#MURASHUI
0742296590
KENYATTA UNIVERSITY
DIGITAL SCHOOL OF VIRTUAL AND OPEN LEARNING
IN COLLABORATION WITH
SCHOOL OF PURE AND APPLIED SCIENCES
DEPARTMENT OF MATHEMATICS AND ACTUARIAL SCIENCE,
SMA 103: ANALYTICAL GEOMETRY
WRITTEN BY:
VETTED BY:
Dr. Ambrose Wahome
Mr. Augustine Ruto
1
INTRODUCTION
Welcome to this module. The module deals with different geometries and their applications.
This is an interactive instructional module that uses both action and collaborative learning styles
that provide you with diverse online learning experiences and effective learning processes. The
key purpose of this module is to expose you albeit theoretically to the operations carried out on
various geometries. This should hopefully equip you with the necessary knowledge and skills
crucial in real life applications of geometry.
Analytical geometry is the study of geometry by application of algebraic methods through the
coordinate system. This involves changing a geometric problem into an algebraic problem so
that it obeys certain rules of algebra. This module offers elementary techniques for solving
geometrical problems analytically for undergraduates in mathematics and engineering.
In this module, we shall analyse different geometries and their applications.
We hope that you will find this module exciting, educative, and engaging.
2
COURSE CONTENT
WEEK
TOPIC
WEEK 1
STRAIGHT LINE
WEEK 2
CIRCLE
WEEK 3
POLAR COORDINATES
WEEK 4
PARABOLA
WEEK 5
WEEK 6
PARABOLA
ELLIPSE
WEEK 7
ELLIPSE
WEEK 8
HYPERBOLA
WEEK 9
HYPERBOLA
WEEK 10
TRIGONOMETRY
WEEK 11
TRIGONOMETRY
WEEK 12
TRIGONOMETRY
WEEK 13&14
EXAMINATION
3
OVERVIEW OF THE COURSE
Week 1: Straight Line
In this lesson, the Cartesian coordinate system is introduced. The distance between points and
points which divide a line segment internally and externally are covered. The general, point
slope and normal form of the equation of the straight line are covered. Also covered is the
relationship between lines.
Week 2: The Circle
In this lesson we study the circle and relationships between circles. A circle is the locus of a
point which moves such that its distance from a fixed point is constant. The fixed point is called
the center of the circle. The distance from the point is called the radius.
Week 3: Polar Coordinates
In this lesson, we study the relationship between Cartesian and polar coordinates.
Week 4: Parabola
In this lesson we study how to determine the line of symmetry, focus, vertex, directrix and length
of latus rectum of a parabola whose center is at the origin and whose center is not at the origin.
Week 5: Parabola
In this lesson we study how to determine the equations of tangent and normal lines to a
parabola.
Week 6: Ellipse
In this lesson we study how to determine the foci, vertices, directrices, lengths of minor/major
axes and lengths of latus rectum of an ellipse whose center is at the origin.
Week 7: Ellipse – Tangent and Normal Tangent and Normal
In this lesson we study how to determine the equations of tangent and normal lines to an
ellipse.
4
Week 8: Hyperbola
In this lesson we study how to determine the foci, vertices, directrices, equations of asymptotes,
lengths of transverse axes and lengths of latus rectum of a hyperbola whose center is at the
origin.
Week 9: Hyperbola – Tangent and Normal Tangent and Normal
In this lesson we study how to determine the equations of tangent and normal lines to a
hyperbola.
Week 10: Trigonometry
In this lesson we study radian measure, trigonometric ratios and trigonometric identities.
Week 11: Trigonometry
In this lesson we study double-angle and factor formulae.
Week 12: Trigonometry
In this lesson we study inverses of angles, t-formula and plotting of trigonometric graphs.
Week 13 & 14: Examination
These two weeks bring together the work you have been doing to an end. This course unit will
be examined and will partially contribute to the award of the degree in the programme that you
are undertaking. The university examinations regulations will apply.
MODULE LEARNING OUTCOMES
By the end of this module, you will be able to:
1. Determine the distance between two points, the mid-point, angle between two lines and the
equation of a line.
2. Determine the equation of a circle and the equation of a tangent from a point to the circle.
3. Convert Cartesian co-ordinates to polar co-ordinates and vice versa.
4. Determine the foci, vertices, latus rectum, directrices and equations of the parabola, ellipse
and hyperbola.
5. Solve trigonometric equations.
COURSE DESCRIPTION
This is a core unit meant for all students who are taking Bachelor of Science, Bachelor of
Education and Bachelor of Economics degrees. The rationale of offering the course is to equip
students with knowledge and skills in Analytical Geometry and how they can apply these skills
in solving problems. You will therefore be required to set aside about 5 hours per lesson to
complete this course successfully.
COURSE REQUIREMENTS
5
This is a blended learning course that will utilize the flex model. This means that learning
materials and instructions will be given online and the lessons will be self-guided with the
lecturer being available briefly for face to face sessions and online support. Your lecturer will be
meeting you face to face to introduce a lesson and put it into perspective and you will actively
participate in your search for knowledge by undertaking several online activities. This means
that some of the 39 instructional hours of the course will be delivered face to face while other
lessons will be taught online through various learner and lecturer activities. It is important for
you to note that one instructional hour is equivalent to two online hours. Three instructional
hours will be needed per week. Out of these, one will be used for face to face contact with your
lecturer (also referred as e-moderator in the online activities) while the other two instructional
hours (translating to four online hours) will be used for online activities otherwise referred to as
e-tivities in the lessons. This will add up to the 5 hours requirement per lesson earlier
mentioned. There are 27 online activities each taking at least two hours and totaling to 54
online hours. You are advised to follow the topic flow-chart given so that you cover at least a
lesson every week.
You will be required to participate and interact online with your peers and the e-moderator who
in this case is your lecturer. Guidelines for the online activities (which we shall keep referring to
as e-tivities) will be provided whenever there is an e-tivity. Please note that since the online etivities are part of the learning process, they may be graded at the discretion of your e-moderator.
Such grading will however be communicated in the e-tivity guidelines and feedback given as
soon as possible after the e-tivity. The e-tivities will include but will not be limited to online
assessment quizzes, assignments and discussions. There are also assessment questions that you
can attempt at the end of every lesson to test your understanding of the lesson. The answers to all
the assessment questions are at the end of the module after lesson 11. All the resource that have
been used in this module in form of books are available under the resources section after the
answers to the questions.
ASSESSMENT
It is important to note that the module has embedded certain learner formative assessment
feedback tools that will enable you gauge your own learning progress. The tools include online
collaborative discussions forums that focus on team learning and personal mastery and will
therefore provide you with peer feedback, lecturer assessment and self- reflection. You will also
be required to do one major assignment/project that is meant to assess the application of the
skills and knowledge gained during the course. The project score in combination with scores for
e-tivities (where graded) will account for 30% of your final examination score with the
remaining 70% coming from a face to face sit-in final written examination that will be guided by
your university examination policy and procedures.
We wish you the very best of experiences in this course.
6
TABLE OF CONTENTS
Introduction
2
Overview of Course
4
Module learning Outcomes
.
5
Course Description
..
5
Course Requirements
..
5
Course Assessment
.
6
Lesson 1: The Straight Line..
..
. 9
1.2.1 Properties of The Straight Line
.9
1.2.1.1 The Distance between Two Points
. 10
1.2.1.2 The Mid-point of a Line
. 10
1.2.1.3 Inclination and Slope of a Line
.. 10
1.2.1.4 Parallel and Perpendicular Lines
....11
1.2.1.5 The Angle between Two Lines
. .12
1.2.1.6 The Equation of a Straight Line
.13
Lesson 2: The Circle.
...…
16
2.2.1 Equation of a Circle
.
.16
2.2.1.1 Standard Equation of a Circle
... 16
2.2.1.2 General Equation of a Circle
.16
2.2.1.3 Distance from a point to a Circle
...17
2.2.2 Equation of the Tangent and Normal
... 20
Lesson 3: Polar Coordinates.
..
24
3.2.1 Converting Cartesian Equations to Polar Equations
... 24
3.2.1.1 Relationship between Polar and Rectangular Coordinates
25
Lesson 4: The Parabola
.
.
29
4.2.1 Standard Equation of a Parabola
..29
4.2.1.1 General Equation of a Parabola
.31
Lesson 5: The Parabola: Tangent and Normal.
34
5.2.1 Tangent and Normal to a Parabola
...34
5.2.1.1 Normal to a Parabola
.36
Lesson 6: The Ellipse
..
40
6.2.1 Equation of an Ellipse
..40
6.2.1.1 General Equation of an Ellipse
... 42
6.2.1.2 Length of the Latus Rectum
..44
Lesson 7: The Ellipse: Tangent and Normal.
47
7.2.1 Tangent and Normal to an Ellipse
47
7.2.1.1 Normal to an Ellipse
..48
Lesson 8: The Hyperbola...
.
51
8.2.1 Equation of a Hyperbola
...51
8.2.1.1 The Asymptotes of a Hyperbola
53
8.2.1.2 Hyperbola with Center not at the origin
54
Lesson 9: The Hyperbola: Tangent and Normal..
.
56
9.2.1 Tangent and Normal to a Hyperbola
.
56
Lesson 10: Trigonometry: Identities...
.
60
10.2.1 Trigonometric Ratios
...60
7
10.2.1.1 Trigonometric Identities
10.2.1.2 Ptolemy’s Identities …………………………………………..60s Identities
Lesson 11: Trigonometry: Factor Formulae
...
11.2.1 Factor Formulae
11.2.1.1 Identities for Negative Angles
11.2.1.2 Double-angle Formulae for Sine and Cosine
11.2.1.3 Factor Formulae
Solutions to Assessment Questions
..
8
...60
..60
64
..64
.64
... 64
64
68
LESSON 1
THE STRAIGHT LINE
1.1 Introduction
In this chapter, the Cartesian coordinate system is introduced. The distance between points,
mid-points, perpendicular/parallel lines and angles between two lines are covered. The general
and point slope form of the equation of the straight line are also covered.
1.2 Lesson Learning Outcomes
By the end of this lesson, you should be able to;
1.2.1 Carry out operations on the Straight Line.
1.2.1
The Straight Line
In this work the geometrical problems considered are those that are presented on two
dimensional plane only and it is therefore important at this point to discuss the plane.
Consider any plane with a point denoted by O, which we call the point of reference or the
origin. Let line X be horizontal and pass through O and line Y be vertical and pass through O.
The horizontal line X is referred to as the x-axis and the vertical line Y is the y-axis. Distance
measured from the y-axis is referred to as an abscissa and it is positive if measured to the right
and negative if measured to the left. A vertical distance from the x-axis is called a y-ordinate
and it is positive if measured above the x-axis and negative if measured below the x-axis. Any
point on the plane can described by its abscissa and its ordinate as P(x, y), for any point P. The
symbol (x, y) represents the coordinates of the point, where x is the abscissa and y the ordinate
and it is also referred to as the Cartesian coordinates of P.
The two axis partitions the plane into four quadrants as shown below. In the first quadrant
both x and y are positive. In the second quadrant x is negative while y is positive. In the third
quadrant both x and y are negative. In the fourth quadrant x is positive and y negative. The
coordinates of the origin are (0, 0).
Quad II
Y
Quad I
x
P(x,y)
(-,+)
(+,+)
y
O
X
(-,-)
Quad III
Quad IV (+,-)
9
Figure 1.1
1.2.1.1 The Distance Between Two Points
The distance between two points
and
Example 1.1
Find the distance between the points
on a straight line is given by;
and
.
Solution
Figure 1.2
The distance between the points is;
1.2.1.2 The Mid-point of a Line
The co-ordinates of the mid-point of a line between two end points
. In figure 1, the mid-point of the straight line joining
and
and
is;
is
given by;
1.2.1.3 Inclination and Slope of the Line
In figure 1.3, the inclination of the line L is the angle measured positively from the positive axis
to the line L.
10
L
Figure 1.3
The slope of the line L is the tangent of the angle of inclination. If m is the slope of L, then
Suppose that
and
are points on the line L in figure 1.4;
Figure 1.4
Considering the right angle triangle, then we have;
Example 1.2
Find the slope and the angle of inclination of the line passing through points
and
.
Solution
1.2.1.4 Parallel and Perpendicular Lines
If two lines
and
are parallel, then their slopes are equal;
11
Figure 1.5
Next, suppose that
and
are perpendicular
Figure 1.6
Given that the slope of is
and the slope of
is
, then the product of their slopes is -1;
1.2.1.5 The Angle between Two Lines
Let that
and
meet at angle . Further, let the slope of
Figure 1.7
12
be
and the slope of
be
Let
be the angle of inclination of
Hence
and
be the angle of inclination of
. Then
but
Example 1.3
The angle between two lines is
. If the slope
of
is , determine the slope of
Solution
1.2.1.6 The Equation of a Straight Line
Suppose that
and
are points on a line.
Figure 1.8
Then the slope of the line is;
Taking the general point
Where
, we get the equation of the line to be;
is the y-intercept
Example 1.4
Find the equation of a line whose
and
intercepts are 5 and 3 respectively.
13
.
Solution
The points are
and
Taking a general point
. Therefore the slope is
we have the equation of the line as;
E-tivity 1.2.1 The Straight Line
Numbering, pacing and sequencing
Lesson 1.2.1
14
Title
The Straight Line
Purpose
To introduce you to operations on the straight line.
Brief summary of overall task
Watch the videos on the straight line
https://www.youtube.com/watch?v=EoI6uig5Lps
https://www.youtube.com/watch?v=DrsCIoLtiBI
and
https://www.youtube.com/watch?v=BoUuEmYy8_k
Spark
Individual contribution


Interaction begins



E-moderator interventions
Schedule and time
Next
Watch the videos on the straight line.
Carry out the various operations on the straight
line.
Post your answers on the discussion forum 1.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind.
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 1st week of
the semester.
The Circle
1.3 Assessment
a) Find the distance between the points A(7,9) and B (4,5) .
b) Determine whether the points (  7,65) and (5,23) lie on the line through point (3, 5)
whose slope is 9.
c) Find the equation of the line with intercept
on -axis and intercept
1.4 References
15
on -axis.
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
https://byjus.com/jee/straight-lines/
http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-strtlines-2009-1.pdf
16
LESSON 2
THE CIRCLE
2.1 Introduction
In this chapter we study the circle and relationships between circles. A circle is the locus of a
point which moves such that its distance from a fixed point is constant. The fixed point is called
the center of the circle and the distance from the point is called the radius.
2.2 Lesson Learning Outcomes
By the of this lesson, you should be able to;
2.2.1 Determine the standard and the general equation of a circle given its:(i)
Radius and the center.
(ii)
Any three points the circle passes through.
2.2.2 Find the tangents and the normal to a circle.
2.2.1
Equation of a Circle
2.2.1.1 Standard Equation of Circle
Consider a circle with center C ( h, k ) and radius r (Fig 2.1)
y
P ( x, y )
r
x
Figure 2.1
From the definition of the circle
___
CP  r  ( x  h) 2  ( y  k ) 2 .
Hence tha equation of the circle is
( x  h) 2  ( y  k ) 2  r 2 .
If the center of the circle is at the origin the equation is
x2  y2  r 2 .
2.2.1.2 General Equation of a circle
Expanding the equation ( x  h) 2  ( y  k ) 2  r 2 we get
x 2  y 2  2hx  2ky  h 2  k 2  r 2  0 .
17
This can be written as
x 2  y 2  Dx  Ey  C  0
where D  2h, E  2k
and C  h 2  k 2  r 2 . Equation () is called the general equation of the
circle.
If the equation of a circle is given in general form the center and the radius can be obtained by
completing the squares as
(x 
D 2
E
D 2  E 2  4C
)  ( y  )2 
.
2
2
4
This is the equation of a circle with center C (
r1
D E
,
) and radius
2
2
D 2  E 2  4C .
2
Note the following
(a) If D 2  E 2  C  0 , then the equation represents a point which coincides with the center
D E
,
).
2
2
(b) If D 2  E 2  C  0 , then the radius is imaginary. The circle is referred to as a virtual circle.
(c) If D 2  E 2  C  0 , then the circle is real.
C(
Example 2.1
Find the center and radius of the circle x 2  y 2  4 x  10 y  2  0 .
Solution
Center
Radius
2.2.1.3 Distance from a point to a circle
Consider point P1 ( a1 , b1 ) outside the circle ( x  h) 2  ( y  k ) 2  r 2 . Let B1 be a point on the
circle such that it lies on the line segment P1C , C being the center of the circle.
P1 (a1 , b1 )
y
B2
B1
P2
C
x
Figure 3.2
____
The distance from P1 (a1 , b1 ) to the circle is P1 B1 which is obtained as
18
____
____
____
P1 B1  P1C  B1C
= (a1  h) 2  (b1  k ) 2  r
Now consider point another P2 ( a 2 , b2 ) inside the circle lying on the line segment CB2 . The
distance from P2 ( a 2 , b2 ) to the circle is
____
P2 B2  r  (a 2  h) 2  (b2  k ) 2 .
In general, the distance d from point P (a, b) to circle ( x  h) 2  ( y  k ) 2  r 2 is
d 
(a  h) 2  (b  k ) 2  r
with d positive if the point is outside the circle and negative if inside the circle.
Example 2.2
Find the distance from the point P (5,8) to the circle x 2  y 2  2 x  6 y  3  0 .
Solution
The center of the circle is C ( 1,3) . The radius of the circle is
r1
D 2  E 2  4C
2
1

16  36  12
2
 7
Distance
E-tivity 2.2.1 – Tangent and Normal Equation of a Circle
Numbering, pacing and sequencing
Lesson 2.2.1
19
Title
Equation of a Circle
Purpose
To determine the standard and general equation of a circle.
Brief summary of overall task
Watch
the
video
on
the
https://www.youtube.com/watch?v=I7MBrjPEyps
https://www.youtube.com/watch?v=x4E1qOTy9zw
Spark
( x  h) 2  ( y  k ) 2  r 2
circle
Individual contribution

Determine how to obtain the equation of a circle.
Interaction begins



Post your answers on the discussion forum 2.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of your
colleagues have posted keeping etiquette in mind




E-moderator interventions
Focused group discussion
Providing feedback/ teaching points
Summarising key points
Closing the discussion
This discussion should be done during the 2 nd week of the
semester.
Equations of tangent and normal
Schedule and time
Next
2.2.2 Equations of the tangent and normal
D1 ( m1 , n1 )
20
P1 ( x1 , y1 )
C(h,k
)
D2 (m 2 , n 2 )
Remark
The chord through D1 and D2 is called the chord of contact of tangents from the point
P1 ( x1 , y1 ) with respect to the given circle.
Example 2.3
Find the tangent points of the circle x 2  y 2  5 from the point (8, 3)
Solution
Let ( m, n) be the points on the circle and on the tangent
Then
mx  ny  5
 8m  3n  5
put
n
5  8m
3
then
m2 
and
(5  8m) 2
5
9
or
9m 2  64m 2  80m  25  45
or
73m 2  80m  20  0
m
If m  1.33, n 
80  80 2  80  73
 1.33 or – Tangent and Normal0.21
146
5  8  1.33
 1.88
3
If m  0.21, n 
5  8  0.21
 2.23
3
Hence the two points on the circle are
D1 (1.33,1.88)
and
D2 ( 0.21,2.23)
21
m2  n2  5
E-tivity 2.2.2 – Tangent and Normal Equation of Tangent and Normal
Numbering, pacing and sequencing
Lesson 2.2.2
22
Title
Equation of Tangent and Normal
Purpose
Brief summary of overall task
To determine the equations of tangent and normal to a
circle
Watch the video on the circle
https://www.youtube.com/watch?v=6l5Bg8rYTq4
https://www.youtube.com/watch?v=xCXCp1WzCYE
Spark
( x  h) 2  ( y  k ) 2  r 2
Individual contribution

Interaction begins



E-moderator interventions
Schedule and time
Next
How do you determine the length of a tangent
from an external point?
Post your answers on the discussion forum 2.2.2
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 2nd week of the
semester.
Polar Coordinates
2.3 Assessment
a) Find the equation of the circle whose centre and radius are as given.
(i) Center (0, 0),
radius r  8
(ii) Center (-3, 4),
radius r  2
b) Find the center and the radius of the circle given by the equation x 2  y 2  12 x  6 y  7  0
c) Find the equation of the circle through the given points.
(1,2) (3,7) (2,-3)
d) Find the equation of the tangents to the given circle from the given external point.
x 2  y 2  10  0
(10, 2)
2.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
23
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
24
LESSON 3
POLAR COORDINATES
3.1 Introduction
In this chapter we introduce the polar coordinate system. In the system, a fixed point is described
by its distance from a fixed point and its direction from a fixed line through the point. The fixed
point is called the pole and the fixed line is called the polar axis.
3.2 Lesson Learning Outcomes
By the end of this lesson, you should be able to;
3.2.1 Convert Cartesian to polar coordinates and vice versa.
3.2.1 Converting a Cartesian equation to polar equation and vice versa.
Let the fixed point be O (Fig 3.1). Suppose point P is at a distance r from O. Let angle XOP be
 . Then the polar coordinates of P are given as an ordered pair (r , )
P ( r , )
r

X
O
Figure 3.1
The angle is measured in an anticlockwise direction from the fixed line OX. A negative value of
 implies the angle is measured clockwise from the fixed line. The distance r is always positive
as long as it is measured from O to P.
P1 ( r , )


P3 ( r , )
P2 ( r , )
Figure 3.2
25
Points P1 ( r , ) , P2 ( r , ) and P3 (r , ) give the various positions when r and  take
negative values.
3.2.1.1 Relationship between polar and rectangular coordinates
y
P ( x, y )
r

x
O
Figure 3.3
Consider a point P ( x, y ) whose polar coordinates are P ( r , ) . From the diagram it is clear that
y
 tan  .
x
x  rCos , and y  rSin
r 2  x2  y2
The above relationships can be used to transform the Cartesian coordinates to polar coordinates
and vice versa.
Example 3.1
Consider point (3, 4) on the x y plane. From the above relationships, we get
4
 53.13
5
Hence the polar coordinates for P (3,4) are P (5,53.13)
r
32  4 2  5
and
  tan 1
Example 3.2
Consider point (7, 60) on the plane. From the above relationships, we get r  7 ,
x  r cos  7 cos 60  7 / 2
y  r sin   7 sin 60  7 3 / 2
Hence the Cartesian coordinates for (7, 60) are (7/2,73/2).
Example 3.3
Write in Cartesian form the equation r  9Cos .
26
Solution
Put
r 2  x2  y2
x
x
and Cos  r  2
x  y2
Hence
x2  y2
=9
x
x  y2
2
 x 2  y 2  9x .
Or
( x  4.5) 2  y 2  4.5 2
which is the equation of the circle with center ( 4.5,0) and radius 4.5 .
Example 3.4
Write in Cartesian form the equation r 
3
.
2  2 cos
Solution
We get
( x 2  y 2 ) (2  2
x
(x  y 2 )
2

2 (x 2  y 2 )  2x  3
)3
.

4( x 2  y 2 )  4 x 2  12 x  9
This reduces to
y 2  3( x  3 / 4) .
27
E-tivity 3.2.1 - Polar Coordinates
Numbering, pacing and sequencing
Lesson 3.2.1
Title
Polar coordinates
Purpose
To introduce you to the method of converting Cartesian
to polar coordinates and vice versa.
Watch the video on polar coordinates https:
https://www.youtube.com/watch?v=RZ9cow954_s
https://www.youtube.com/watch?v=fHfxw12BTOg
Brief summary of overall task
Spark
Individual contribution


Interaction begins



E-moderator interventions
Schedule and time
Next
Watch the videos on polar coordinates.
Discuss on how one can convert from one
system to the other.
Post your answers on the discussion forum 3.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 3rd week of
the semester.
Parabola
3.3 Assessment
Convert the Cartesian coordinates to polar coordinates
(1, 1)
3.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
28
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
29
LESSON 4
PARABOLA
4.1 Introduction
In the next three lessons we study the conic sections; the parabola, the ellipse and the hyperbola.
A conic section is the locus of a point described such that the ratio of its distance from a fixed
point to that from a fixed line is a constant. This ratio is called the eccentricity, denoted by e .
The fixed line is called the directrix and the fixed point the focus.
Definition: A parabola is the locus of a point which moves such that the ratio of its distance from
a fixed point to the distance from a fixed line is one.
4.2 Lesson Learning Outcomes
By the end of the lesson, you should be able to:4.2.1 Find the standard and general equation of a parabola given its focus and directrix.
4.2.1 Standard equation of a parabola
Let the focus be denoted by F and the directrix be the line DM. The line passing through the
focus and perpendicular to the directrix is called the axis of the parabola (Fig 4.1). The point V
where the parabola intersects with the axis is called its vertex.
y
M
P(x, y)
x
D
V
F (a, 0)
Figure 4.1
To get the equation of the parabola, suppose the axis coincides with the x-axis and let the vertex
V be at the origin. Let the distance from the vertex to the focus be a . The focus is F ( a , 0).
From the definition of the parabola,
PF
1 ,
MP
where P( x, y ) is any other point on the parabola. The distances PF and PM are
PF  ( x  a ) 2  y 2 and PM  x  a
Thus
( x  a) 2  y 2  x  a
30
which gives
( x  a) 2  y 2  ( x  a ) 2
or
x 2  2ax  a 2  y 2  x 2  2ax  a 2
or
y 2  4ax
.
This is the standard equation of a parabola whose axis is horizontal and the vertex is at the
origin. A chord (a line segment whose end points are points on the parabola) passing through the
focus F is referred to as the focal chord. The focal chord perpendicular to the axis is called the
latus rectum.
The distance of any point P(x,y) on the curve to the focus is referred to as its focal distance. The
length of the latus rectum is equal to 4a.
Example 4.1
Find the equation of the parabola whose focus is F (2,0) and the directrix is x = -2.
Solution
Since the focus is F (2,0) and the directrix is x = -2, then the vertex is at the origin and its axis is
the x-axis. Hence a=2. The equation is
y 2  4ax  8 x
Example 4.2
Find the focus, the equation of the directrix and the length of latus rectum for the parabola
5 y 2  24 x.
Solution
Write the equation in standard form to get
y2 
24
x.
5
This implies
4a 
24
6
a .
5
5
Hence the focus is F (6/5,0).
The equation of the directrix is
6
x  a   .
5
The length of the latus rectum is
l  4a 
24
5
4.2.1.1 General equation of a parabola
31
Let the focus of the parabola be F (h,k) and the equation of the directrix be Ax' By 'C  0
(Fig 4.2)
y-axis
P(x,y)
M
axis
F(h,k)
V
D
x-axis
O
Figure 4.2
FP 
( x  h) 2  ( y  k ) 2
.
The distance from point (x, y) to line Ax+By+C=0 is MP 
( Ax  By  C )
A2  B 2
The general equation of the parabola is therefore given by
( Ax  By  C ) 2
( x  h) 2  ( y  k ) 2 
A2  B 2
Or
for axis parallel to the x-axis in the positive orientation
for axis parallel to the x-axis in the negative orientation
for axis parallel to the y-axis in the positive orientation
for axis parallel to the y-axis in the negative orientation
4.2.1 E-tivity – Standard and General Equation of a Parabola
32
Numbering, pacing and sequencing
Lesson 4.2.1
Title
Equation of a Parabola
Purpose
To introduce you to the parabola conic section.
Brief summary of overall task
Watch
the
video
on
the
parabola
https://www.youtube.com/watch?v=u5V-sytPAig
https://www.youtube.com/watch?v=LNouPJaNf5Y
Spark
Individual contribution


Interaction begins



E-moderator interventions
Schedule and time
Next
Watch the videos on the parabola.
Discuss the equation of a parabola whose vertex
is not at the origin.
Post your answers on the discussion forum 4.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 4th week of
the semester.
Parabola: Tangent and Normal
4.3 Assessment
Find the vertex, the focus, the equation of the directrix and the length of the latus rectum for the
parabola.
x 2  8 x  2 y  10  0
4.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
33
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
http://www.yiminmathcentre.com/Homework/Year12/
Year12_4Unit_Homework_Conics_Part3.pdf
https://doubtnut.com/question-answer/equation-of-normal-of-ellipse-1339792
34
LESSON 5
PARABOLA: TANGENT AND NORMAL
5.1 Introduction
The parabola is defined as the conic section whose eccentricity is one.
5.2
Lesson Learning Outcome
At the end of the chapter, you should be able to
5.2.1 Determine the tangent and normal equations to a parabola.
5.2.1 Tangent and Normal to a parabola
We consider the case when the axis is horizontal and when it is vertical.
For the horizontal case, let the equation of the parabola be ( y  c 2 ) 2  4a ( x  c1 ) . To get the
equation of the tangent to the parabola at point P ( x1 , y1 ) , consider another point Q ( x 2 , y 2 )
(Fig 5.2)
y
P
Q
F
x
O
Figure 5.2
The slope of PQ is
y  y1
m PQ  2
.
x 2  x1
As Q approaches P, x 2 approaches x1 and y 2 approaches y1 . The limit of the slope of PQ is
therefore
y  y1
0
Lim(m PQ )  Lim( 2
)
Q P
Q P x  x
0
2
1
which is undefined. To get the limit, consider the equations
( y 2  c 2 ) 2  4a ( x 2  c 2 ) -----------------------------------------(5.1)
35
( y1  c 2 ) 2  4a ( x1  c1 ) ------------------------------------------(5.2)
Subtracting (5.2) from (5.1) gives
( y 2  c 2 ) 2  ( y1  c 2 ) 2  4a( x 2  x1 )
or
[( y 2  c 2 )  ( y1  c 2 )][( y 2  c 2 )  ( y1  c 2 )]  4a ( x 2  x1 )

y 2  y1
4a

x 2  x1 y1  y 2  2c 2
Hence as Q approaches P the limit of the slope is
 y  y1 
2a
 
mt  Lim 2
.
Q P x  x
y1  c 2
1 
 2
Let R(x, y) be any other point on the tangent. Then the equation of the tangent is
y  y1
2a

x  x1
y1  c 2
or
( y  y1 )( y1  c 2 )  2a ( x  x 1 ) .
This can be restructured as below.
[( y  c 2 )  ( y1  c 2 )]( y1  c 2 )  2a[( x  c 1 )  ( x 1  c1 )]

( y  c 2 )( y1  c 2 )  ( y1  c 2 ) 2  2a( x  c1 )  2a( x1  c1 )

( y  c 2 )( y1  c 2 )  4a ( x1  c1 )  2a( x  c1 )  2a( x1  c1 )

( y  c 2 )( y1  c 2 )  2a( x  x1  2c1 ) .
When the vertex is at the origin, the equation of the tangent at point P ( x1 , y1 ) is
yy1  2a ( x  x1 ) .
Example 5.1
Find the equation of the tangent to the parabola y 2  2 y  12 x  23  0 at point P (-7/4, 10).
Solution
Write the equation in standard form as
y 2  2 y  1  12 x  24
( y  1) 2  ( x  2)
Hence c  2, c 2  1 and a  3 . The equation of the tangent at point P (-7/4, 10)
is therefore
1
( y  1)(10  1)  2  3( x  7 / 4  4)
or
6 x  6 y  17  0 .
36
Example 5.2
Find the equations of the tangents at the end points of the latus rectum of the parabola
y 2  6 y  20 x  49  0 .
Solution
Write the equation in standard form to get
( y  3) 2  20( x  2) .
This implies 4a  20,  a  5 . The axis of symmetry is y=-3. The vertex is V(2,-3), the focus is
F(7,-3) and the endpoints of the latus rectum are P1 (7,7) and P2 (7,13) .
The equation of the tangent at P1 (7,7) is
( y  3)(7  3)  10( x  7  14)
or
x  y  10  0
The equation of the tangent at P2 (7,13) is
( y  3)(13  3)  10( x  7  14)
5.2.1.1 Normal to a parabola
The normal to a parabola at a given point on the parabola is the line perpendicular to the tangent
at the point. The slope of the normal is therefore the negative reciprocal of the slope of the
tangent. Suppose the equation of the parabola is ( y  c 2 ) 2  4a ( x  c1 ) . The slope of the tangent
at point P ( x1 , y1 ) on the parabola was found to be
mt 
2a
.
y1  c 2
The slope of the normal is therefore
 ( y1  c 2 )
mn 
.
2a
Let R(x,y) be any other point on the normal. The equation of the normal is then
 ( y1  c 2 )
y  y1 
( x  x1 ) .
2a
Example 5.3
Find the equations of the tangent and normal to the parabola
y 2  10 y  2 x  41  0
at the point (-10,3).
Solution
The standard form is
( y  5) 2  2( x  8) .
This implies 4a  2,  a  1 / 2 . The axis of symmetry is y=5. The vertex is V (-8, 5), the focus is
F (-5, -4.5). The slope of the tangent at (-10, 3) is
mt 
 2a
 1/ 2 .
y1  5
The equation of the tangent is therefore
37
( y  5)(3  5)  2  1 / 2( x  10  16)
or
x  2 y  16  0 .
The equation of the normal at (-10, 3) is
y  3  2( x  10)
or
2 x  y  23  0 .
5.2.1 E-tivity – Tangent and Normal Tangent and normal to a parabola
38
Numbering, pacing and sequencing
Lesson 5.2.1
Title
Tangent and normal to a parabola
Purpose
To introduce you to tangent and normal to a parabola
Brief summary of overall task
Watch the following videos; https://www.youtube.com/
watch?v=RWiDocMwta4
Spark
Individual contribution


Interaction begins
E-moderator interventions
Schedule and time
Next
Watch the video and read on tangent and normal
to a parabola.
Write the parametric equation of a parabola.



Post your answers on the discussion forum 5.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 5 th week of
the semester.
Ellipse
5.3 Assessment
Find the equation of the tangent and the normal to the following parabola at the given point.
y 2  4 y  4 x  8  0 , P (5/4, 1)
5.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
39
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
http://www.yiminmathcentre.com/Homework/Year12/
Year12_4Unit_Homework_Conics_Part3.pdf
https://doubtnut.com/question-answer/equation-of-normal-of-ellipse-1339792
40
LESSON 6
ELLIPSE: EQUATION
6.1 Introduction
The ellipse is defined as the conic section whose eccentricity is less than one. This curve is also
the locus of a point that moves such that the sum of the distance from two fixed points is fixed.
6.2 Lesson Learning Outcomes
At the end of this lesson, you should be able to:6.2.1 Determine the standard and the general equation of an ellipse given its directrix and focus
or focus and eccentricity.
6.2.1 The equation of an ellipse
Let the focus of the ellipse be F and the directrix be DM. See figure below.
M
D
D’s Identities …………………………………………..60
V
V’s Identities …………………………………………..60
F
Figure 6.1
From the definition
PF= e PM
Between D and F, there is a point V such that
FV=e VD
Which lies on the ellipse.
Also further away from D along line DF there is another point V, which satisfies
FV=e VD
Let point C be the midpoint between V and V. Also let the distance CV= CV= a
Hence VV= 2a
Then
FV+ FV=e (VD+VD) =e DD
where VD=VD
also DD=2CD
Hence
VV=2e CD=2a
or
CD=
a
e
Also write
FV-FV=e (VD-VD)=e VV= 2ae
41
But
FV-FV=FV+FV-2FV
=2CV-2FV=2(CV-FV)
=2CF
Hence
CF= ae
Let C be placed at the origin on the XY plane, CD be along the x - axis and the vertical line
through C the y - axis.
Notice that there now can be another point F as the focus and directrix DM that would have
given the same results as above.
If C is the origin, the focus is F( ae, o ) and F ( ae, o ) and the equation of the directrix is x=
a
e
or
x=
a
e
Now
FP=
( x  ae) 2  y 2
and
PM=
a
-x
e
From FP=e PM, we get
( x  ae ) 2 + y
2
2


=   x e2
a
e
x 2 - 2aex+a 2 e 2 +y 2 =e 2
= a
This gives

2
 a 2 2a

 2 
x  x 2 
e
e

- 2aex + e 2 x 2
2
y
x2
1
+ 2
2
a (1  e 2 )
a
The curve has two axis. The longer one is the major axis which has length 2a.The shorter axis
obtained for the case above when x=0 as 2 a 2 (1  e 2 ) is the minor axis.
x2 y2
Let b 2 =a 2 (1 - e 2 ) Then the equation is 2  2  1
which is in standard form.
a
b
(1-e 2 ) x 2 +y 2 =a 2 - a 2 e 2
or
Example 6.1
Find the directrix, eccentricity and focus of the ellipse given by 4x 2 +9y 2 =36.
Solution
Writing in standard form
x2 y2

1
9
4
To get the eccentricity, we have
b
2
=a
2
a 2 =9
a=3, and b 2 = 4 implying b=2
(1-e 2 )
9(1-e 2 )=4
or
ae =9 5 =35
and
3
42
5
5
=
3
3
9
27
a
=
3=
5
5
e
e=
Hence the focus is F (3
5,
0) and the directrix is x=
27
5
Example 6.2
2
center at origin and directrix
3
Find the equation of and sketch the ellipse whose eccentricity is
x  6 and focus (1, 0)
Solution
y
P ( x, y )
D
x
F(1,0)
Figure 6.2
Since
a
2
=6, then a  6e  6   4.
e
3
4  16  5 80

b 2  a 2 (1  e) 2  161   

9
9
9


x2
y2

1
Hence the equation is 16  80 
 
 9 
6.2.1.1 The general equation of an ellipse
Using the general equation we can obtain the equation of an ellipse whose axis is not necessarily
as in the illustration in section 6.2.
Consider the ellipse whose focus is F (h, k ) and the directrix is any line ax  by  c  0 . Let
P ( x, y ) be any point on the ellipse.
y
M
43
D
P ( x, y )
F
x
O
Figure 6.3
Now
FP=
and
( x  h) 2  ( y  k ) 2
PM=
ax  by  c
a2  b2
Hence the general equation is
( x - h ) 2 +( y - k ) 2 = e
2
(ax  by  c) 2
a2  b2
Example 6.3
Find the equation of the ellipse whose focus is
2
and the directrix is y =9.
3
Solution
From the general equation we get
(x-o) 2  ( y-4) 2 =
4
(y-9) 2
9
or
5
9
x 2  y 2  20
4 2
y  8 y  36
9
x2 y2

0
20 36
x 2  y 2 -8y+16which is
or
This is an ellipse whose major axis is along the y-axis and is of length 2b=2
axis along the y-axis and length 2a=2 20 =4 5 .
We have seen that when the center of the ellipse is ( o, o ), then the equation is
If the center is shifted to say C ( m, n ), the equation of the ellipse becomes
( x  m) ( y  n )

1
a2
b2
36
=12 and minor
x2 y2

1
a
b
Example 6.4
Find the focus, the center, the eccentricity and directrix of the ellipse whose equation is
x 2 4 y 2  4 x  24 y  24  0
Solution
Writing in the standard form gives
44
x 2 4 y 2  4 x  24 y  24
(x+2) 2 4( y  3) 2  24  4  36  16
or
Hence
( x  2) 2 ( y  3) 2

 1 Where
16
4
a 2  16  a=4
b 2  4  b=2
The center is C (-2, 3) eccentricity is given by b 2  a 2 (1  e) 2  4  16(1  e 2 ) or e 
Since the centre is C (-2, 3) the abscissa for the foci are  2  ae  2 
3.4
2
Here they are
F1 ( 2  2 3 , 3)
F2 ( 2  2 3 , 3)
and
The directrix is at
x  2 
a
8
 2 
.
e
3
6.2.1.2 The length of the latus rectum.
Consider the origin-centered ellipse,
P
M
D
O
F
Figure 6.4
From the definition of an ellipse we have for the length l through the focus
l
e
a
 ae
e
a
 l  e(  ae)
e
or
l  a (1  e) 2
2
Hence the length of the latus rectum of the ellipse is 2l  2a (1  e) 2 , thus l  2b
a
6.2.1 E-tivity – Tangent and Normal Equation of an Ellipse
Numbering, pacing and sequencing
Lesson 6.2.1
45
3
2
Title
The Equation of an Ellipse
Purpose
To introduce you to the equation of an ellipse
Brief summary of overall task
Watch
the
following
videos;
https://www.youtube.com/watch?v=HO-XMY1g8Hc
https://www.youtube.com/watch?v=IMA92FjipqI
Spark
Individual contribution


Interaction begins



E-moderator interventions
Schedule and time
Next
Watch the videos and read on the ellipse.
Write the equation of an ellipse whose center is
not at the origin.
Post your answers on the discussion forum 6.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 6th week of
the semester.
Ellipse: Tangent and Normal
6.3 Assessment
Find the foci, the center, the eccentricity and directrix of the given ellipse.
x2
y2

1
169 25
6.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
46
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
http://www.yiminmathcentre.com/Homework/Year12/
Year12_4Unit_Homework_Conics_Part3.pdf
https://doubtnut.com/question-answer/equation-of-normal-of-ellipse-1339792
LESSON 7
ELLIPSE: TANGENT AND NORMAL
7.1 Introduction.
47
The ellipse is defined as the conic section whose eccentricity is less than one. This curve is also
the locus of a point that moves such that the sum of the distance from two fixed points is fixed.
7.2 Lesson Learning Outcomes
At the end of this lesson, you should be able to
7.2.1 Determine the tangent and normal equations to an ellipse.
7.2.1 Tangent and Normal to an ellipse
( x  m) 2 ( y  n) 2

1
a2
b2
We wish to get the equation of the tangent at point say P1 ( x1 y1 ) on the ellipse.
Consider the ellipse whose equation is
P1 ( x1 y1 )
y
Q
y
O
Figure 7.1
Let Q ( x 2 , y 2 ) be any other point on the ellipse which approaches P1 . The gradient of P1Q1
y 2  y1
. As Q approaches P1 we get the limit from: since Q and P 1 are on the
is m.
x 2  y1
ellipse, then we have
( x1  m) 2 ( y1  n) 2
1

1
a2
b2
And
( x 2  m) 2 ( y  m) 2
2

1
a2
b2
Subtracting 1 from 2 we get
( x 2  m) 2  ( x1  m 2 ) ( y 2  m) 2  ( y1  m)
+
=0
a2
b2
 a2
( x 2  x1 )( x 2  m  x1  m)  2 ( y 2  y1 )( y1  n  y 2  n)
b
48
y 2  y1  b 2 ( x1  m  x 2  m
 2
x 2  x1
a ( y1  n  y 2  n)

With the limit
Lim
(
y 2  y1
 b2 x  m
) 2 ( 1
)
x 2  x1
y1  n
a
QP
Hence the equation of the tangent is for any point P ( x, y ) on the tangent
y  y1  b 2 ( x1  m)
 2
x  x1
a ( y1  n)
 b2
( y  y1 )( y1  n)  2 ( x  x1 )( x1  m)
a
Or
This can be written as
 y  n  ( y1  n)( y1  n)   b2  x  m  ( x1  m)( x1  m)
2
a
 b ( x  m)( x1  m)  a ( y  n)( y1  n)  b 2 ( x1  m) 2  a 2 ( y1  n) 2  a 2 b 2
2
2
Or
( x  m)( x1  m) ( y  n)( y1  n)

1
a2
b2
This is the equation of the tangent.
Example 7.1
Find the equation of the tangent at point P (1,2) of the ellipse ( x  1) 2  4( y  1) 2  4 .
Solution
Write the equation in the standard form to get
( x  1) 2 ( y  1) 2

4
4
1
The tangent passes through (-1, 2)  x1  1 and y1  2 . We get the equation of the tangent as
( x  1)( 1  1) ( y  1)(2  1)

 4  y =5
4
1
7.2.1.1 Normal to an Ellipse
Consider the ellipse whose equation is
x2
a2

y2
b2
1
We wish to get the equation of the tangent at point say P1 ( x1 y1 ) on the ellipse. The equation
for the normal line is given by;
a2 x b2 y

 a2  b2
x1
y1
7.2.1
E-tivity – Tangent and Normal Ellipse: Tangent and Normal
49
Numbering, pacing and sequencing
Lesson 7.2.1
Title
Tangent and normal to an ellipse
Purpose
To introduce you to tangent and normal to an ellipse
Brief summary of overall task
Watch
the
following
videos;
https://www.youtube.com/watch?v=5NkdbxfQTn0
https://www.youtube.com/watch?v=OQRX8Ijg7yY
Spark
Individual contribution

Watch the videos and read on tangent and
normal to an ellipse.
Write the parametric equation of an ellipse.

Interaction begins
E-moderator interventions
Schedule and time
Next



Post your answers on the discussion forum 7.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 7th week of
the semester.
Hyperbola
7.3 Assessment
Find the equation of the tangent and the normal at the given point of the given ellipse.
4 x 2  9 y 2  144,
(3,2 3)
7.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
50
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
http://www.yiminmathcentre.com/Homework/Year12/
Year12_4Unit_Homework_Conics_Part3.pdf
https://doubtnut.com/question-answer/equation-of-normal-of-ellipse-1339792
LESSON 8
HYPERBOLA: EQUATION
8.1 Introduction
The hyperbola is the locus of a point that moves such that the ratio of its distance from a fixed
point to the distance from a fixed line is greater than one.
51
8.2 Lesson Learning Outcomes
At the end of this lesson, you should be able to:8.2.1 Determine the standard and the general equation of a hyperbola.
8.2.1 Equation of a hyperbola
The standard equation can be developed as follows:
Consider a fixed point F and a fixed line D as in the diagram below.
M
P
D
F’s Identities …………………………………………..60
V’s Identities …………………………………………..60
D’s Identities …………………………………………..60
F
V
Figure 8.1
P ( x, y ) is on the hyperbola such that
PF
 e 1
MP
or
PF  eMP
Let the axis of the hyperbola be the horizontal line perpendicular to the directrix through the
focus and vertex V be the point of intersection of the hyperbola and the axis,
Hence
FV  eVD.
There also exists another point V1 along the axis such that FV  = eV  D . Infact there is a
whole set of points and also the virtual focus F and directrix D .
Let the distance VV = 2a and let C be the midpoint of VV .
Then
FV   FV  e(V D  VD)  e( D D  e 2CD
But
FV   FV  V V  2a
 2eCD  2a  CD 
Also
a
e
FV   FV  e(V D  VD)eV V  2ae
But
52
FV   FV  F F  2CF  2CF  2ae  CF  ae
Let C be at the origin. Then the foci are F (ae,0) and F ' (  ae,0) . The directrix is x 
From PF  ePM ,
PF 
( x  ae) 2  y 2
, PM  x 
a
.
e
a
, we get
e
a
( x  ae) 2  y 2  e 2 ( x  ) 2 ,
e
or
x 2 (e 2  1)  y 2  a 2 (e 2  1) ,
or
x2 y2

 1,
a2 b2
where b 2  a 2 (e 2  1) . This is the standard equation of a hyperbola.
The curve has two lines of symmetry, the x-axis and the y-axis. VV ' is called the transverse
axis and BB ' the conjugate axis.
The difference between PF ' and PF is a constant equal to 2a as shown below.
a
)  ex  a
e
a
PF '  eM ' P  e( x  )  ex  a
e
PF  e( x 
PF ' PF
 2a . an alternative
This
provides
definition of a hyperbola as the
locus of a point that moves such that the difference between the distances from two fixed points
a constant.
Example 8.1
Determine the equation of the hyperbola whose eccentricity is 3/2 and the foci are F ' (2,0)
and F (2,0) .
Solution
Since ae  2, and e = 3/2, then a = 4/3. This implies
b 2  a 2 (e 2  1) 
16 9
20
(  1) 
.
9 4
9
Hence the equation is
x2
y2

 1.
(16 / 9) ( 20 / 9)
The latus rectum is the chord perpendicular to the transverse axis and through the focus. The
length of the latus rectum is obtained as follows. Let the length be l  2l1 , where l1 is the
distance from the focus along the latus rectum to the curve. Then
l1
 e,  l1  a (e 2  1) .
ae  a / e
Hence the length of the latus rectum is
53
l  2l1  2a (e 2  1) 
2b 2
.
a
Example 8.2
Find the coordinates of the foci, the eccentricity, the length of the latus rectum and the directrix
of the hyperbola
x2 y2

1.
36
4
Solution
From the equation a=6 and b=2. The eccentricity is obtained from b 2  a 2 (e 2  1) as e  13 10 .
The foci are therefore F ' ( 3 10 ,0) and F (3 10 ,0) . The length of the latus rectum is
10
L  2a (e 2  1)  2  9(  1)  2
9
8.2.1.1 The Asymptotes of a hyperbola
Consider the hyperbola
x2 y2

 1 and let y  mx  C be any line. The line and hyperbola meet at
a2 b2
points where
x 2 y 2 x 2 (mx  C ) 2



0
a2 b2 a2
b2
 (b 2  m 2 a 2 ) x 2  2a 2 mCx  a 2 (b 2  c 2 )  0
The points of intersection are at infinity if the coefficients of x 2 and
This implies b 2  m 2 a 2  0
m
b
a
and
x are zero.
2a 2 mc  0  c  0.
Hence the asymptotes of the hyperbola are y  
b
x
a
8.2.1.2 Hyperbola with center not at the origin
x2 y2
Suppose the center of the hyperbola 2  2  1 is shifted to position C ( h, k ) . The equation
a
b
becomes
( x  h) 2 ( y  k 2

1
a2
b2
If the transverse axis is along the y-axis, then the equation becomes
( y  k ) 2 ( x  h) 2

1
a2
b2
54
Example 8.3
Analyze the hyperbola 9 x 2  16 y 2  18 x  64 y  199  0 .
Solution
Complete the squares and write the equation in standard form to get
9( x  1) 2  16( y  2) 2  144
or
( x  1) 2 ( y  2) 2

1.
16
9
This is the equation of the hyperbola with center C (1,2) . Also a = 4 and b = 3. The
eccentricity is
b2
9
5
e
1 
 1   ae  5 .
2
16
4
a
The vertices are V ' ( 3,2) and V (5,2) , foci F ' ( 4,2) and F (6,2) , asymptotes
y2
3
( x  1) .
4
8.2.1 E-tivity – Tangent and Normal Equation of Hyperbola
Numbering, pacing and sequencing
Lesson 8.2.1
55
Title
Equation of Hyperbola
Purpose
To introduce you to the equation of a hyperbola
Brief summary of overall task
Watch
the
following
videos;
https://www.youtube.com/watch?v=F44WUGEfZQk
https://www.youtube.com/watch?v=yb-dvQEp3Ec
https://www.youtube.com/watch?v=c_xEqSviccY
Spark
Individual contribution


Interaction begins



E-moderator interventions
Schedule and time
Next
Watch the videos and read on the hyperbola.
Write the equation of a hyperbola whose center is
not at the origin.
Post your answers on the discussion forum 8.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of your
colleagues have posted keeping etiquette in mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 8th week of the
semester.
Hyperbola: Tangent and Normal
8.3 Assessment
Find the foci, the vertices, the eccentricity, and the equation of the directrix of the given
hyperbola.
25 y 2  4 x 2  100
8.4 References
http://www.cse.salford.ac.uk/physics/gsmcdonald/pp/PPLATOResources/h-flap/M2_3t.pdf
LESSON 9
HYPERBOLA: TANGENT AND NORMAL
9.1 Introduction
56
The hyperbola is the locus of a point that moves such that the ratio of its distance from a fixed
point to the distance from a fixed line is greater than one.
9.2 Lesson Learning Outcomes
At the end of this lesson, you should be able to
9.2.1 Determine the tangent and normal to the hyperbola.
9.2.1 Tangent and normal to a hyperbola
Consider the hyperbola
x2 y2

1
a2 b2
xx1 yy1
 2 1
a2
b
Note that the normal at any point on the curve can be easily obtained since it is the line
perpendicular to the tangent at that point.
The equation of the tangent to the hyperbola at point P ( x1 , y1 ) is
Example 9.1
Determine the eccentricity, the foci, the length of the latus rectum, the equation of the
asymptotes, the equation of the tangent and the normal at point P (3,
12
) of the hyperbola
5
x2
y2

 1
25
9
Solution
From the equation, a  5 ,
e  1
9

25
34
, and ae 
5
b  3 Hence from b 2  a 2 (e 2  1) we get the eccentricity as
34
.
Hence the foci are F ' ( 34 ,0) and
The length of the latus rectum is
F ( 34 ,0) .
34
18
 1) 
.
25
5
b
3
3
The asymptotes are y   x . Hence y  x and y   x .
a
5
5
12
The equation of the tangent at P (3, ) is
5
3
12
3
15
x
y  1,  y 
x
.
25
45
10
4
 10
The normal has therefore slope m 
. Since it passes through P, then the equation is
3
y  12 / 5  10
57 10

, y 

x.
x3
3
25 3
L  2a (e 2  1)  2  5(
57
9.2.1 E-tivity – Tangent and Normal Hyperbola: Tangent and Normal
Numbering, pacing and sequencing
Lesson 9.2.1
58
Title
Tangent and normal to a hyperbola
Purpose
To introduce you to tangent and normal to a hyperbola
Brief summary of overall task
Watch the following videos; https://www.youtube.com/
watch?v=KmZOGPMdero
https://www.youtube.com/watch?v=RWiDocMwta4
Spark
Individual contribution


Interaction begins
E-moderator interventions
Schedule and time
Next
Watch the videos and read on tangent and normal
to a hyperbola.
Write the parametric equation of a hyperbola.



Post your answers on the discussion forum 9.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of
your colleagues have posted keeping etiquette in
mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 9th week of the
semester.
Trigonometry
9.3 Assessment
Find the equation of the tangent and the normal lines at the given point on the hyperbola.
25 y 2  x 2  25,
(5, 2 )
9.4 References
L. Bostock, S. Chandler, and C. Rourke. Further Pure Mathematics. Stanley Thornes
(Publishers) Ltd, 1982.
R.E Larson and R.P. Hostetler. Precalculus. Houghton Mifflin Company, 1997.
59
P.K. Jain and Khalil Ahmad. Analytical Geometry of Two Dimensions. Wiley Eastern Limited,
1983.
https://www.masterjeeclasses.com/wp-content/uploads/2019/01/11.-HYPERBOLATHEORY.pdf
LESSON 10
TRIGONOMETRY: IDENTITIES
10.1 Introduction
60
In this lesson we study trigonometric ratios and identities.
10.2 Lesson Learning Outcomes
At the end of the chapter, you should be able to;
10.2.1 Evaluate trigonometric ratios and their identities.
10.2.1 Trigonometric ratios
These are defining relations for tangent, cotangent, secant, and cosecant in terms of sine and
cosine.
10.2.1.1 Trigonometric identities
10.2.1.2 Ptolemy’s Identities …………………………………………..60s identities
Example 10.1
If sin A 
4
12
and cos B 
, evaluate cos A  B 
5
13
Solution
cos A  B   cos A cos B  sin A sin B
13
5
5
4
A
3
B
61
12
By Pythagorus theorem,
cos A  B  
3 12 4 5 36  20 16

 


5 13 5 13
65
65
Example 10.2
If tan x 
3
, determine the value of sin 3 x
4
Solution
5
3
x
4


sin 3 x  sin  x  2 x   sin x cos 2 x  cos x sin 2 x  sin x cos2 x  sin 2 x  cos x 2 sin x cos x 
sin 3 x  sin x  2 x   sin x cos 2 x  cos x sin 2 x 
3  16 9  4  3 4  21
48  27




   
5  25 25  5  5 5  125 125 125
10.2.1 E-tivity – Tangent and Normal Trigonometry: Ratios and Identities
Numbering, pacing and Lesson 10.2.1
sequencing
62
Title
Trigonometry: Ratios and Identities
Purpose
To introduce you to ratios and identities
Brief summary of overall Watch the following videos; https://www.youtube.com/watch?
task
v=21z6gkvhlbQ
https://www.youtube.com/watch?v=RjvYKVmV2EQ
Spark
Individual contribution

Interaction begins



Watch the videos and read on trigonometry.
Schedule and time
Post your answers on the discussion forum 10.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of your
colleagues have posted keeping etiquette in mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 10th week of the semester.
Next
Trigonometry: Factor formulae
E-moderator
interventions
10.3 Assessment
Show that sin 3  3 sin 
 4 sin 3 
10.4 References
Backhouse, J.K. & Houldsworth, S.P.T (2000). Pure Mathematics I & II. London: Longman
Group.
Algebra and Trogonometry with Analytic Geometry, 2 nd edition by Skowkowski E. (2010)
Brooks/Cole Publishing Co.
https://www.govst.edu/uploadedFiles/Academics/Colleges_and_Programs/CAS/
Trigonometry_Short_Course_Tutorial_Lauren_Johnson.pdf
https://www.webassign.net/resources/bmcc/Ch11.pdf
63
LESSON 11
TRIGONOMETRY: FACTOR FORMULAE
11.1 Introduction
In this lesson we study the factor formulae as applied to trigonometry.
64
11.2 Lesson Learning Outcomes
At the end of this lesson, you should be able to
11.2.1 Apply the factor formulae to solve trigonometric problems.
11.2.1 Factor Formulae
11.2.1.1 Identities for negative angles.
Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even
functions.
11.2.1.2 Double angle formulae for sine and cosine.
Note that there are three forms for the double angle formula for cosine. You only need to know
one, but be able to derive the other two from the Pythagorean formula.
11.2.1.3 Factor Formulae
This group of identities allows you to change a sum or difference of sines or cosines into a
product of sines and cosines. Look up the cosines of the sum α + β. and the difference α – β.
Average those two cosines. You get the product xy! Three table look-ups, and computing a sum,
a difference, and an average rather than one multiplication.
The factor formulae are given by;
65
A B
A B
cos
2
2
A B
A B
sin A  sin B  2 cos
sin
2
2
A B
A B
cos A  cos B  2 cos
cos
2
2
A B
A B
cos A  cos B  2 sin
sin
2
2
sin A  sin B  2 sin
Example 11.1
Show that
i)
Cos 2 A  Cos 2 B
 Cot  A  B  Cot  A  B 
Cos 2 B  Cos 2 A
ii)
Sin 2 A Sec 2 A  2TanA
Solution
i)
2 cos
 2 A  2 B  cos  2 A  2 B 
cos A  B  cos A  B 
2
2

 2 B  2 A cos  2 B  2 A  sin  B  A sin  B  A
 2 sin
2
2
From negative angles we have sin   x    sin x , therefore;
Cos 2 A  Cos 2 B

Cos 2 B  Cos 2 A
cos A  B  cos A  B 
cos A  B  cos A  B 
cos A  B  cos A  B 


 sin  A  B  sin[ A  B  ]  sin  A  B    sin  A  B 
sin  A  B  sin  A  B 
 cot( A  B) cot( A  B)

2
ii) Sin2 A Sec A  2 SinACosA 
1
2
Cos A
2
11.2.1 E-tivity – Tangent and Normal Trigonometry: Factor Formulae
Numbering,
sequencing
pacing
and Lesson 11.2.1
66
SinA
 2TanA
CosA
Title
Trigonometry: Factor formulae
Purpose
To introduce you to factor formulae
Brief summary of overall task
Watch
the
following
https://www.youtube.com/watch?v=EcYtHfkEPnk
https://www.youtube.com/watch?v=uUh2B9hnI_U
https://www.youtube.com/watch?v=DKhN4j7WrJg
videos;
Spark
Individual contribution

Interaction begins



E-moderator interventions
Schedule and time
Next
11.3 Assessment
Prove the following identity
Watch the videos and read on trigonometry.
Post your answers on the discussion forum 11.2.1
Read what your colleagues have posted.
In a sentence or two, comment on what two of your
colleagues have posted keeping etiquette in mind
 Focused group discussion
 Providing feedback/ teaching points
 Summarising key points
 Closing the discussion
This discussion should be done during the 11th week of the
semester.
Solutions to Assessment Questions
sin x
1  cos x

1  cos x
sin x
67
11.4 References
Backhouse, J.K. & Houldsworth, S.P.T (2000). Pure Mathematics I & II. London: Longman
Group.
Algebra and Trogonometry with Analytic Geometry, 2 nd edition by Skowkowski E. (2010)
Brooks/Cole Publishing Co.
https://www.govst.edu/uploadedFiles/Academics/Colleges_and_Programs/CAS/
Trigonometry_Short_Course_Tutorial_Lauren_Johnson.pdf
https://www.webassign.net/resources/bmcc/Ch11.pdf
SOLUTIONS TO ASSESSMENT QUESTIONS
Assessment 1.3
1.
68
2.
3.
Assessment 2.3
1. a)
b)
2. Center
, radius =
3.
Assessment 2.4
Equation of tangent:
Equation of normal:
Assessment 3.4
Assessment 4.3
,
Equation of directrix;
Length of latus rectum = 2
Assessment 5.3
Equation of tangent:
Equation of normal:
Assessment 6.3
Equations of directrices:
Assessment 7.3
69
Equation of tangent:
Equation of tangent:
Assessment 8.3
,
Equation of directrix;
Assessment 9.3
Equation of tangent:
Equation of normal:
Assessment 10.3
Assessment 11.3
70
71