the difference between its distances from two fixed
foci is a constant~
481
482
C/mptcr~)
u-~
'.f::::'....i
Quadratic Helarions and SysteIl1S
~
'lsI
~
Figure 9-4e illustrates this definition. Let a and b stand for the semi-trans­
verse and semi-conjugate axes, respectively. The values of a and b will be
the x- and y-radii. If the hyperbola opens in the x-direction, then a = rx •
If it opens in the y-direction, then a = r y •
For hyperbolas, the hypotenuse of the right triangle shown on the right in
Figure 9-4e is equal to the focal radius. Letting c stand for focal radius,
the Pythagorean Theorem gives
c2 = a2
+ b2•
So in this case,
c 2 = 42
:. c
+ 32
25
= 5.
The discussion above has been for a hyperbola centered at the origin. If
the center is at (h, k), the x in the equation is replaced by (x - h), and the
v
I~f
1='1
I C
Id 1
"x
I
Ai
i
d 2 is constant
v
/'
/'
'-....
"
/'
/'
c ~ a2 + b 2
'"
Figurc9-4c ____________________________________________
__
2
9-4
Hyperbolas
Y is replaced by (y - k), as for the circle and ellipse_ The following is a
summary of the above information_
CONCLUSIONS
The general equation of a hyperbola is
I
(~)' (~)' ~ 1
-
or
The hyperbola opens in the x-direction if the sign in front of the
term containing x is plus_
The hyperbola opens in the y-direction if the sign in front of the
term containing y is plus_
.
(y-radius)
The slope of the asymptotes IS always ±. (
di) .
x-ra us
If:
a is the length of semi-major axis, b is the length of semi-minor axis, c is the focal radius, and d 1 and d2 are the distances from a point (x, y) on the hyperbola to the two foci, Then:
I
d1
-
d2 1
= 2a = length of transverse axis.
c2 = a2 + b 2
If the center is at the origin, then (h, k)
duce to
(~y - (~y = 1
or
= (0,
-
0), and these equations re­
(~r + (~r = I
EXAMPLE 1
Sketch a graph of
9x 2
4y2 + 90x + 32y + 197 = 0.
483
4·H'+
C11dpter q
~~
Qll<1clratic Relations and Systems
y
'/
\
8
1
I,
I
'
13
1
\
/
7
6
'
:
3
5
')(-~ ,
1
\ '3
I
4
/:
13
3
I
2
/;
/;
figurc9-4f _______________________________________________
Solution:
First transform the equation by completing the square.
9(x 2 + lOx
) - 4(y2
8y
) = -197
9(x 2 + lOx + 25)
9(x
4( y2 - 8y + 16)
+ 5)2 -
= -197 +
9 . 25 - 4 . 16
4(y - 4)2 = -36
To make the right member equal to + 1, divide each member by
ting
(x
-
+ 5)2
22
+
36, get­
(y - 4)2 _ 1
32
­
after simplification.
The graph is a hyperbola, opening in the y-direction, centered at (-5,4).
The x-radius is asymptotes of slopes ±t and vertices which are ±3 units
from the center in the y-direction.
9-4
Hyperbolas
Using this information, you should first locate the center, then draw in the
asymptotes with the proper slope. The only actual points you need plot are
the two vertices. With this information, you can sketch a remarkably good
hyperbola, as shown in Figure 9-4f.
•
EXAMPLE 2 Find and plot the foci of the hyperbola in Example 1. Solution: Since the focal radius, c, is given by c 2
foci. c2
=
a2 +
b\ you can locate the = 22 + 32
=13
:. C
=
vl3 =
3.61
The foci are ±3.61 units from the center, as shown in Figure 9-4f.
•
The following exercise is designed to give you practice in sketching hyper­
bolas rapidly.
EXERCISE 9-4
Do These Quickly
The following problems are intended to refresh your skills. You should be
able to do all 10 in less than 5 minutes.
Tell what figure the graph will be.
Ql.
3x 2 + 4y2
Q2.
3x 2
Q3.
3x 2 + 3y2 = 17
Q4.
17
3x + 3y
2
Y 3x + 17
Q5.
17
4y2 = 17
Answer the questions.
V32.
Q6.
Evaluate
Q7.
Simplify: - 3x (x
Q8.
Multiply: (3x
+ 5) - 17
+ 4y)(3x - 4y)
485
486
Chapter 0
+ 3)2
Q9. Factor: (x
QI0.
[g][B
Quadratic Relations and Systems
16
What kind of function has the general equation y
=
a . b?
X
For Problems 1 through 18,
a. Complete the square (if necessary), find the center, draw the asymp­
totes, plot the vertices, then sketch the graph.
b. Calculate the focal radius, and plot the foci.
c. Confirm your sketch by plotting the original equation, using PLOT
CONIC on the accompanying disk, or similar program.
I. 25x 1
16y2
100x
96y - 444
9y2 + 16x + 108y - 344
2. 4x 2
0
0
126y + 684 = 0
- 9y2 + 300x
2
4. 4x - 36y2 - 40x + 216y - 80 = 0 69 = 0
5. x 2
y2 + 4x + 16y
3. 25x 2
6. 7. 9x
8. y2 - 14x - 8y + 37 = 0
Xl
2
-
4y2
54x
16y - 79
4y2 + 4x + 32y - 96
Xl -
9. 9x 2
0
0
- 90x + 4y + 302 = 0
25x 2 - 4y2 + 200x - 8y + 796
0
I l. 16x 2 - 9y2 + 144
12. 25x 2 - 144y2 - 3600 = 0
10.
13. 4x 2
5y2
15. 16x 2
17. 9x
D-:}
2
-
16
=
3y2
y2
II
0
0
14.
27x 2
16.
5x 2
18.
2
4x
-
4y2
7y2
y2
=
=
=
-36
17
0
PARABOLAS
You recall that a quadratic equation has the general equation
y = ax 2 + bx + c.
The graph is a parabola with its axis of symmetry vertical, as shown in
the left of Figure 9-5a. If the x and yare reversed,
x = ay 2
+
by
+ c,
the parabola has a horizontal axis of symmetry. The right part of Figure
9-5a shows how the graph might look.
9·5
y
Parabolas
y
x
x
x= ay2+by+c
y=ax 2 +bx+c
Figure 9-5a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
All the graphing techniques you learned in Chapter j apply, but the roles
of x and y are switched. Vertex form looks like this:
x - h = a(y - k)2
The equation can be transformed to this form by completing the square.
The quick way of finding the vertex applies, but to y instead of x. The
y -coordinate, k, of the vertex is given by
b
k == - 2a'
In this section you will refresh your memory about graphs of parabolas by
sketching some that open in the x-direction, as well as the more familiar
y-direction.
Objective:
Given the equation of a parabola, find the vertex and intercepts, sketch the
graph, and check by computer graphics.
EXAMPLE
For x == -2y 2 + 12y - 10
a. Find the vertex and intercepts.
b. Sketch the graph.
c. Confirm your results by computer graphics.
Solution:
a. The y -coordinate of the vertex is
k
12
2(-2) = 3.
487
488
Cilapler q
Quadratic Relations and Systems
!b~1
II •
I
Substituting 3 for y gives x
= -2(9) + 12(3)
- 10
=8
So the vertex is at (8, 3). Setting y = 0 gives -10 for the x-intercept. Setting x
0 and solving the quadratic equation gives 5 and I for the
two y -intercepts.
b. The graph is shown in Figure 9-5b.
v
5
18.3)
3
8
-10
x
x -2v2+12V-10
Figure9-5b ____________________________________________
c. To use the computer program PLOT CONIC on the accompanying
disk, you must first have all terms on the left side and 0 on the right.
_2y2 + 12y
x
2y2
+x
12y
10
+ 10
0
When the program calls for the coefficients, you enter
0,0,2, I,
12,10
2
because the x and xy terms have coefficients of O. The graph looks
similar to Figure 9-5b, thus confirming the sketch.
•
You can tell just by looking at the equation if the graph is a parabola. The
equation will have only one squared term. If the Xl term is missing, the
graph opens in the x-direction. If the y2 term is missing, it opens in the
y -direction.
In the next section you will learn how to tell what the graph of any
quadratic relation is, just by looking at its equation. You will also learn
that parabolas, too, have a geometrical definition involving a focus.
9-5 Parabolas
EXERCISE 9-5
Do These Quickly
The following problems are intended to refresh your skills. You should be
able to do all 10 in less than 5 minutes.
What will the graph be?
x2 + yl
25 2
y2 = 25
Q2. x
Ql.
Q3.
Xl
y2
-25 Q4.
x2 + y2
-25 Q5_
x+y
Q6.
x-y=25 25 Answer the questions.
What percent of 200 is 30? Q7.
Q8 _ If you go 50 miles in
21
hours, what is your average speed? Q9 _ Divide 90 by 1 QlO.
For
a. b. c. Evaluate the complex number
,7. Problems 1 through 10, Find the vertex and intercepts.
Sketch the graph.
Confirm your sketch using PLOT CONIC from the accompanying
disk, or another suitable graphing program.
Find the vertex and intercepts of the following parabolas, and sketch the
graph. Find decimal approximations for any radicals you encounter.
/'"
1. .'.1.; X = y2
3,j) X
s:
3y2
.
1 2
5.' x = '2?
1.. 7/
y
4y
=
+
3
12y
5
+ 3y + 4
-4x 2
+ 20x
16
2.
X = y2
4.
x
6.
x = _y2 - 2y - 9
3
8.
y
=
+ 2y - 3
-5y2
+ 30y +
1
1
= -x 2 +
5
2x - ­
11
5
11
489
4~)(
)
Chapter 9
9.
Quaciratic Relations anci Systems
1
x =-y
4
2
10.
~lE
x = - 10
For Problems 11 through 14, sketch the graph quickly.
11. Y=X2
13. x =
y2
12.
Y
14.
x
-x 2
= y2
From the graphs in Problems 11
through 14, what feature of the equation of a parabola tells' you that
it opens
a. in the positive y -direction?
b. in the positive x-direction?
c. in the negative y-direction?
d. in the negative x-direction?
15. Parabola Conclusions Problem
(Ii)
I
EQL',\TJO:\'S
FRO~l
GEOMETRICAL OEFJ:"ITJO:"S
You have learned geometrical definitions for circles, ellipses, and hyperbo­
las in Sections 9-2, 9-3, and 9-4. In this section you will learn a geomet­
rical definition for a parabola. You will also learn that there are other geo­
metrical definitions for all four kinds of graphs. Using these definitions,
you will derive equations for the graphs.
( )i)jecti\c:
GiVen the geometrical definition of a set of points. derive an equation
relating x and \'. If the equation is quadratic. tell whether the graph is a
circle. ellip~e. hyperbola, or parabola.
Circles, ellipses, hyperbolas, and parabolas have afamily name that comes
from their relationship to slices ("sections") of a cone. As shown in Figure
9-6a, a complete cone has two halves, called "nappes." If a plane cuts the
cone parallel to one of its elements, the section is a parabola. If the plane
exceeds the parallel position, it will cut both nappes, forming a hyperbola
(from the Greek work "hyperbole," meaning "to exceed"). If the plane
does not tilt far enough, one of the nappes will be left out, and the graph
will be an ellipse (from the Greek word "ellipsis," meaning "to leave
out"). A circle is formed by a cutting plane perpendicular to the axis of
the cone.
Because of the properties stated above, circles, ellipses, hyperbolas, and
parabolas are given the name conic sections. Perhaps you have seen vad­
9-6
Equations from Geometrical Definitions
Conic sections
y
...... 1,<
) x
Parabola
I) x
Ellipse
y
y
I ) x
Circle
'*'.
)
x
Hyperbola
Figure 9-6a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
ous conic sections when the cone of light from a lamp is cut by the plane
of a wall or ceiling (Figure 9-6b).
The second part of the objective may now be stated, "If the equation is
quadratic, tell which conic section the graph is." In order to do this, you
should pull together what you have learned about the coefficients of the
491
.+t).2
(JlI(\( irdlit HcJdli()l1<'; <llH I S\'c;t('tns
\ 11d\lWr 'I
---
-
-
~[ESJ
-..,.
Hyperbolic shadow
on wall
Lamp shade
j.
WIn, 'i-hh ................__..
~ __.... _..........__..
squared terms in the equation of a quadratic relation. The following table
should help refresh your memory.
RECOGNITION
EQUATIONS
Circle:
Ellipse:
Hyperbola:
Parabola:
OF CONIC SECTIONS FROM THEIR
x 2 and y2 terms have equal coefficients.
x 2 and y2 terms have unequal coefficients, but
the same sign.
x 2 and y2 terms have opposite signs.
Equation has only one squared term. If the x 2 term
is missing, the parabola opens in the x-direction.
If the y2 term is missing, the parabola opens in
the y-direction (a quadratic function).
Note: All of the above conclusions assume that there is no xy-term.
If there is an xy-term, you must use the discriminant to identify the
conic section, as explained in Exercise 9-7, Problem 5.
9-6
Equations from Geornetrical Definitions
EXA);IPLE
A parabola is defined to be a set of coplanar points, each of which is the
same distance from a fixed focus as it is from a fixed straight line (called
the "directrix"). Find the equation of the parabola whose focus is (2, 3)
and whose directrix is the line y = -7. Show that the equation has the
ax 2 + bx + c.
form y
y
bl'
><..
'(I
x
d,
y ; -7
~D~e~trIx
(x,
-7)
Figure 9·6c _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
The first step is to draw a picture showing the given focus and directrix
(Figure 9-6c). Then pick a point (any point!) that could be on the graph,
and label it (x, y). By the definition of parabola, above,
d1
d2 •
The point where the segment labeled d 1 meets the line y
-7 has coordi­
nates (x, -7). So d 1 and d2 can be found using the Distance Formula.
Substituting the appropriate coordinates into the equation d 1 = d2 gives
V(x - x)2 + (y + 7)2 = V(x - 2)2 + (y - 3)2,
which reduces to
Iy
+
7/ = V(x
- 2)2 + (y - 3)2.
Squaring both members transforms this to a polynomial equation,
y2 + 14y + 49
(x - 2)2 + (y
3)2.
Expanding the binomial squares on the right gives
y2 + 14y + 49 = x 2 ­ 4x + 4 + y2 - 6y + 9.
Note that the y2 terms will cancel, leaving only one squared term. The
equation can be transformed to
20y = x 2
-
4x - 36,
.+93
494
Chapter 9
Quadratic Relations and Systems
....
r~1 ~ ~.
~~
or
]
2
I
9
y = -x - -x - ­
20
5
5'
Since this last equation has the form y = ax· + bx + c, you can see that
this set of points described geometrically fits your old definition of a
•
parabola as the graph of a quadratic function.
The exercise which follows is designed to give you practice finding equa­
tions and identifying graphs from given geometrical definitions.
EXERCISE 9-6
For Problems 1 through 12, find a polynomial equation with integer
coefficients for the set of coplanar points described. Tell whether or not the
graph is a conic section and, if it is, tell which conic section.
1. For each point, its distance from the fixed point (- 3, 0) is twice its
distance from the fixed point (3, 0).
2. For each point, its distance from the fixed point (0, -5) is ~ of its distance from the fixed point (0, 5). 3. For each point, its distance from the fixed point (4, 3) is 3 times its
distance from the fixed point (-1, 2).
4. For each point, its distance from the fixed point (-I, -3) is ~ of its
distance from the fixed point (5, -8).
5. Each point is twice as far from the line x = 3 as it is from the line y == -2. 6. Each point is three times as far from the line y = 5 as it is from the
line x = I.
7. Each point is equidistant from the point (3, -4) and the line y = 2.
8. Each point is equidistant from the point (2, 5) and the line x
- 3.
9. For each point, its distance from the point (- 3, I) is half its distance
from the line y = 4.
10. For each point, its distance from the point (-4, - 1) is ~ times its
distance from the line x = - 5 .
11. For each point, its distance from the point (0, 3) is ~ times its dis­
tance from the line y = 3.
9-7
Quadratic Relations-xy-Term
12. For each point, its distance from the point (4, 0) is twice its distance
from the line x = -4.
13. Use the geometric definition of an ellipse (Section 9-3) to show that
the ellipse with foci (4, 0) and (-4, 0) and major axis 10 units long
is
9x 2
+ 25y2
=
225.
14. Use the geometric definition of hyperbola (Section 9-4) to show that
the hyperbola with foci (5,0) and (-5, 0) and transverse axis (i.e.,
the distance between vertices) 8 units long is
9x 2
16y2 = 144.
15. Show that another definition of a circle is a set of points for each of
which its distance from the fixed point (j, 0) is c times its distance
from the fixed point
j, 0).
16. Find a polynomial equation of a set of points for each of which the
distance from the fixed point (j, 0) is e times its distance from the
vertical line x = - j, where e stands for a constant. Show that the
graph will be an ellipse if 0 < e < 1; a parabola if e = 1; and a hy­
perbola if e > 1.
17. The geometrical definitions in Problem 16 are called the "focus­
directrix" definitions of the conic sections. The straight line is called
the "directrix," the fixed point is called the "focus," and the constant
e is called the "eccentricity." What would be the eccentricity for a
circle? Where would the directrix be for a circle?
18. A parabola may be thought of as having a second focus, just like el­
lipses and hyperbolas do. Where do you suppose this second focus
would be? (Remember the cones and the lamp shade of Figures 9-6a
and 9-6b.)
9- 7
I QUADRATIC RELl\.TIONS-xy-TERM
In your investigation of quadratic relations, you have not yet encountered
one for which the equation had a non-zero xy-term. In this section you
will find out that the graphs are still conic sections, but the xy-term rotates
the graph and affects its shape.
Objective:
Determine by actual plotting what effects the xy-term in the equation of a
quadratic relation has on the graph.
495
i:ll!rd!i(
HelaTiol1s and Systems
~~
The following exercise is designed to accomplish this objective.
+ y2 = 9.
1. Plot a graph of
2. The following equations differ from the one in Problem 1 only by the addition of increasingly larger xy-terms: i. x 2 + xy T y2 = 9.
ii. x 2 + 4xy + y2
9.
a. For each relation, select values of x, calculate the corresponding
values of y, and plot the graph.
b. For each graph, tell which of the conic sections is plotted.
c. Does the xy-term affect the x and y-intercepts? Explain.
3. i Uf1illCS for xy-Term Problem
Use the program PLOT
CONIC from the accompanying disk to draw the graphs of:
a. x 2 + Y 2
lOx - 8y T 16 = 0
b. x 2 + xy + y2
lOx - 8y + 16 = 0
2
C. x T 2xy + y2
lOx - 8y T 16 = 0 d. x 2
T 4xy + y2
lOx - 8y + 16 = 0
4 . " l>roiJIem
Tell which conic section each graph in
Problem 3 will be.
5. !'roh/em Use your graphs from Problems I, 2, and 3
to answer the following questions:
a. When there is an xy-term in the equation, can you still tell by
looking at the coefficients of x 2 and y2 just which of the conic
sections the graph will be?
b. The equation AX2 + Bxy + Cy2 + Dx + Ey + F
0 has a
discriminant, which equals
B2 4AC.
For each of the equations in Problems 1 and 3, calculate the dis­
criminant.
c. Show that the sign of the discriminant tells the following about
the kind of conic section:
Circle or Ellipse-discriminant
Parabola-discriminant = O.
Hyperbola-discriminant> O.
6. < O.
Prove that the graph of an inverse variation function is a hyperbola.
9·8
9·8
Systems of Quadratics
I SYSTEMS OF QUADRATICS
In Chapter 4 you learned how to solve systems of two linear equations
with two variables. The answer tells the point where the two straight-line
graphs cross. There are situations in the real world where it is important
to know where the graphs of quadratic relations cross. For example, the
paths of spaceships, planets, comets, and so forth, which travel under the
action of gravity, turn out to be circles, ellipses, hyperbolas, or parabolas.
Finding the point where a spaceship crosses the path of the planet it is ap­
proaching is of fundamental importance! Ships at sea use the LORAN sys­
tem (for LOng RAnge Navigation) to find their location by receiving radio
signals which locate them on several hyperbolas. The intersection point of
the hyperbolas tells the ship's position (See Figure 9-8b).
Planet (ell iplical
path)
{Parabolic or
hyperbolic path)
Figure 9-8a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
/
LORAN System
Figure 9-8b _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
497
~P)~.
i
I)<tl)i('l ()
,)lldtird\!(
~~
j{1'ldli()] Ie, diHi ~\,'-,I('lll'-,
L----J
;1 );(' Ii \ : '
In t\\,) \~lIl~ibk:\. \\here at lea'! Olle equa­
',.il\LTI ;: ,\ '.[, '1,,)1 : ,\.) equall,)I"
'Jl;d,j!·,:~
1 1 ,':1
"
ll'.':1' ~I;'~
<!~I'r d\..)~ll~L\l.
he' ahlc
tl):
lill'UletiC lh,: \,)iU!!()1l '.l't oj the \\ -tCI11
S\1,·\\ lhal '-,(lUI' clll"\l?r 1\ !,,?Cl\Ulwhie h\ graphIng,
As shown in Figure 9-8c, the graphs of a linear and a quadratic equation
may cross in two places and the graphs of two quadratics may cross in four
places. The technique for finding the solutions is the same as for systems
of linears-eliminate one variable, solve the resulting equation for the
other, then substitute back to find the corresponding value of the first vari­
able. However, the algebraic operations get much more complicated. At
worst, solving these systems may take just about every algebraic skill you
have ever learned!
v
v
Line
/
f
", "), ........
X
\
L,near·quadratlc
two solutions
hgllL' <}·S~
\
17ttT
•
Quadratlc'quadratlc
four solutions
~
...------ .. - - , - - - - - - - - - - - - - - - ­
I. \.. ,,, IPI. F
Solve the system
9x 2
3x
+
2 -
32 y 2 = 324
CD,
3
(2).
y2
=
Solution:
Since both equations have only squared terms, the problem is almost as
easy as solving a system of two linears. You would multiply both members
of Equation (2) by - 3 and add the resulting equation to Equation CD:
x
9·8
Systems of Quadratics
-9x 2 + 3y2 = -9
9X2 + 32y2 = 324
35y2 = 315
y2 = 9
y
±3.
These values of y may be substituted into either of the equations one at a
time. Substituting into Equation ® gives
If y
-3,
3x 2 - 9 = 3
3x 2 = 12
x2 = 4
x = ±2
If y
3,
3x 2 - 9 = 3
3x 2 = 12
x2 = 4
x = ±2
:. S = {(2, 3), (2, -3), (-2, 3), (-2, -3)}.
The solution set can be checked graphically. By sketching or computer
graphics, you find that Equation <D gives an ellipse, centered at the origin,
with x-radius =
6, and y-radius =
= vlo = 3.2. Equa­
tion ® gives a hyperbola, centered at the origin, opening in the x-direc­
tion, with asymptotes of slope ±Y3 = ± 1.7, and vertices 1 unit from
the center. The graphs are shown in Figure 9-8d. You can see that the
four points you have calculated are actually the points of intersection.
vW
y
-6
t
I
I
-2
j
I
16 .,x
I
2
•
Figure 9-8d
EXAMPLE 2
Solve the system using the "substitution" technique.
7x 2
-
5y2 + 20y = 3
2lx 2
+ 5y2
= 209
499
500
Cilapter 9
Quadratic Relations and Systems
gj[B
Solution:
The linear combination technique used in Example I works only in special
cases. A more general technique of eliminating a variable, called
"substitution," works for any system of quadratics no matter how compli­
cated. Starting with
7x 2
+ 20y
5y2
-
= 3
- - (1),
209 - - ®,
21x + 5y2 +
2
you could solve Equation ® for x in terms of y, getting
21x 2 = 209 - 5y2
Xl
=
1
21(209 - 5y2)
@.
5y2) for x 2 in Equation (1) gives
Substituting 2\ (209
1
7 . 21 (209
+ 20y
5y2) - 5y2
= 3,
which reduces to
y2 - 3y
10 = O.
Factoring and solving for y gives
(y
y = 5
5)( y
or
+ 2)
0,
y = -2.
These values of y may now be substituted, one at a time, into Equation @
to find x.
If y = 5,
If y
Xl
;1 (209 - 125)
x2
x2
4
x2
x = ±2
= ;1 (209
- 20)
9
±3 x
:. S = {(2, 5),
-2,
2,5), (3, -2), (-3, -2)}. EXAMPLE 3
Check the answer to Example 2 by graphing.
Solution:
The check is most easily done by computer graphics. Otherwise, by com­
pleting the square, Equation (1) can be transformed into
x2
(y
2)2
c;) + c;)
=
1.
•
9·8
Systems of Quadratics
This is a hyperbola, centered at (0, 2), opening vertically, with asymptotes
V¥
J
of slope ± ~
vlJ.
7"
= ±v1 = ±L2.
Equation ® can be transformed to
x2
y2
e~:) + e~9)
= 1.
This graph is an ellipse with y-radius of ~ =
x-radius of ~ = \IiO "'" 3.2.
v42 = 6.5, and
The graphs are shown in Figure 9-8e. As you can see, the intersection
points are (2, 5),
2,5), (3, -2), and (-3, -2).
y
I:'
lI0.cp-~
l JI
..
~
/.
~
x
0..
Figure 9·8e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
•
EXAMPLE 4
Solve the system
x2
+
y2
4x
4x
+
+
2y
20,
3y = 5.
Solution:
If one of the two equations is linear, then the linear combination technique
seldom works and you must use substitution.
x 2 + y2 - 4x + 2y = 20
4x
+
3y
=
5
CD
®
501
;',I,llil
1{(']dti()I1S
,mel Systems
[g][B
Solving ® for y in terms of x gives
y
Substituting
~ (5
x
2
+
1
= - (5
3
- 4x)
@
- 4x) for y in Equation
[~(5
-
CD gives
4X)r - 4x + 2[~(5 -
4X)] = 20,
Simplifying and multiplying each member by 9 to eliminate the fractions
gives
9x 2
+ 25 - 40x + 16x 2
-
36x
25x
2
-
+ 30 - 24x
=
180
100x - 125
=
0
x2
4x - 5 = 0
-
(x - 5)(x
x=5
+ 1) = 0
or
x=-l.
Substituting these answers one at a time into the linear equation, or better
still, into Equation @, gives
If x
y
:. S
= 5,
1
= -
3
Ifx=-l,
(5 - 20)
=
-5
1
y = - (5 + 4) = 3
3
= {(5, -5), (-1, 3)}.
By completing the square, you can find that the graph of Equation CD is a
circle of radius 5, centered at (2, -1). The second graph is a straight
line of slope -5 and y-intercept ~. The graphs are shown in Figure 9-8f.
v
'(
'\1
V
lax
9·8
Systems of
Ql"cJI(;,:;c
If both equations have squared and also linear terms, or xy-terms, the job
of solving the system becomes much more complicated. Solving one equa­
tion for y in terms of x involves a clever use of the quadratic formula. The
resulting equation will be fourth degree in x, and will have four solutions.
Problem 30 in the following exercise leads you stepwise through the solu­
tion of one such system.
EXERCISE 9-8
Do These Quickly
The following problems are intended to refresh your skills. You should be
able to do all lOin less than 5 minutes.
What will the graph be?
Ql. x 2 + y2
Q2.
Q3.
=
100
x 2 + 9y2 = 100 x 2 + 5xy + 9y2 = 100 + 6xy +
Q5. Xl + txy +
Q6. x 2 + 5xy -
Q4.
X2
9y 2
=
100 9y2
100 9y2
100 Sketch the graph:
Q7.
A system of independent linear equations Q8.
A system of inconsistent linear equations Q9.
An increasing exponential function QlO.
x = y2
For Problems 1 through 26,
a. calculate the solution set, and
b. demonstrate that your solutions are right by drawing the graphs,
preferably by computer.
l. 2X2 + 5y2 =
2X2 - y2 =
3. 98
2.
x 2 - y2
8X2 - 3y2
4.
4X2
4x
2
x 2 + y2 = 25
y- X2 = -5
+
16 -3 y2 = 100 y2 = -20 304
Chapter 9
Quadratic Relations and Systems
5. 3x 2 + 7y2
3x 2 - 7y
7. x 2 + 2y2
x 2 + y2
=
187
47
8.
33
+ 2x
9. 3x 2 - 5y2
,
3X2 - v-
9x 2 + y2
2x 2 3y2
5x 2 3y2
5x 2 6y2
x 2 + y2
2x 2 3y2
6.
19
[g]
=
=
85 6
=
-22 + 12y = -85 + 6x = 16
22
6x = 8
10.
II. Xl + y2 64
100
x 2 + lOy
13. x 2 + Y 2 + 8x = - 15
9x 2 + 25)'2
225
12.
+ = 100
8x 2 + 13y2 = 1405
14.
- 6y
34
+ y2 = 25
20x 2 3y2 + 12y
20x 2 + 3y2 = 128
15. 5x 2
x2
+ 9)' = 161
4y
16.
4
17. Xl - 5y2
xy
-24
-44
25. x 2
X
20.
lOx
8
2
7x + y2 = 64
4
X+ y
- y2
2x
3y
22.
24.
3
+ 9y2
26. 3x 2
x - y
16
xy
+ 8x + 9
3y
=
+ 4y2 = 68
18.
3y2 = -II
19. 16x2
8x
11
Y
21. Xl + 2y2
33
11
3x + 2y
23. y = X 2
x
y
= 24
=7
-7
2X2 + 8x + 27
y
2x + y = 15
+ 36y = 20
2
+ 30x + 6y =
= -7
)'2
-63
27. Given the equations
a. b. c. d. - y2 - 2y = 17
l
7x + 3)'2 - 24y = 139
CD
x-y=9 @,
®
solve the system of CD and ®,
solve the system of CD and @,
solve the system of ® and @,
graph all three equations on the same Cartesian coordinate sys­
tem, showing that your answers for a. b, and c are reasonable.
28. Repeat Problem 27 for the equations
8x 2
-
3y2 + 30y
32x 2 + 3y2
=
=
10
8x - 3y
=
140
80
CD
®
@.
9-8
Systems of Quadratics
29_ Meteorite Tracking Problem Suppose that you have been hired by
the Palomar Observatory near San Diego. Your assignment is to track
incoming meteorites to find out whether or not they will strike the
Earth. Since the Earth has a circular cross-section, you decide to set
up a Cartesian coordinate system with its origin at the center of the
Earth _The equation of the Earth's surface is
x2
+
y2
40,
where x and y are distances in thousands of kilometers.
y
I
... x
a. The first meteorite you observe is moving along the parabola
whose equation is
18x -
y2
144.
Will this meteorite strike the Earth's surface? If so, where? If
not, how do you tell?
505
-~~
It "
Clld]lI('r(j
()tl,Hlr,lll(
;!('i,!1: I
~!"Ill"
""
'
12] ~~
~
___ ~ __,
L' _____I
b. The second meteorite is coming in from the lower left along one
branch of the hyperbola
4x 2 - y2 - 80x = -340,
(see sketch). Will it strike the Earth's surface? If so, where? If
not, how do you tell?
c. To the nearest 100 kilometers, what is the radius of the Earth?
30. The solution of a general
system of quadratics is extremely complicated. It involves just about
all of the algebraic techniques you know! This problem leads you
stepwise through the solution of a system which is general except that
the equations contain no xy-terms. You must be able to use substitu­
tion, the Quadratic Formula, solving a radical equation, squaring a
trinomial, factoring a quartic polynomial by the Factor Theorem,
getting decimal approximations of radicals, and discarding extrane­
ous solutions by drawing a graph.
Cella,,' (fliiliimllc--(j<;artruuc ,y,l[un
Given the system
5x 2 + y2
x 2 + 3y2
+ 30x - 6y = -9
+ 4x - 6y = 21
CD
@,
do the following:
a. By considering x to be a "constant," write Equation CD as a
quadratic equation in y. Then use the Quadratic Formula to show
that
y
=3
±
V- 5x
2 -
30x.
b. Substitute this value of y into Equation @ and carry out the indi­
cated operations. (Be careful squaring the value of y!) Then col­
lect like terms and simplify as much as possible. You should find
that 2 is a common factor of each term.
c. Isolate the remaining radical on one side of the equation, then
square both members; all terms should now be polynomial.
(Don't worry if some of the coefficients are in the thousands!)
d. Transform the equation so that all terms are on the left and 0 is
on the right. You should have a quadratic (fourth degree) poly­
nomial on the left, beginning with 49x 4 and ending with 36.
e. Use the Factor Theorem to show that (x + 1) and (x + 6) are
factors of this polynomial. Then find the other polynomial by di­
vision.
f. Show that the remaining quadratic factor is prime.
g. Solve the equation in part d. Use the Quadratic Formula, where
needed, and get decimal approximations for the radicals.
h. Find the values of y corresponding to each value of x from part
g (4 values!).
9·9
Chapter Reyjew and Test
i. Sketch the graphs of the two equations. Then discard the y­
values from part h which do not correspond to crossing points.
J. Write the solution set of the system.
9·9
I CHAPTER REVIE\\, ,,,,"'\'D TEST
In this chapter you have extended your knowledge of quadratics to rela­
tions that have a y2 term in the equation. You have found that the graph
could be a circle, ellipse, or hyperbola, as well as the familiar parabola.
For each of these conic sections you learned properties that allowed you to
sketch the graph quickly. By analysis of geometric properties ("analytic
geometry") you were able to derive equations of relations from geometric
definitions. Finally, you extended the system-solving of Chapter 4 to sys­
tems of quadratic equations.
The Review Problems, below, give you a chance to try accomplishing
these objectives one at a time. The Concepts Problems are designed to see
if you are familiar enough with the concepts of the chapter to be able to
use them in problems you have never seen before! This Chapter Test is
more like a normal classroom test.
REVIEW PROBLEMS
The following problems are numbered according to the three objectives
above.
Rl.
a. Tell which conic section the graph or the boundary of the re­ gion will be. i. Xl + y2 = 36
ii. x 2 + y2 < 36
2
iii. x + y2 2: 36
iv. x 2 - y2
36
2
v. x - y2 = -36
vi. x 2 + y2 = -36
vii. Xl + 4yZ = 36
viii. 4x 2 + y2 = 36
2
ix. 4x + Y = 36
x. 4x + y2
36
xi. x 2 - 9y2 + lOx + 54y - 47 = 0
xii. x 2 + y2 + 16x - 22y + 85 = 0
xiii. 9x 2 + 4y2
54x + 16y 479
0
xiv. X - yZ + 6y
3 = 0
b. Transform each equation in part a to a standard form (if neces­
sary), and sketch the graph.
c. Find the focal radius for Equations xi and xiii.
R2.
a. Write the geometric definition of
i. a circle,
ii. an ellipse,
307
508
Chapter 9
Quadratic Relations and Systems
~[B
m.
a hyperbola,
iv. a parabola.
b. From the geometric definition, derive a polynomial equation of
the ellipse with foci (3, 0) and (- 3, 0), and major axis 10
units long.
R3. a.
Solve the following systems:
i. 9x 2 + y2
2y
80
+ y2 lOy 0
ii. 9x 2 16y2 + 36x
32y
195
3x + 4y = 15
b. Show by graphing that your solutions for part a are correct.
CO;\lCEPTS
PROBLE~lS
C 1. Sketch graphs that illustrate how a system of two quadratic equa­
tions with two variables can have
a. exactly four solutions,
b. exactly three solutions,
c. exactly two solutions,
d. exactly one solution,
e. no solutions.
C2. You have learned how to find graphs of quadratic relations from
their equations. Using the same properties, reverse the process and
write equations or inequalities for the following graphs:
a.
v
1_
I
.!',(~'!{}'!{)'Il
~.
x
g.g
Chapter ReYic'
y
b.
x
c.
y
\
It
-3 -2 -1 10 1
-1
-2
...
3
4
5 6
\
~
~
7
X
c
0.::'
SIO
Chapter~)
Quaciratic Relations and Systems
~
v
d.
_,
... x
,..",,~
C3. It is possible to write the equation of a parabola in terms of the dis­
tance, p, between the focus and the directrix (see sketch).
a. Draw x- and y-axes so that the vertex is at the origin. Then
use the definition of a parabola to show that the equation is
J
Focus
p
Directrix
b. For the parabola y = 3x 2 + 5x - 7, how far is the focus from
the directrix? Justify your answer.
C4. You have graphed inequalities with circles or ellipses as boundaries,
but never with hyperbolas. In this problem you will discover what
such a graph looks like.
a. The boundary for the graph of 4x 2 - y2 ~ -16 is a hyper­
bola. Draw this hyperbola.
b. To find which side of the boundary line the region lies on, sub­
stitute 0 for x, and solve the resulting inequality for y. Re­
member the properties of order, and the fact that
Y(number)2
1number I·
c. Shade the correct region on your graph.
9·9
Chapter Review and Test
C5. Recalling the definitions of x-intercept and y-intercept, find the in­
tercepts of
Xl
+ yl + 16x - 22y + 85
= O.
C6. Suppose that you are aboard a spaceship. The Earth is at the origin
of a Cartesian coordinate system, and the path of your spaceship is
the graph of
9Xl
a. b. c. d. + 25y2 - 72x
= 81.
Which conic section is this path, and how do you tell?
Sketch the graph of this path.
Show that the Earth is at the focus of this conic section.
To determine your position at a particular time, you tune in to
the LORAN station and find that you are also on the graph of
9x 2 - 15y2 = 9.
Which conic section is this graph, and how do you tell?
e. By solving the system formed by the equations of the two
conic sections above, find the three possible points at which
your spaceship could be located.
f. Draw the graph of the conic section in part d on the same
Cartesian coordinate system as in part b, and thus show that
your answer to part e is correct.
g. To ascertain at which one of the three points you are located,
you find that you are also on the graph of
3x - 5y
27.
At which of the points are you located? (There is a clever way
to do this, as well as the longer way.)
CHAPTER TEST
For Problems T1 through T5, transform the equation by completing the
square. Then sketch the graph.
TI.
T2.
4x 2 + y2 - 40x + 6y
x 2 ­ 4y2 - 8x + 16y
+ 93
0
36 = 0 + 90x + 32y + 353
T4. Xl + yl + 6x - l6y + 48 :5 0 T3.
9Xl - l6yl
T5.
x = O.2yl
2y - 3 = 0
511
, 1'···..,
For Problems T6 through T8, solve the system.
T6. 7x 2
5x 2
-
9y2
31 + 9y2 161 TI. x 2 + y2 - 4x = 21 x 2 + 5y2 = 49 T8. x 2 ­
x - y
Tq
T9.
3y2 =
=
11 -I Sketch graphs that iIlustrate the following:
a. A circle and an ellipse that intersect at four points.
b. A parabola and an ellipse that intersect at exactly three distinct
points.
c. A hyperbola and a circle that intersect at exactly two distinct
points.
d. An ellipse and a hyperbola that intersect at only one point.
e. Two parabolas that intersect at no points.