Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one
of the foci is the star around which it revolves. The planet is closest to the star
when it is at one vertex. It is farthest from the star when it is at the other vertex.
Suppose the closest and farthest distances of the planet from this star, are 420
million kilometers and 580 million kilometers, respectively. Find the equation of
the ellipse, in standard form, with center at the origin and the star at the x-axis.
Assume all units are in millions of kilometers.
Solution. In the figure above, the orbit is drawn as a horizontal ellipse with
center at the origin. From the planet’s distances from the star, at its closest
and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000
(million kilometers), so a = 500. If we place the star at the positive x-axis,
then it is c = 500 − 420 = 80 units away from the center. Therefore, we get
b2 = a2 − c2 = 5002 − 802 = 243600. The equation then is
y2
x2
+
= 1.
250000 243600
The star could have been placed on the negative x-axis, and the answer would
still be the same.

Seatwork/Homework 1.3.3
 1. The arch of a bridge is in the shape of a semiellipse, with its major axis at the
water level. Suppose the arch is 20 ft high in the middle, and 120 ft across its
major axis. How high above the water level is the arch, at a point 20 ft from
the center (horizontally)? Round off to 2 decimal places. Refer to Example
1.3.6.
Answer: 18.86 ft
45
Exercises 1.3
1. Give the coordinates of the center, vertices, covertices, and foci of the ellipse
with the given equation. Sketch the graph, and include these points.
y2
x2
+
=1
169 25
y2
x2
+
=1
(b)
144 169
(c) 4x2 + 13y 2 = 52
(a)
(x + 7)2 (y − 4)2
+
=1
16
25
(e) 9x2 + 16y 2 + 72x − 96y + 144 = 0
(d)
(f) 36x2 + 20y 2 − 144x + 120y − 396 = 0
Answer:
Item
Center
Vertices
Covertices
Foci
(a)
(0, 0)
(±13, 0)
(0, ±5)
(±12, 0)
(b)
(0, 0)
(±12, 0)
(0, ±5)
(c)
(0, 0)
(0, ±13)
√
(± 13, 0)
(0, ±2)
(±3, 0)
(d)
(−7, 4)
(−7, −1)
(−11, 4)
(−7, 1)
(−7, 9)
(−3, 4)
(−8, 3)
(−4, 0)
(−7, 7)
√
(−4 ± 7, 3)
(0, 3)
(−4, 6)
√
(2 ± 2 5, −3)
(2, −7)
(e)
(f)
(−4, 3)
(2, −3)
(2, −9)
(2, −3)
(2, 3)
(2, 1)
(a)
(b)
46
(c)
(d)
(e)
(f)
2. Find the standard equation of the ellipse which satisfies the given conditions.
(a) foci (−7, 6) and (−1, 6), the sum of the distances of any point from the
(x + 4)2 (y − 6)2
+
=1
foci is 14
Answer:
49
40
(b) center (5, 3), horizontal major axis of length 20, minor axis of length 16
(x − 5)2 (y − 3)2
Answer:
+
=1
100
64
(c) major axis of length 22, foci 9 units above and below the center (2, 4)
(x − 2)2 (y − 4)2
Answer:
+
=1
40
121
(d) covertices (−4, 8) and (10, 8), a focus at (3, 12)
(x − 3)2 (y − 8)2
+
=1
Answer:
49
65
Solution. The midpoint of the covertices is the center, (3, 8). From this
point, the given focus is c = 4 units away. Since b = 7 (the distance from
the center to a covertex), then a2 = b2 + c2 = 65. The ellipse then has
(x − 3)2 (y − 8)2
equation
+
= 1.
49
65
(e) focus (−6, −2), covertex (−1, 5), horizontal major axis
(x + 1)2 (y + 2)2
Answer:
+
=1
74
49
Solution. Make a rough sketch of the points to see that the center is to
the right of the given focus, and below the given covertex. The center is
thus (−1, −2). It follows that c = 5, b = 7, so a2 = b2 + c2 = 74. The
(x + 1)2 (y + 2)2
ellipse then has equation
+
= 1.
74
49
3. A semielliptical tunnel has height 9 ft and a width of 30 ft. A truck that is
about to pass through is 12 ft wide and 8.3 ft high. Will this truck be able to
pass through the tunnel?
Answer: No
47
4. A truck that is about to pass through the tunnel from the previous item is 10
ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel?
Answer: Yes
5. An orbit of a satellite around a planet is an ellipse, with the planet at one
focus of this ellipse. The distance of the satellite from this star varies from
300, 000 km to 500, 000 km, attained when the satellite is at each of the two
vertices. Find the equation of this ellipse, if its center is at the origin, and the
vertices are on the x-axis. Assume all units are in 100, 000 km.
Answer:
x2
16
6. The orbit of a planet around a star is described by the equation
+
y2
15
=1
x2
640,000
+
y2
630,000
= 1, where the star is at one focus, and all units are in millions of
kilometers. The planet is closest and farthest from the star, when it is at the
vertices. How far is the planet when it is closest to the sun? How far is the
planet when it is farthest from the sun?
Answer: 700 million km, 900 million km
Solution. The ellipse has center at the origin, and major axis on the x-axis.
Since a2 = 640, 000, then a = 800, so √
the vertices√are V1 (−800, 0) and
V2 (800, 00). Since b2 = 630, 000, then c = a2 − b2 = 10, 000 = 100. Suppose the star is at the focus at the right of the origin (this choice is arbitrary,
since we could have chosen instead the focus on the left). Its location is then
F (100, 0). The closest distance is then V2 F = 700 (million kilometers) and
the farthest distance is V1 F = 900 (million kilometers).
7. A big room is constructed so that the ceiling is a dome that is semielliptical
in shape. If a person stands at one focus and speaks, the sound that is made
bounces off the ceiling and gets reflected to the other focus. Thus, if two
people stand at the foci (ignoring their heights), they will be able to hear each
other. If the room is 34 m long and 8 m high, how far from the center should
each of two people stand if they would like to whisper back and forth and hear
each other?
Answer: 15 m
48
Solution. We could put a coordinate system with the floor of the room on
the x-axis, and the center of the room at the origin, as shown in the figures.
The major axis has length 34, and the height of the room is half of the minor
y2
x2
axis. The ellipse that contains the ceiling then has equation 17
2 + 82 = 1. The
√
√
distance of a focus from the center is c = a2 − b2 = 172 − 82 = 15. Thus,
the two people should stand 15 m away from the center.
8. A whispering gallery has a semielliptical ceiling that is 9 m high and 30 m
long. How high is the ceiling above the two foci?
Answer: 5.4 m
Solution. As in the previous problem, put a coordinate system with the floor
of the room on the x-axis, and the center of the room at the origin. The major
axis has length 30, and half the minor axis is 9. The ellipse that contains the
y2
x2
ceiling then has equation 15
2 + 92 = 1. The distance of a focus from the center
√
√
is c = a2 − b2 = 152 − 92 = 12. If we put x = 12 in the equation of the
2
2
ellipse, we get 12
+ y92 = 1. Solving for y > 0 yields y = 27
= 5.4. The height
152
5
of the ceiling above each focus is 5.4 m.
4
Lesson 1.4. Hyperbolas
Time Frame: 3 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) define a hyperbola;
(2) determine the standard form of equation of a hyperbola;
(3) graph a hyperbola in a rectangular coordinate system; and
(4) solve situational problems involving conic sections (hyperbolas).
Lesson Outline
(1) Definition of a hyperbola
(2) Derivation of the standard equation of a hyperbola
(3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas
49
Introduction
A hyperbola is one of the conic sections that most students have not encountered formally before, unlike circles and parabolas. Its graph consists of two
unbounded branches which extend in opposite directions. It is a misconception
that each branch is a parabola. This is not true, as parabolas and hyperbolas
have very different features. An application of hyperbolas in basic location and
navigation schemes are presented in an example and some exercises.
1.4.1. Definition and Equation of a Hyperbola
Consider the points F1 (−5, 0) and F2 (5, 0) as shown in Figure 1.21. What is the
absolute value of the difference of the distances of A(3.75, −3) from F1 and
 from
F2 ? How about the absolute value of the difference of the distances of B −5, 16
3
from F1 and from F2 ?
|AF1 − AF2 | = |9.25 − 3.25| = 6
!
!
! 16 34 !
|BF1 − BF2 | = !! − !! = 6
3
3
There are other points P such that |P F1 − P F2 | = 6. The collection of all such
points forms a shape called a hyperbola, which consists of two disjoint branches.
For points P on the left branch, P F2 − P F1 = 6; for those on the right branch,
P F1 − P F2 = 6.
Figure 1.21
Figure 1.22
Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 differ by a certain constant, is called a
hyperbola. The points F1 and F2 are called the foci of the hyperbola.
In Figure 1.22, given are two points on the x-axis, F1 (−c, 0) and F2 (c, 0), the
foci, both c units away from their midpoint (0, 0). This midpoint is the center
50
Figure 1.23
Figure 1.24
of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute
value of the difference of the distances of P from F1 and F2 , be 2a (the coefficient
2 will make computations simpler). Thus, |P F1 − P F2 | = 2a, and so
!
!"
"
!
2 + y2 −
2 + y 2 ! = 2a.
(x
+
c)
(x
−
c)
!
!
Algebraic manipulations allow us to rewrite this into the much simpler
x2 y 2
− 2 = 1,
a2
b
where b =
√
c 2 − a2 .
√
When we let b = c2 − a2 , we assumed c > a. To see why this is true, suppose
that P is closer to F2 , so P F1 − P F2 = 2a. Refer to Figure 1.22. Suppose also
that P is not on the x-axis, so !P F1 F2 is formed. From the triangle inequality,
F1 F2 + P F2 > P F1 . Thus, 2c > P F1 − P F2 = 2a, so c > a.
Now we present a derivation. For now, assume P is closer to F2 so P F1 > P F2 ,
Teaching Notes and P F1 − P F2 = 2a.
If it is assumed
that P is closer to
F1 , then the same
equation will be
obtained because
of symmetry.
P F1 = 2a + P F2
"
(x + c)2 + y 2 = 2a + (x − c)2 + y 2
$2 #
$2
#"
"
2
2
2
2
= 2a + (x − c) + y
(x + c) + y
"
cx − a2 = a (x − c)2 + y 2
$2
# "
(cx − a2 )2 = a (x − c)2 + y 2
"
(c2 − a2 )x2 − a2 y 2 = a2 (c2 − a2 )
b2 x2 − a2 y 2 = a2 b2
x2 y 2
− 2 =1
a2
b
by letting b =
51
√
c 2 − a2 > 0
We collect here the features of the graph of a hyperbola with standard equation
x2 y 2
− 2 = 1.
a2
b
√
2
2
Let c = a + b .
(1) center : origin (0, 0)
(2) foci : F1 (−c, 0) and F2 (c, 0)
• Each focus is c units away from the center.
• For any point on the hyperbola, the absolute value of the difference of
its distances from the foci is 2a.
(3) vertices: V1 (−a, 0) and V2 (a, 0)
• The vertices are points on the hyperbola, collinear with the center and
foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1 V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = ab x and y = − ab x, the lines
1
and
2
in Figure 1.24
• The asymptotes of the hyperbola are two lines passing through the center which serve as a guide in graphing the hyperbola: each branch of
the hyperbola gets closer and closer to the asymptotes, in the direction
towards which the branch extends. (We need the concept of limits from
calculus to explain this.)
• An aid in determining the equations of the asymptotes: in the standard
2
2
equation, replace 1 by 0, and in the resulting equation xa2 − yb2 = 0, solve
for y.
• To help us sketch the asymptotes, we point out that the asymptotes
1 and 2 are the extended diagonals of the auxiliary rectangle drawn
in Figure 1.24. This rectangle has sides 2a and 2b with its diagonals
intersecting at the center C. Two sides are congruent and parallel to
the transverse axis V1 V2 . The other two sides are congruent and parallel
to the conjugate axis, the segment shown which is perpendicular to the
transverse axis at the center, and has length 2b.
Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola
with equation
x2 y 2
−
= 1.
9
7
Sketch the graph, and include these points and lines, the transverse and conjugate
axes, and the auxiliary rectangle.
52
2
2
Solution. With
√ a = 9 and
√ b = 7, we have
a = 3, b = 7, and c = a2 + b2 = 4.
foci: F1 (−4, 0) and F2 (4, 0)
vertices: V1 (−3, 0) and V2 (3, 0)
√
√
asymptotes: y = 37 x and y = − 37 x
The graph is shown at the right. The conju√
gate axis drawn has its endpoints b = 7 ≈
2.7 units above and below the center.

Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are
F1 (−5, 0) and F2 (5, 0), such that for any point on it, the absolute value of the
difference of its distances from the foci is 6. See Figure 1.21.
Solution. We have 2a = 6 and c = 5, so a = 3 and b =
x2 y 2
hyperbola then has equation
−
= 1.
9
16
√
c2 − a2 = 4. The

Seatwork/Homework 1.4.1
1. Determine foci, vertices, and asymptotes of the hyperbola with equation
x2
y2
−
= 1.
16 20
Sketch the graph, and include these points and lines, along with the auxiliary
rectangle.
Answer: foci
F1 (−6, 0) and
F2 (6, 0), vertices V1 (−4, 0) and V2 (4, 0), asymp√
√
5
5
totes y = 2 x and y = − 2 x
√
2. Find the equation
in standard form of the hyperbola whose foci are F1 (−4 2, 0)
√
and F2 (4 2, 0), such that for any point on it, the absolute value of the
y2
x2
−
=1
difference of its distances from the foci is 8.
Answer:
16 16
53
1.4.2. More Properties of Hyperbolas
The hyperbolas we considered so far are “horizontal” and have the origin as their
centers. Some hyperbolas have their foci aligned vertically, and some have centers
not at the origin. Their standard equations and properties are given in the box.
The derivations are more involved, but are similar to the one above, and so are
not shown anymore.
Center
Corresponding Hyperbola
(0, 0)
x2 y 2
− 2 =1
a2
b
y 2 x2
− 2 =1
a2
b
(x − h)2 (y − k)2
−
=1
a2
b2
(y − k)2 (x − h)2
−
=1
a2
b2
transverse axis: horizontal
transverse axis: vertical
conjugate axis: vertical
conjugate axis: horizontal
(h, k)
54
√
In all four cases above, we let c = a2 + b2 . The foci F1 and F2 are c units
away from the center C. The vertices V1 and V2 are a units away from the center.
The transverse axis V1 V2 has length 2a. The conjugate axis has length 2b and is
perpendicular to the transverse axis. The transverse and conjugate axes bisect
each other at their intersection point, C. Each branch of a hyperbola gets closer
and closer to the asymptotes, in the direction towards which the branch extends.
The equations of the asymptotes can be determined by replacing 1 in the standard
equation by 0. The asymptotes can be drawn as the extended diagonals of the
auxiliary rectangle determined by the transverse and conjugate axes. Recall that,
for any point on the hyperbola, the absolute value of the difference of its distances
from the foci is 2a.
In the standard equation, aside from being positive, there are no other restrictions on a and b. In fact, a and b can even be equal. The orientation of the
hyperbola is determined by the variable appearing in the first term (the positive
term): the corresponding axis is where the two branches will open. For example,
if the variable in the first term is x, the hyperbola is “horizontal”: the transverse
axis is horizontal, and the branches open to the left and right in the direction of
the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
(y + 2)2 (x − 7)2
−
=1
25
9
(2) 4x2 − 5y 2 + 32x + 30y = 1
(1)
2
2
Solution.
(1) From
√ a = 25 and b = 9, we have a = 5, b = 3, and c =
√
a2 + b2 = 34 ≈ 5.8. The hyperbola is vertical. To determine the asymp2
(x−7)2
−
= 0, which is equivalent to y + 2 = ± 53 (x − 7).
totes, we write (y+2)
25
9
We can then solve this for y.
center: C(7, −2)
foci: F1 (7, −2 −
√
34) ≈ (7, −7.8) and F2 (7, −2 +
vertices: V1 (7, −7) and V2 (7, 3)
asymptotes: y = 53 x −
41
3
and y = − 35 x +
√
34) ≈ (7, 3.8)
29
3
The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.
55
(2) We first change the given equation to standard form.
4(x2 + 8x) − 5(y 2 − 6y) = 1
4(x2 + 8x + 16) − 5(y 2 − 6y + 9) = 1 + 4(16) − 5(9)
4(x + 4)2 − 5(y − 3)2 = 20
(x + 4)2 (y − 3)2
−
=1
5
4
√
√
We have a = 5 ≈ 2.2 and b = 2. Thus, c = a2 + b2 = 3. The hyperbola
2
2
is horizontal. To determine the asymptotes, we write (x+4)
− (y−3)
= 0
5
4
2
√
which is equivalent to y − 3 = ± 5 (x + 4), and solve for y.
center: C(−4, 3)
foci: F1 (−7, 3) and F2 (−1, 3)
√
√
vertices: V1 (−4 − 5, 3) ≈ (−6.2, 3) and V2 (−4 + 5, 3) ≈ (−1.8, 3)
asymptotes: y =
√2 x
5
+
√8
5
+ 3 and y = − √25 x −
√8
5
+3
The conjugate axis drawn has its endpoints b = 2 units above and below
the center.
56
Example 1.4.4. The foci of a hyperbola are (−5, −3) and (9, −3). For any point
on the hyperbola, the absolute value of the difference of its of its distances from
the foci is 10. Find the standard equation of the hyperbola.
Solution. The midpoint (2, −3) of the foci is the center of the hyperbola. Each
focus is c = 7 units away from the center. From the given difference, 2a = 10 so
a = 5. Also, b2 = c2 − a2 = 24. The hyperbola is horizontal (because the foci are
horizontally aligned), so the equation is
(x − 2)2 (y + 3)2
−
= 1.
25
24

Example 1.4.5.
√ A hyperbola has vertices (−4, −5) and (−4, 9), and one of its
foci is (−4, 2 − 65). Find its standard equation.
Solution. The midpoint (−4, 2) of the vertices is the center of the hyperbola,
which is vertical (because the vertices are vertically aligned).
Each vertex is
√
a = 7 units away from the center. The given focus is c = 65 units away from
the center. Thus, b2 = c2 − a2 = 16, and the standard equation is
(y − 2)2 (x + 4)2
−
= 1.
49
16
57

Seatwork/Homework 1.4.2
1. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with equation 9x2 − 4y 2 − 90x − 32y = −305. Sketch the graph, and
include these points and lines, along with the auxiliary rectangle.
√
√
Answer: center C(5, −4), foci F1 (5, −4−2 13) and F2 (5, −4+2 13), vertices
V1 (5, −10) and V2 (5, 2), asymptotes y = − 32 x + 27 and y = 32 x − 23
2
2. A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is (−2, 9). Find
(x − 7)2 (y − 9)2
−
=1
its standard equation.
Answer:
36
45
1.4.3. Situational Problems Involving Hyperbolas
We now give an example on an application of hyperbolas.
Example 1.4.6. An explosion is heard by two stations 1200 m apart, located at
F1 (−600, 0) and F2 (600, 0). If the explosion was heard in F1 two seconds before
it was heard in F2 , identify the possible locations of the explosion. Use 340 m/s
as the speed of sound.
Solution. Using the given speed of sound, we deduce that the sound traveled
340(2) = 680 m farther in reaching F2 than in reaching F1 . This is then the
difference of the distances of the explosion from the two stations. Thus, the
explosion is on a hyperbola with foci are F1 and F2 , on the branch closer to F1 .
58
We have c = 600 and 2a = 680, so a = 340 and b2 = c2 − a2 = 244400.
The explosion could therefore be anywhere on the left branch of the hyperbola
y2
x2
− 244400
= 1.

115600
Seatwork/Homework 1.4.3
 1. Two stations, located at M (−1.5, 0) and N (1.5, 0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of
0.33 km/s. If the signal from N was received by the ship four seconds before
the signal it received from M , find the equation of the curve containing the
y2
x2
possible location of the ship.
Answer: 0.4356
− 1.8144
= 1 (right branch)
Exercises 1.4
1. Give the coordinates of the center, foci, vertices, and the asymptotes of the
hyperbola with the given equation. Sketch the graph, and include these points
and lines.
x2
y2
−
=1
36 64
x2
y2
−
=1
(b)
25 16
(c) (x − 1)2 − y 2 = 4
(a)
(y + 2)2 (x + 3)2
−
=1
15
10
(e) 3x2 − 2y 2 − 42x − 16y = −67
(d)
(f) 25x2 − 39y 2 + 150x + 390y = −225
59
Answer:
Item
Center
Vertices
Foci
(a)
(0, 0)
(±6, 0)
(b)
(0, 0)
(0, ±5)
(c)
(1, 0)
(d)
(−3, −2)
(−1, 0), (3, 0)
√
(−3, −2 ± 15)
(±10, 0)
√
(0, ± 41)
√
(1 ± 2 2, 0)
(e)
(7, −4)
(3, −4), (11, −4)
(f)
(−3, 5)
(−3, 0), (−3, 10)
(−3, −7), (−3, 3)
√
(7 ± 2 10, −4)
(−3, −3), (−3, 13)
Item
Asymptotes
(a)
y = ± 43 x
(b)
y = ± 54 x
(c)
y = x − 1, y = −x + 1
%
%
y = ± 32 x ± 3 32 − 2
%
%
y = ± 32 x ∓ 7 32 − 4
(d)
(e)
(f)
y = ± √539 x ±
(a)
√15
39
+5
(b)
60
(c)
(d)
(e)
(f)
2. Find the standard equation of the hyperbola which satisfies the given conditions.
(a) foci (−4, −3) and (−4, 13), the absolute value of the difference of the
distances of any point from the foci is 14
(y − 5)2 (x + 4)2
−
=1
Answer:
49
15
(b) vertices (−2, 8) and (8, 8), a focus (12, 8)
(x − 3)2 (y − 8)2
Answer:
−
=1
25
56
(c) center (−6, 9), a vertex (−6, 15), conjugate axis of length 12
(y − 9)2 (x + 6)2
−
=1
Answer:
25
36
(d) asymptotes y = 43 x + 31 and y = − 43 x + 41
, a vertex (−1, 7)
3
(x − 5)2 (y − 7)2
−
=1
Answer:
36
64
Solution. The asymptotes intersect at (5, 7). This is the center. The
distance of the given vertex from the center is a = 6. This vertex and
center are aligned horizontally, so the hyperbola has equation of the form
2
(x−h)2
− (y−k)
= 1. The asymptotes consequently have the form y − k =
a2
b2
61
± ab (x − h), and thus, have slopes ± ab . From the given asymptotes,
Since a = 6, then b = 8. The standard equation is then
b
a
= 43 .
(x − 5)2 (y − 7)2
−
= 1.
36
64
and y = − 13 x + 73 , a focus (1, 12)
(y − 2)2 (x − 1)2
Answer:
−
=1
10
90
Solution. The asymptotes intersect at (1, 2). This is the center. The
distance of the given focus from the center is c = 10. This focus and
center are aligned vertically, so the hyperbola has equation of the form
2
(y−k)2
− (x−h)
= 1. The asymptotes consequently have the form y − k =
a2
b2
a
± b (x − h), and thus, have slopes ± ab . From the given asymptotes, ab = 31 ,
so b = 3a.
c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2
(e) asymptotes y = 13 x +
5
3
Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is
(y − 2)2 (x − 1)2
−
= 1.
10
90
3. Two control towers are located at points Q(−500, 0) and R(500, 0), on a
straight shore where the x-axis runs through (all distances are in meters).
At the same moment, both towers sent a radio signal to a ship out at sea, each
traveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)
before the message from R.
(a) Find the equation of the curve containing the possible location of the
x2
y2
ship.
Answer:
−
= 1 (left branch)
202500 47500
(b) Find the coordinates (rounded off to two decimal places) of the ship if it
is 200 m from the shore (y = 200).
Answer: (−610.76, 200)
Solution. Since the time delay between the two signals is 3 µs, then the difference between the distances traveled by the two signals is 300 · 3 = 900 m. The
ship is then on a hyperbola, consisting of points whose distances from Q and R
(the foci) differ by 2a = 900. With a = 450 and c = 500 (the distance of each
focus from the center, the origin), we have b2 = c2 − a2 = 5002 − 4502 = 47500.
y2
x2
Since a2 = 202500, the hyperbola then has equation 202500
− 47500
= 1. Since
the signal from Q was received first, the ship is closer to Q than R, so the
ship is on the left branch of this hyperbola. Using y = 200, we then solve
x2
2002
− 47500
= 1 for x < 0 (left branch), and we get x ≈ −610.76.
202500
4
62
Lesson 1.5. More Problems on Conic Sections
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) recognize the equation and important characteristics of the different types of
conic sections; and
(2) solve situational problems involving conic sections.
Lesson Outline
(1) Conic sections with associated equations in general form
(2) Problems involving characteristics of various conic sections
(3) Solving situational problems involving conic sections
Introduction
Inspecting the equation can lead us to the right conic section for its graph,
and set us on the right step towards analyzing it. We will also look at problems
that use the properties of the different conic sections, allowing us to synthesize
what has been covered so far.
1.5.1. Identifying the Conic Section by Inspection
The equation of a circle may be written in standard form
Ax2 + Ay 2 + Cx + Dy + E = 0,
that is, the coefficients of x2 and y 2 are the same. However, it does not follow
that if the coefficients of x2 and y 2 are the same, the graph is a circle.
(A)
(B)
General Equation
2x2 + 2y 2 − 2x + 6y + 5 = 0
x2 + y 2 − 6x − 8y + 50 = 0
Standard Equation


2
2
x − 21 + y + 32 = 0
(x − 3)2 + (y − 4)2 = −25
graph
point
empty set
For a circle with equation (x − h)2 + (y − k)2 = r2 , we have r2 > 0. This is
not the case for the standard equations of (A) and (B).
In (A), because the sum of two squares can only be 0 if and only if each square
3
1
is
 1 0, it3 follows that x − 2 = 0 and y + 2 = 0. The graph is thus the single point
, −2 .
2
In (B), no real values of x and y can make the nonnegative left side equal to
the negative right side. The graph is then the empty set.
63
Let us recall the general form of the equations of the other conic sections. We
may write the equations of conic sections we discussed in the general form
Ax2 + By 2 + Cx + Dy + E = 0.
Some terms may vanish, depending on the kind of conic section.
(1) Circle: both x2 and y 2 appear, and their coefficients are the same
Ax2 + Ay 2 + Cx + Dy + E = 0
Example: 18x2 + 18y 2 − 24x + 48y − 5 = 0
Degenerate cases: a point, and the empty set
(2) Parabola: exactly one of x2 or y 2 appears
Ax2 + Cx + Dy + E = 0 (D $= 0, opens upward or downward)
By 2 + Cx + Dy + E = 0 (C $= 0, opens to the right or left)
Examples: 3x2 − 12x + 2y + 26 = 0 (opens downward)
− 2y 2 + 3x + 12y − 15 = 0 (opens to the right)
(3) Ellipse: both x2 and y 2 appear, and their coefficients A and B have the same
sign and are unequal
Examples: 2x2 + 5y 2 + 8x − 10y − 7 = 0 (horizontal major axis)
4x2 + y 2 − 16x − 6y + 21 = 0 (vertical major axis)
If A = B, we will classify the conic as a circle, instead of an ellipse.
Degenerate cases: a point, and the empty set
(4) Hyperbola: both x2 and y 2 appear, and their coefficients A and B have different signs
Examples: 5x2 − 3y 2 − 20x − 18y − 22 = 0 (horizontal transverse axis)
− 4x2 + y 2 + 24x + 4y − 36 = 0 (vertical transverse axis)
Degenerate case: two intersecting lines
The following examples will show the possible degenerate conic (a point, two
intersecting lines, or the empty set) as the graph of an equation following a similar
pattern as the non-degenerate cases.
2
2
(1) 4x + 9y − 16x + 18y + 25 = 0
=⇒
=⇒
(2) 4x2 + 9y 2 − 16x + 18y + 61 = 0
=⇒
=⇒
64
(x − 2)2 (y + 1)2
+
=0
32
22
one point: (2, −1)
(x − 2)2 (y + 1)2
+
= −1
32
22
empty set
(3) 4x2 − 9y 2 − 16x − 18y + 7 = 0
=⇒
=⇒
(x − 2)2 (y + 1)2
−
=0
32
22
2
two lines: y + 1 = ± (x − 2)
3
A Note on Identifying a Conic Section
by Its General Equation
It is only after transforming a given general equation to standard
form that we can identify its graph either as one of the degenerate
conic sections (a point, two intersecting lines, or the empty set) or as
one of the non-degenerate conic sections (circle, parabola, ellipse, or
hyperbola).
Seatwork/Homework 1.5.1
The graphs of the following equations are (nondegenerate) conic sections. Identify
the conic section.
(1) 5x2 − 3y 2 + 10x − 12y = 22
Answer: hyperbola
(2) 2y 2 − 5x − 12y = 17
Answer: parabola
(4) 3x2 + 6x + 4y = 18
Answer: parabola
(3) 3x2 + 3y 2 + 42x − 12y = −154
Answer: circle
(5) 7x2 + 3y 2 − 14x + 12y = −14
2
Answer: ellipse
2
(6) −4x + 3y + 24x − 12y = 36
Answer: hyperbola
1.5.2. Problems Involving Different Conic Sections
The following examples require us to use the properties of different conic sections
at the same time.
Example 1.5.1. A circle has center at the focus of the parabola y 2 + 16x + 4y =
44, and is tangent to the directrix of this parabola. Find its standard equation.
Solution. The standard equation of the parabola is (y + 2)2 = −16(x − 3). Its
vertex is V (3, −2). Since 4c = 16 or c = 4, its focus is F (−1, −2) and its directrix
is x = 7. The circle has center at (−1, −2) and radius 8, which is the distance
from F to the directrix. Thus, the equation of the circle is
(x + 1)2 + (y + 2)2 = 64.

Example 1.5.2. The vertices and foci of 5x2 − 4y 2 + 50x + 16y + 29 = 0 are,
respectively, the foci and vertices of an ellipse. Find the standard equation of
this ellipse.
65
Solution. We first write the equation of the hyperbola in standard form:
(x + 5)2 (y − 2)2
−
= 1.
16
20
For this hyperbola, using the notations ah , bh , and ch to refer to a, b, and √
c of
the standard
equation of the hyperbola, respectively, we have ah = 4, bh = 2 5,
"
2
2
ch = ah + bh = 6, so we have the following points:
center: (−5, 2)
vertices: (−9, 2) and (−1, 2)
foci: (−11, 2) and (1, 2).
It means that, for the ellipse, we have these points:
center: (−5, 2)
vertices: (−11, 2) and (1, 2)
foci: (−9, 2) and (−1, 2).
In this case, ce = 4 and ae = 6, so that be =
equation of the ellipse is
"
√
a2e − c2e = 20. The standard
(x + 5)2 (y − 2)2
+
= 1.
36
20

Seatwork/Homework 1.5.2
1. Find the standard equation of all circles having center at a focus of 21x2 −
4y 2 + 84x − 24y = 36 and passing through the farther vertex.
Answer: (x + 7)2 + (y + 3)2 = 49, (x − 3)2 + (y + 3)2 = 49
2. Find the standard equation of the hyperbola one branch of which has focus and
vertex that are the same as those of x2 − 6x + 8y = 23, and whose conjugate
axis is on the directrix of the same parabola.
(y − 6)2 (x − 3)2
Answer:
−
=1
4
12
Exercises 1.5
1. The graphs of the following equations are non-degenerate conic sections. Identify the conic section.
(a) 5x2 + 7y 2 − 40x − 28y = −73
Answer: ellipse
(b) 5y 2 + 2x − 30y = −49
Answer: parabola
(c) 3x2 − 3y 2 + 12x − 12y = 5
Answer: hyperbola
(d) 3x2 + 3y 2 + 12x + 12y = 4
Answer: circle
66
(e) 2x2 + 24x − 5y = −57
Answer: parabola
2. The graphs of the following equations are degenerate conic sections. What are
the specific graphs?
(a) x2 + 3y 2 − 4x + 24y = −52
Answer: point: (2, −4)
Answer: lines: y + 2 = ± 23 (x + 1)
(b) 9x2 − 4y 2 + 18x − 16y = 7
(c) 3x2 + 5y 2 − 6x − 20y = −25
Answer: empty set
3. An ellipse has equation 25x2 + 16y 2 + 150x − 32y = 159. Find the standard
equations of all parabolas whose vertex is a focus of this ellipse and whose
focus is a vertex of this ellipse.
Answer: (x + 3)2 = −8(y + 2), (x + 3)2 = 32(y + 2), (x + 3)2 = −32(y − 4),
and (x + 3)2 = 8(y − 4)
Solution. The standard equation of the ellipse is
(x + 3)2 (y − 1)2
+
= 1.
16
25
Its center is (−3, 1). Since a = 5 and b = 4, we get c = 3, so the vertices are
P (−3, −4) and S(−3, 6), while its foci are Q(−3, −2) and R(−3, 4). We then
get four parabolas satisfying the conditions of the problem. The focal distance
indicated below is the distance from the vertex to the focus.
vertex
focus
focal distance
standard equation
Q(−3, −2)
P (−3, −4)
2
(x + 3)2 = −8(y + 2)
Q(−3, −2)
S(−3, 6)
8
(x + 3)2 = 32(y + 2)
R(−3, 4)
P (−3, −4)
8
(x + 3)2 = −32(y − 4)
R(−3, 4)
S(−3, 6)
2
(x + 3)2 = 8(y − 4)
4. Find the standard equation of the hyperbola whose conjugate axis is on the
directrix of the parabola y 2 + 12x + 6y = 39, having the focus of the parabola
as one of its foci, and the vertex of the parabola as one of its vertices.
(x − 7)2 (y + 3)2
−
=1
Answer:
9
27
Solution. The standard equation of the parabola is (y + 3)2 = −12(x − 4), so
its vertex is V (4, −3), and it opens to the left. With 4c = 12, or c = 3, its
focus is F (1, −3), and its directrix is x = 7. The hyperbola has its center on
67
x = 7, its conjugate axis, and a vertex at (4, −3). Its center is then C(7, −3).
The conjugate axis is vertical so the hyperbola is horizontal, with constants
ah = CV = 3 and ch = CF = 6, so b2h = c2h − a2h = 27. The standard equation
of the required hyperbola is
(x − 7)2 (y + 3)2
−
= 1.
9
27
5. Find the standard equation of the parabola opening to the left whose axis
contains the major axis of the ellipse x2 + 4y 2 − 10x − 24y + 45 = 0, whose
focus is the center of the ellipse, and which passes through the covertices of
this ellipse.
Answer: (y − 3)2 = −4(x − 6)
Solution. The standard form of the ellipse is
(x − 5)2 (y − 3)2
+
= 1.
16
4
Its center (5, 3) is the focus of the parabola. Since b = 2, its covertices are
W1 (5, 1) and W2 (5, 5). The vertex of the parabola, c units to the right of (5, 3),
is (5 + c, 3). Its equation can be written as (y − 3)2 = −4c(x − (5 + c)). Since
(5, 5) is a point on this parabola, we have (5 − 3)2 = −4c(5 − (5 + c)). Solving
this equation for c > 0 yields c = 1. Therefore, the standard equation of the
required parabola is (y − 3)2 = −4(x − 6).
6. Find the standard equation of the ellipse whose major and minor axes are the
transverse and conjugate axes (not necessarily in that order) of the hyperbola
(x − 2)2 (y + 3)2
+
=1
4x2 − 9y 2 − 16x − 54y = 29.
Answer:
9
4
Solution. The standard equation of the hyperbola is
(y + 3)2 (x − 2)2
−
= 1,
4
9
with center (2, −3), and constants ah = 2 and bh = 3. Since its conjugate axis
(which is horizontal and has length 2bh = 6) is longer than its transverse axis
(length 2ah = 4), the ellipse is horizontal. Its major axis has length 2ae = 6
and its minor axis has length 2be = 4, so ae = 3 and be = 2. The ellipse shares
the same center as the hyperbola. Thus, the standard equation of the required
ellipse is
(x − 2)2 (y + 3)2
+
= 1.
9
4
7. If m $= −3, 2, find the value(s) of m so that the graph of
(2m − 4)x2 + (m + 3)y 2 = (m + 3)(2m − 4)
is
68
(a) a circle,
(b) a horizontal ellipse,
(c) a vertical ellipse,
(d) a hyperbola (is it horizontal or vertical?), or
(e) the empty set.
Answer: (a) m = 7, (b) 2 < m < 7, (c) m > 7, (d) −3 < m < 2 (horizontal),
(e) m < −3
Solution. It might be helpful to observe that the equation is equivalent to
y2
x2
+
= 1.
m + 3 2m − 4
(a) The graph is a circle if m + 3 = 2m − 4 > 0 (positive, so the graph is not
a point or the empty set). This happens if m = 7.
(b) We require 0 < 2m − 4 < m + 3. Thus, 2 < m < 7.
(c) We require 0 < m + 3 < 2m − 4. Thus, m > 7.
(d) We need m + 3 and 2m − 4 to have different signs. We consider two cases.
i. If m + 3 < 0 < 2m − 4, then m < −3 AND m > 2, which cannot
happen.
ii. If 2m − 4 < 0 < m + 3, then −3 < m < 2. In this case, the equation
can be written, with positive denominators, as
y2
x2
−
= 1.
m + 3 4 − 2m
The hyperbola is horizontal.
(e) The remaining case is when m < −3. In this case, m + 3 < 0 and
2m − 4 < 0. This makes the expression
x2
y2
+
m + 3 2m − 4
negative, and never equal to 1. The graph is then the empty set.
4
69
Lesson 1.6. Systems of Nonlinear Equations
Time Frame: 4 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate systems of nonlinear equations;
(2) determine the solutions of systems of nonlinear equations using techniques
such as substitution, elimination, and graphing; and
(3) solve situational problems involving systems of nonlinear equations.
Lesson Outline
(1) Review systems of linear equations
(2) Solving a system involving one linear and one quadratic equation
(3) Solving a system involving two quadratic equations
(4) Applications of systems of nonlinear equations
Introduction
After recalling the techniques used in solving systems of linear equations in
Grade 8, we extend these methods to solving a system of equations to systems
in which the equations are not necessarily linear. In this lesson, the equations
are restricted to linear and quadratic types, although it is possible to adapt the
methodology so systems with other types of equations. We focus on quadratic
equations for two reasons: to include a graphical representation of the solution
and to ensure that either a solution is obtained or it is determined that there is
no solution. The latter is possible because of the quadratic formula.
Teaching Notes
1.6.1. Review of Techniques in Solving Systems of Linear
Equations
Recall that the
task of solving a
system of
equations is
equivalent to
finding points of
intersection.
Recall the methods we used to solve systems of linear equations. There were Teaching Notes
Systems of linear
three methods used: substitution, elimination, and graphical.
equations and
solving them were
Example 1.6.1. Use the substitution method to solve the system, and sketch introduced and
studied in Grade 8
the graphs in one Cartesian plane showing the point of intersection.
at the last part of

Quarter I.
 4x + y = 6
 5x + 3y = 4
Solution. Isolate the variable y in the first equation, and then substitute into the
second equation.
70
4x + y = 6
=⇒ y = 6 − 4x
5x + 3y = 4
5x + 3(6 − 4x) = 4
−7x + 18 = 4
x=2
y = 6 − 4(2) = −2
Example 1.6.2. Use the elimination method to solve the system, and sketch the
graphs in one Cartesian plane showing the point of intersection.

 2x + 7 = 3y
 4x + 7y = 12
Solution. We eliminate first the variable x. Rewrite the first equation wherein
only the constant term is on the right-hand side of the equation, then multiply
it by −2, and then add the resulting equation to the second equation.
2x − 3y = −7
(−2)(2x − 3y) = (−2)(−7)
−4x + 6y = 14
−4x + 6y = 14
4x + 7y = 12
13y = 26
y=2
1
x=−
2
Seatwork/Homework 1.6.1
Use either substitution or elimination method to solve the system, and sketch the
graphs in one Cartesian plane showing the point of intersection.

 x − 3y = 5
1.
 2x + 5y = −1
71
Answer: (2, −1)

 5x + 3y = 4
2.
 3x + 5y = 9

Answer: 12 , 23
1.6.2. Solving Systems of Equations Using Substitution
We begin our extension with a system involving one linear equation and one
quadratic equation. In this case, it is always possible to use substitution by
solving the linear equation for one of the variables.
Example 1.6.3. Solve the following system, and sketch the graphs in one Cartesian plane.

 x−y+2=0
 y − 1 = x2
72
Solution. We solve for y in terms of x in the first equation, and substitute this
expression to the second equation.
x−y+2=0
y − 1 = x2
(x + 2) − 1 = x2
x2 − x − 1 = 0
√
1± 5
x=
2
Solutions:
=⇒
y =x+2
√
√
√
1+ 5
5+ 5
1+ 5
x=
=⇒ y =
+2=
2√
2√
2√
1− 5
5− 5
1− 5
=⇒ y =
+2=
x=
2
2
2
)
)
√
√ *
√ *
√
1+ 5 5+ 5
1− 5 5− 5
,
,
and
2
2
2
2
The first equation represents a line with x-intercept −2 and y-intercept 2,
while the second equation represents a parabola with vertex at (0, 1) and which
opens upward.
Seatwork/Homework 1.6.2
Solve each system, and sketch the graphs in one Cartesian plane showing the
point(s) of intersection.

 x2 + y 2 = 16
1.
 x−y =4
Answer: (4, 0) and (0, −4)
Solution. Solving for x in the second equation, we get x = y + 4. Substitute
73
this expression into the first equation.
x2 + y 2 = 16 =⇒
(y + 4)2 + y 2 = 16
y 2 + 8y + 16 + y 2 = 16
2y 2 + 8y = 0
y = 0 or y = −4
y = 0 =⇒ x = 4
and
y = −4 =⇒ x = 0
Solutions: (4, 0) and (0, −4)

 y = x2
2.
 x = y2
Answer: (0, 0) and (1, 1)
Solution. Since the equations represent parabolas, we can use either of them
to isolate one variable. This is in fact the form in which both equations are
given. Substituting y = x2 into x = y 2 , we get
x = y 2 =⇒
x = (x2 )2
x4 − x = 0
x(x3 − 1) = 0
x = 0 or x = 1
74
Teaching Notes
We substitute each
value of y (0 and
−4) to the second
equation x − y = 4
(or x = y + 4).
x = 0 =⇒ y = 0
and
x = 1 =⇒ y = 1
Solutions: (0, 0) and (1, 1)
1.6.3. Solving Systems of Equations Using Elimination
Elimination method is also useful in systems of nonlinear equations. Sometimes,
some systems need both techniques (substitution and elimination) to solve them.
Example 1.6.4. Solve the following system:

 y 2 − 4x − 6y = 11
 4(3 − x) = (y − 3)2 .
Solution 1. We expand the second equation, and eliminate the variable x by
Teaching Notes adding the equations.
The variable y
could also be
4(3 − x) = (y − 3)2
eliminated first by
subtracting the
second equation
from the first.
=⇒ 12 − 4x = y 2 − 6y + 9 =⇒ y 2 + 4x − 6y = 3

 y 2 − 4x − 6y = 11
 y 2 + 4x − 6y = 3
Adding these equations, we get
2y 2 −12y = 14 =⇒ y 2 −6y−7 = 0 =⇒ (y−7)(y+1) = 0 =⇒ y = 7 or y = −1.
Teaching Notes
We may actually
substitute y = 7
and y = −1 (one at
a time) into any of
the two given
equations, and
then solve for x.
Solving for x in the second equation, we have
x=3−
y = 7 =⇒ x = −1
(y − 3)2
.
4
and
y = −1 =⇒ x = −1
Solutions: (−1, 7) and (−1, −1)
75

The graphs of the equations in the preceding example with the points of
intersection are shown below.
Sometimes the solution can be simplified by writing the equations in standard
form, although it is usually the general form which is more convenient to use in
solving systems of equations. Moreover, the standard form is best for graphing.
We solve again the previous example in a different way.
Solution 2. By completing the square, we can change the first equation into standard form:
y 2 − 4x − 6y = 11 =⇒ 4(x + 5) = (y − 3)2 .

 4(x + 5) = (y − 3)2
 4(3 − x) = (y − 3)2
Using substitution or the transitive property of equality, we get
4(x + 5) = 4(3 − x) =⇒ x = −1.
Substituting this value of x into the second equation, we have
4[3 − (−1)] = (y − 3)2 =⇒ 16 = (y − 3)2 =⇒ y = 7 or y = −1.
The solutions are (−1, 7) and (−1, −1), same as Solution 1.
Example 1.6.5. Solve the system and graph the curves:

 (x − 3)2 + (y − 5)2 = 10
 x2 + (y + 1)2 = 25.
76

Solution. Expanding both equations, we obtain

 x2 + y 2 − 6x − 10y + 24 = 0
 x2 + y 2 + 2y − 24 = 0.
Teaching Notes Subtracting these two equations, we get
Because the
equation
x + 2y − 8 = 0 is
−6x − 12y + 48 = 0 =⇒ x + 2y − 8 = 0
obtained by
combining the two
x = 8 − 2y.
equations (through
substraction), this
equation also
We can substitute x = 8 − 2y to either the first equation or the
contains the
solutions of the For convenience, we choose the second equation.
original system. In
fact, this is the line
x2 + y 2 + 2y − 24 = 0
passing through
the common points
(8 − 2y)2 + y 2 + 2y − 24 = 0
of the two circles.
2
second equation.
y − 6y + 8 = 0
y = 2 or y = 4
y = 2 =⇒ x = 8 − 2(2) = 4
and
The solutions are (4, 2) and (0, 4).
y = 4 =⇒ x = 8 − 2(4) = 0
√ The graphs of both equations are circles. One has center (3, 5) and radius
10, while the other has center (0, −1) and radius 5. The graphs with the points
of intersection are show below.
77
Seatwork/Homework 1.6.3
Solve the system, and graph the curves in one Cartesian plane showing the
point(s) of intersection.

 x2 + y 2 = 25
1.
2
2
 x +y =1
18 32
Answer: (3, 4), (−3, 4), (3, −4), and (−3, −4)

 x2 + 2y − 12 = 0
2.
 x2 + y 2 = 36
 √
 √
Answer: (0, 6), 2 5, −4 , and −2 5, −4
78

 (x − 1)2 + (y − 3)2 = 10
3.
 x2 + (y − 1)2 = 5
Answer: (−2, 2) and (2, 0)
1.6.4. Applications of Systems of Nonlinear Equations
As we expect, systems of equations are important in applications. In this session,
we consider some of them.
 Example 1.6.6. The screen size of television sets is given in inches. This
indicates the length of the diagonal. Screens of the same size can come in different
shapes. Wide-screen TV’s usually have screens with aspect ratio 16 : 9, indicating
the ratio of the width to the height. Older TV models often have aspect ratio
4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the
width of the screen.
Solution. Let w represent the width and h the height of the screen. Then, by
Pythagorean Theorem, we have the system

 w2 + h2 = 402 =⇒ w2 + h2 = 1600
 w = 16 =⇒ h = 9w
h
9
16
79
2
2
2
w + h = 1600 =⇒
w +
+
9w
16
,2
= 1600
337w2
= 1600
256
409 600
w=
≈ 34.86
337
h=
19(34.86)
19x
≈
= 19.61
16
16
Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and
19.61 inches high.

Seatwork/Homework 1.6.4
1. From a circular piece of metal sheet with diameter 20 cm, a rectangular
piece with perimeter 28 cm is to be cut as shown. Find the dimensions of
the rectangular piece.
Answer: 6 cm × 8 cm
Exercises 1.6
1. Solve the system, and graph the curves.

 y = 2x + 4
(a)
 y = 2x2
Answer: (−1, 2) and (2, 8)
80

 x2 + y 2 = 25
(b)
 2x − 3y = −6
 63 16
Answer: (3, 4) and − 13
, − 13

 x2 + y 2 = 12
(c)
 x2 − y 2 = 4
 √
 √
 √
 √
Answer: 2 2, 2 , −2 2, 2 , 2 2, −2 , and −2 2, −2
81

 x2 − 4y 2 = 200
(d)
 x + 2y = 100

Answer: 51, 49
2

 1 (x + 1)2 − (y + 2)2 = 1
4
(e)
 (y + 2)2 = − 1 (x − 1)
4
#
Answer: (1, −2), −4, −2 +
√
5
2
$
#
, and −4, −2 −
82
√
5
2
$
 2. A laptop has screen size 13 inches with aspect ratio 5 : 4. Find the length and
the width of the screen.
Answer: 10.15 in × 8.12 in
 3. What are the dimensions of a rectangle whose perimeter is 50 cm and diagonal
18 cm?
Answer: 14.9 cm × 10.1 cm
2
4. The graph of 2xy−y +5x+20 = 0 is a rotated hyperbola. Find the intersection
of this hyperbola with the graph of 3x + 2y = 3. (The graph is not required.)
 71 25
Answer: (−1, 3), 21
,− 7
5. For what values of a will the system

 x2 + y 2 + 2x − 1 = 0
have only one solution?
 x−y+a=0
Answer: a = −1 or a = 3
4
83
Unit 2
Mathematical Induction
https://commons.wikimedia.org/wiki/File%3ABatad rice terraces in Ifugao.jpg
By Ericmontalban (Own work)
[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],
via Wikimedia Commons
Listed as one of the UNESCO World Heritage sites since 1995, the two-millenniumold Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimony
of mankind’s creative engineering to adapt with physically-challenging environment in nature. One of the five clusters of terraces inscribed in the UNESCO list
is the majestic Batad terrace cluster (shown above), which is characterized by its
amphitheater-like semi-circular terraces with a village at its base.
Lesson 2.1. Review of Sequences and Series
Time Frame: 1 one-hour session
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) illustrate a series; and
(2) differentiate a series from a sequence.
Lesson Outline
(1) Sequences and series
(2) Different types of sequences and series (Fibonacci sequence, arithmetic and
geometric sequence and series, and harmonic series)
(3) Difference between sequence and series
Introduction
Pose the following problem to the class:
Jason’s classroom is on the second floor of the school. He can take
one or two steps of the stairs in one leap. In how many ways can
Jason climb the stairs if it has 16 steps?
Get students to suggest strategies they can use to solve this problem. Lead
or encourage them to try out smaller number of steps and find a pattern. Work
with the class to complete the following table (on the board):
85
Number of Steps
in the Stairs
Number of Ways
to Climb the Stairs
1
1
2
2
Teaching Notes
This is equivalent
3
3
to the number of
ways to express a
number (number of
4
5
steps in the stairs)
as a sum of 1’s and
5
8
2’s. For example,
we can write 3 as a
..
..
sum of 1’s and 2’s
.
.
in three ways:
2 + 1, 1 + 2, and
1 + 1 + 1. In 2 + 1,
The students should be able to recognize the Fibonacci sequence. Ask the it means Jason
leaps 2 steps first,
students to recall what Fibonacci sequences are and where they had encountered then 1 step to
finish the
this sequence before.
three-step stairs.
In this lesson, we will review the definitions and different types of sequences
and series.
Lesson Proper
Recall the following definitions:
A sequence is a function whose domain is the set of positive integers
or the set {1, 2, 3, . . . , n}.
A series represents the sum of the terms of a sequence.
If a sequence is finite, we will refer to the sum of the terms of the
sequence as the series associated with the sequence. If the sequence has
infinitely many terms, the sum is defined more precisely in calculus.
A sequence is a list of numbers (separated by commas), while a series is a
sum of numbers (separated by “+” or “−” sign). As an illustration, 1, − 12 , 13 , − 41
7
is its associated series.
is a sequence, and 1 − 12 + 31 − 41 = 12
The sequence with nth term an is usually denoted by {an }, and the associated
series is given by
S = a1 + a2 + a3 + · · · + an .
86
Example 2.1.1. Determine the first five terms of each defined sequence, and
give their associated series.
(1) {2 − n}
(3) {(−1)n }
(2) {1 + 2n + 3n2 }
(4) {1 + 2 + 3 + · · · + n}
Solution. We denote the nth term of a sequence by an , and S = a1 + a2 + a3 +
a4 + a5 .
(1) an = 2 − n
First five terms: a1 = 2 − 1 = 1, a2 = 2 − 2 = 0, a3 = −1, a4 = −2, a5 = −3
Associated series: S = a1 + a2 + a3 + a4 + a5 = 1 + 0 − 1 − 2 − 3 = −5
(2) an = 1 + 2n + 3n2
First five terms: a1 = 1 + 2 · 1 + 3 · 12 = 6, a2 = 17, a3 = 34, a4 = 57, a5 = 86
Associated series: S = 6 + 17 + 34 + 57 + 86 = 200
(3) an = (−1)n
First five terms: a1 = (−1)1 = −1, a2 = (−1)2 = 1, a3 = −1, a4 = 1,
a5 = −1
Associated series: S = −1 + 1 − 1 + 1 − 1 = −1
(4) an = 1 + 2 + 3 + · · · + n
First five terms: a1 = 1, a2 = 1+2 = 3, a3 = 1+2+3 = 6, a4 = 1+2+3+4 =
10, a5 = 1 + 2 + 3 + 4 + 5 = 15
Associated series: S = 1 + 3 + 6 + 10 + 15 = 35

An arithmetic sequence is a sequence in which each term after the first
is obtained by adding a constant (called the common difference) to the
preceding term.
If the nth term of an arithmetic sequence is an and the common difference is
d, then
an = a1 + (n − 1)d.
The associated arithmetic series with n terms is given by
Sn =
n(a1 + an )
n[2a1 + (n − 1)d]
=
.
2
2
87
A geometric sequence is a sequence in which each term after the first
is obtained by multiplying the preceding term by a constant (called
the common ratio).
If the nth term of a geometric sequence is an and the common ratio is r, then
an = a1 rn−1 .
The associated geometric series with n terms is given by


 na1
if r = 1
n
Sn =
a1 (1 − r )

if r =
$ 1.

(1 − r)
The proof of this sum formula is an example in Lesson 2.3.
When −1 < r < 1, the infinite geometric series
a1 + a1 r + a1 r2 + · · · + a1 rn−1 + · · ·
has a sum, and is given by
S=
a1
.
1−r
If {an } is an arithmetic sequence, then the sequence with nth term
bn = a1n is a harmonic sequence.
Seatwork/Homework
1. Write SEQ if the given item is a sequence, and write SER if it is a series.
(a) 1, 2, 4, 8, . . .
Answer: SEQ
(b) 2, 8, 10, 18, . . .
Answer: SEQ
(c) −1 + 1 − 1 + 1 − 1
Answer: SER
(d)
1 2 3 4
, , , ,...
2 3 4 5
2
Answer: SEQ
(e) 1 + 2 + 2 + 23 + 24
Answer: SER
(f) 1 + 0.1 + 0.001 + 0.0001
Answer: SER
2. Write A if the sequence is arithmetic, G if it is geometric, F if Fibonacci, and
O if it is not one of the mentioned types.
(a) 3, 5, 7, 9, 11, . . .
Answer: A
88
Teaching Notes
The proof of the
fact that the
infinite geometric
series
a1 + a1 r + · · · has
a sum when |r| < 1
is beyond the scope
of Precalculus, and
can be found in
university
Calculus.
(b) 2, 4, 9, 16, 25, . . .
(c)
(d)
(e)
1
,
4
1
,
3
1
,
5
Answer: O
1 1
, , 1 ,...
16 64 256
2 3 4
, , ,...
9 27 81
1 1 1 1
, , , ,...
9 13 17 21
Answer: G
Answer: O
Answer: A
(f) 4, 6, 10, 16, 26, . . .
√ √ √ √
(g) 3, 4, 5, 6, . . .
Answer: F
Answer: O
(h) 0.1, 0.01, 0.001, 0.0001, . . .
Answer: G
3. Determine the first five terms of each defined sequence, and give their associated series.
(a) {1 + n − n2 }
Answer: a1 = 1, a2 = −1, a3 = −5, a4 = −11, a5 = −19
Associated series: 1 − 1 − 5 − 11 − 19 = −35
(b) {1 − (−1)n+1 }
Answer: a1 = 0, a2 = 2, a3 = 0, a4 = 2, a5 = 0
Associated series: 0 + 2 + 0 + 2 + 0 = 4
(c) a1 = 3 and an = 2an−1 + 3 for n ≥ 2
Answer: a1 = 3, a2 = 9, a3 = 21, a4 = 45, a5 = 93
Associated series: 1 − 1 − 5 − 11 − 19 = −35
(d) {1 · 2 · 3 · · · n}
Answer: a1 = 1, a2 = 1 · 2 = 2, a3 = 1 · 2 · 3 = 6, a4 = 24, a5 = 120
Associated series: 1 + 2 + 6 + 24 + 120 = 153
4. Identify the series (and write NAGIG if it is not arithmetic, geometric, and
infinite geometric series), and determine the sum (and write NO SUM if it
cannot be summed up).
(a) 4 + 9 + 14 + · · · + 64
(b) 81 + 27 + 9 + · · · +
Answer: Arithmetic, 442
1
81
Answer: Geometric,
(c) 1 + 3 + 6 + 10 + 15 + 21 + · · · + 55
(d) −10 − 2 + 6 + · · · + 46
1
2
Answer: NAGIG, 220
Answer: Arithmetic, 144
(e) 10 + 2 + 0.4 + 0.08 + · · ·
(f)
9841
81
Answer: Infinite geometric, 12.5
+ 13 + 51 + 71 + · · ·
Answer: NAGIG, NO SUM
(g) 1 − 0.1 + 0.01 − 0.001 + · · ·
Answer: Infinite geometric,
10
11
4
89
Lesson 2.2. Sigma Notation
Time Frame: 2 one-hour sessions
Learning Outcomes of the Lesson
At the end of the lesson, the student is able to use the sigma notation to
represent a series.
Lesson Outline
(1) Definition of and writing in sigma notation
(2) Evaluate sums written in sigma notation
(3) Properties of sigma notation
(4) Calculating sums using the properties of sigma notation
Introduction
The sigma notation is a shorthand for writing sums. In this lesson, we will
see the power of this notation in computing sums of numbers as well as algebraic
expressions.
2.2.1. Writing and Evaluating Sums in Sigma Notation
Mathematicians use the sigma notation to denote a sum. The uppercase Greek
letter Σ (sigma) is used to indicate a “sum.” The notation consists of several
components or parts.
Let f (i) be an expression involving an integer i. The expression
f (m) + f (m + 1) + f (m + 2) + · · · + f (n)
can be compactly written in sigma notation, and we write it as
n
/
f (i),
i=m
which is read “the summation of f (i) from i = m to n.” Here, m
and n are integers with m ≤ n, f (i) is a term (or summand ) of the
summation, and the letter i is the index, m the lower bound, and n
the upper bound.
90
Teaching Notes
Emphasize that
the value of i starts
at m, increases by
1, and ends at n.
Example 2.2.1. Expand each summation, and simplify if possible.
n
4
/
/
(3)
ai
(2i + 3)
(1)
i=1
i=2
(2)
5
/
2
√
6
/
n
(4)
n+1
n=1
i
i=0
Solution. We apply the definition of sigma notation.
(1)
4
/
(2i + 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27
i=2
(2)
5
/
2i = 20 + 21 + 22 + 23 + 24 + 25 = 63
i=0
(3)
n
/
ai = a1 + a2 + a3 + · · · + an
i=1
√
√
√
√
√
6
/
n
2
3 2
5
6
1
(4)
= +
+
+ +
+
n+1
2
3
4
5
6
7
n=1
Example 2.2.2. Write each expression in sigma notation.
1
1 1 1
+ + + ··· +
2 3 4
100
(2) −1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − 9 + · · · − 25
(1) 1 +
(3) a2 + a4 + a6 + a8 + · · · + a20
(4) 1 +
1 1 1
1
1
1
1
+ + +
+
+
+
2 4 8 16 32 64 128
100
Solution. (1) 1 +
/1
1 1 1
1
+ + + ··· +
=
2 3 4
100 n=1 n
(2) −1 + 2 − 3 + 4 − 5 + · · · − 25
= (−1)1 1 + (−1)2 2 + (−1)3 3 + (−1)4 4
+ (−1)5 5 + · · · + (−1)25 25
=
25
/
(−1)j j
j=1
91

(3) a2 + a4 + a6 + a8 + · · · + a20
= a2(1) + a2(2) + a2(3) + a2(4) + · · · + a2(10)
=
10
/
a2i
i=1
7
/ 1
1
1
1
1
1 1 1
+
+
+
=
(4) 1 + + + +
2 4 8 16 32 64 128 k=0 2k

The sigma notation of a sum expression is not necessarily unique. For example, the last item in the preceding example can also be expressed in sigma
notation as follows:
8
/ 1
1 1 1
1
1
1
1
+
+
+
=
.
1+ + + +
2 4 8 16 32 64 128 k=1 2k−1
However, this last sigma notation is equivalent to the one given in the example.
Seatwork/Homework 2.2.1
1. Expand each summation, and simplify if possible.
(a)
5
/
k=−1
(b)
n
/
(2 − 3k)
Answer: −28
xj
Answer: x + x2 + x3 + · · · + xn
j=1
(c)
6
/
j=3
(d)
4
/
(j 2 − j)
Answer: 68
(−1)k+1 k
Answer: −2
k=1
(e)
3
/
n=1
(an+1 − an )
Answer: a4 − a1
2. Write each expression in sigma notation.
2
3
4
(a) x + 2x + 3x + 4x + 5x
5
Answer:
5
/
kxk
k=1
(b) 1 − 2 + 3 − 4 + 5 − 6 + · · · − 10
92
Answer:
10
/
k=1
(−1)k+1 k
Teaching Notes
Equivalent answer:
1+3+5+· · ·+101 =
51

(2k − 1)
k=1
(c) 1 + 3 + 5 + 7 + · · · + 101
Answer:
50
/
(2k + 1)
k=0
(d) a4 + a8 + a12 + a16
Answer:
4
/
a4k
k=1
(e) 1 −
1 1 1 1
+ − +
3 5 7 9
Answer:
4
/
k=0
(−1)k
2k + 1
2.2.2. Properties of Sigma Notation
We start with finding a formula for the sum of
n
/
i = 1 + 2 + 3 + ··· + n
i=1
in terms of n.
The sum can be evaluated in different ways. A simple, though informal,
approach is pictorial.
Teaching Notes
This illustration
can be done with
manipulatives, and
allow the students
to guess.
n
/
i = 1 + 2 + 3 + ··· + n =
i=1
n(n + 1)
2
Another way is to use the formula for an arithmetic series with a1 = 1 and
an = n:
n(a1 + an )
n(n + 1)
S=
=
.
2
2
We now derive some useful summation facts. They are based on the axioms
of arithmetic addition and multiplication.
93
n
/
cf (i) = c
i=m
n
/
f (i),
c any real number.
i=m
Proof.
n
/
Teaching Notes
Some proofs can be
skipped. However,
it is helpful if they
are all discussed in
class.
cf (i) = cf (m) + cf (m + 1) + cf (m + 2) + · · · + cf (n)
i=m
= c[f (m) + f (m + 1) + · · · + f (n)]
n
/
=c
f (i)

i=m
n
/
[f (i) + g(i)] =
n
/
f (i) +
i=m
i=m
n
/
g(i)
i=m
Proof.
n
/
[f (i) + g(i)]
i=m
= [f (m) + g(m)] + · · · + [f (n) + g(n)]
= [f (m) + · · · + f (n)] + [g(m) + · · · + g(n)]
n
n
/
/
=
f (i) +
g(i)
i=m

i=m
n
/
i=m
c = c(n − m + 1)
Proof.
n
/
i=m
c = c0 + c + c12+ · · · + 3c
n−m+1 terms
= c(n − m + 1)
94