Precalculus

Quarter 1 – Module 1
ANALYTIC GEOMETRY

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Pre-calculus – Grade 11
Alternative Delivery Mode
Quarter 1 – Module 1: Analytic Geometry
First Edition, 2020

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 Precalculus
MODULE 1:
Analytic Geometry

This instructional material was collaboratively developed and reviewed

by educators from public and private schools, colleges, and/or universities. We
encourage teachers and other education stakeholders to email their feedback,
comments, and recommendations to the Department of Education at
action@deped.gov.ph.

We value your feedback and recommendations.

Department of Education • Republic of the Philippines

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 Table of Contents

What This Module is About 1

What I need to know 2
How to learn from this Module 3
Icons of this Module 3
What I Know (Pretest) 4
Lesson 1 – Introduction of Conic Sections and the Circle
What I Need to Know 9
What’s In 9
What’s New 10
What Is It 11
What’s More 18
What I Have Learned 22
What I Can Do 23
Lesson 2 – The Parabola
What I Need to Know 28
What’s In 28
What’s New 29
What Is It 30
What’s More 37
What I Have Learned 42
What I Can Do 42
Lesson 3 – The Ellipse
What I Need to Know 45
What’s In 45
What’s New 47
What Is It 47
What’s More 54
What I Have Learned 59
What I Can Do 60
Lesson 4 – The Hyperbola
What I need to Know 61
What’s In 61
What’s New 63
What Is It 64
What’s More 72
What I Have Learned 72
What I Can Do 73
Lesson 5 – Equation and Important Characteristics of Different
Types of Conic Sections
What I need to Know 75
What’s In 75
What’s New 77
What Is It 78
What’s More 79

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 What I Have Learned 82
What I Can Do 83
Lesson 6 – Solving Situational Problems Involving Conic Sections
What I Need to Know 84
What’s In 84
What’s New 84
What Is It 85
What’s More 89
What I Have Learned 90
What I Can Do 91
Summary 92
Assessment (Post-test) 93
Key to Answers 99
References 113

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 What This Module is About

The Precalculus course bridges basic mathematics and calculus. This course
completes your foundational knowledge on algebra, geometry, and trigonometry. It
provides you with conceptual understanding and computational skills that are
prerequisites for Basic Calculus and future STEM courses.

Based on the Most Essential Curriculum Competencies (MELC) for Precalculus

of the Department of Education, the primary aim of this Learning Manual is to give you
an adequate stand-alone material that can be used for the Grade 11 Precalculus
course.

The Module is divided into two units: analytic geometry and mathematical
induction. Each unit is composed of lessons that bring together related learning
competencies in the unit. Each lesson is further divided into sub-lessons that focus on
one or two competencies for effective learning.

At the end of each lesson, more examples are given reinforce the ideas and
skills being developed in the lesson. You have the opportunity to check your
understanding of the lesson by solving the Supplementary Problems.

We hope that you will find this Learning Module helpful and convenient to use.
We encourage you to carefully study this Module and solve the exercises yourselves
with the guidance of your teacher. Although great effort has been put into thisModule
for technical correctness and precision, any mistake found and reported to the Team
is a gain for other students. Thank you for your cooperation.

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 Module Content

The lessons that we will tackle are the following:

✓ Lesson 1 – Introduction of Conic Sections and the Circle
✓ Lesson 2 – The Parabola
✓ Lesson 3 – The Ellipse
✓ Lesson 4 – The Hyperbola
✓ Lesson 5 – Equation and Important Characteristics of Different
Types of Conic Sections
✓ Lesson 6 – Solving Situational Problems Involving Conic
Sections

What I Need to Know

Once you are done with this module, you should be able to:
✓ (STEM_PC11AG-Ia-1) illustrate the different types of conic sections: parabola,
ellipse, circle, hyperbola, and degenerate cases;
✓ (STEM_PC11AG-Ia-2) define circle;
✓ (STEM_PC11AG-Ia-3) determine the standard form of equation of a circle;
✓ (STEM_PC11AG-Ia-5) define a parabola;
✓ (STEM_PC11AG-Ib-1) determine the standard form of equation of a parabola;
✓ (STEM_PC11AG-Ic-1) define an ellipse;
✓ (STEM_PC11AG-Ic-2) determine the standard form of equation of an ellipse;
✓ (STEM_PC11AG-Id-1) define a hyperbola;
✓ (STEM_PC11AG-Id-2) determine the standard form of equation of a hyperbola;
✓ (STEM_PC11AG-Ie-1) recognize the equation and important characteristics of
the different types of conic sections, and
✓ (STEM_PC11AG-Ie-2) solves situational problem involving conic sections.

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How to Learn From This Module

To complete the objectives of this module, you must DO THE FOLLOWING:

• Patiently read the text carefully and understand every sentence. Do not
proceed to the next part of the module without fully understanding the
previous text.
• Read the directions of each activity carefully. You will be guided as to the steps
in answering the exercises and activities of this module.
• Do not proceed to the next part without completing the previous activities.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.

What I know This is an assessment as to your level of

knowledge to the subject matter at hand,
meant specifically to gauge prior related
Knowledge
What’s In This part connects previous lesson with that
of the current one.

What’s New An introduction of the new lesson through

various activities, before it will be presented
to you

What is It These are discussions of the activities as a

way to deepen your discovery and under-
standing of the concept.

What’s More These are follow-up activities that are in-

tended for you to practice further in order to
master the competencies.

What I Have Activities designed to process what you

Learned have learned from the lesson

What I can do These are tasks that are designed to show-

case your skills and knowledge gained, and
applied into real-life concerns and situations.

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 What I Know (Pre-Test)
Multiple Choice. Encircle the letter of the best answer.

1. It is the intersection of a plane and a double-napped cone.

a. locus b. conic section c. circle d. radius

2. A collection of points satisfying a geometric property can also be

referred to as a _ of points.
a. locus b. conic section c. circle d. radius

3. A is the set of all points (x, y) in a plane that are

equidistant from a fixed point, called the center.
a. locus b. conic section c. circle d. radius

4. A/n is the set of all points in the plane that are

equidistant from a fixed point F(called the focus) and a fixed line
l (called the directrix).
a. circle b. ellipse c. parabola d. hyperbola

5. A/n _ is the set of all points in the plane the sum of

whose distances from two fixed points F1 and F2 is a constant.
a. circle b. ellipse c. parabola d. hyperbola

6. A/n is the set of all points in the plane, the difference

of whose distances from two fixed points F1 and F2 is a constant.
a. circle b. ellipse c. parabola d. hyperbola

7. 6𝑥2 + 12𝑥 + 6𝑦2 − 8𝑦 = 100 is an example of a

a. Circle b. parabola c. ellipse d. hyperbola

8. 4𝑥2 + 12𝑥 + 𝑦2 − 8𝑦 = 64 is an example of a

a. Circle b. parabola c. ellipse d. hyperbola

9. 6𝑦 = 3𝑥2 − 15 is an example of a
a. Circle b. parabola c. ellipse d. hyperbola

10. 2𝑥2 + 87𝑥 − 25 = 𝑦2 + 4𝑦 is an example of a

a. Circle b. parabola c. ellipse d. hyperbola
11. 6𝑥 − 15𝑥 = 3𝑦 + 4𝑦 + 13 is an example of a
2 2

a. Circle b. parabola c. ellipse d. hyperbola

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12. A hyperbola is the only conic that has
a. asymptotes b. focus c. vertex d. a minor axis
2 𝑦2
13. 𝑥 − = 1 has a major axis of length
4 25
a. 2 b. 3 c. 4 d. 5

(𝑥+3)2 (𝑦−2)2
Use − = 1 to answer the next 4 questions.
9 4

14. The center is at _?

a. (-3, 2) b. (-3, -2) c. (3, 2) d. (3, -2)

15. The vertices are ?

a. (0, 2), (-6, 2) b. (-3, 5), (-3, -1) c. (-1, 2), (-5, 2) d. (-3, 4), (-3, 0)

16. The foci are at _?

a. (−3 ± √7, 2) 𝑏. (−3 ± √13, 2) c. (−3, 2 ± √13) d. (−3, 2 ± √7)

17. The length of the major axis is _ _?

a. 3 b. 6 c. 9 d. 12

18. The standard form of 9𝑥2 − 4𝑦2 = 36 is

2 𝑦2 2 𝑦2 2 𝑦2 2 𝑦2
a. 𝑥 − = 1 b. 𝑥 − = 1 c. 𝑥 − = 1 d. 𝑥 − =1
4 9 36 9 25 4 5 20

19. The earth’s orbit is an ellipse with the sun at one of the foci. If the farthest
distance of the sun from the earth is 105.5 million km and the nearest
distance of the sun from the earth is 78.25 million km, find the eccentricity
of the ellipse.
a. 0.15 b. 0.25 c. 0.35 d. 0.45
2 𝑦2
20. 𝑥 − = 1 will have a minor and major axis with length (In that order)
9 16
a. 3, 4 b. 6, 8 c. 9, 16 d. 18, 32

21. Which of the following shows the correct graph of the circle?
2 2
a. x +y =4
2 2
b. y =x + 16
2 2
c. x +y = 16
2 2
d. x +y =1

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 22. Which graph represents the equation 𝑥2 − 10𝑥 + 𝑦2 = −9?

a. c.

d.

23. What is the focus and vertex of the parabola (𝑦 − 2)2 + 16(𝑥 − 3) = 0?
a. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−3,14)
b. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−3, −18)
c. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−7, −2)
d. Vertex: 𝑉(3,2) , Focus: 𝐹(−1,2)
24. Find the equation of the parabola with vertex at (5, 4) and focus at (-3, 4).

2 2
a (y-4) = - 32( x - 5 ) c. (y+4) = - 32( x - 5 )
2 2
b (y-4) = 32( x - 5 ) d. (y-4) = 8( x - 5 )
2 (𝑦+9)2
25. Find the center and foci of the ellipse (𝑥+5) + .
5 9
a. center: (5,9) , foci: (5,7),(5,11)
b. center (-5,-9) , foci: (-5,-11), (-5,-7)
c. center: (-5,-9) , foci: (-7,-9), (-3,-9)
d. center: (5,9) , foci: (3,-9), (7,-9)

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26. Find the center and vertices of the ellipse 4x2 + 9y2 - 24x + 72y + 144 = 0
a. center: (-4,3) , vertices: (-7,3), (-1,3)
b. center: (-3,4) , vertices: (-5,4), (-1,4)
c. center: (3,-4) , vertices: (1,-4), (5,-4)
d. center: (3,-4) , vertices: (0,-4), (6,-4)
27. Which of the following shows the correct graphical representation of the ellipse
2
𝑥 2 + 𝑦 =1?
16 4

a. c.

b. d.

28. Find the center and vertices of the hyperbola 11x2 - 25y2 + 22x + 250y - 889 = 0.
a. center: (1,-5) , vertices: (1,-10), (1,0)
b. center: (-1,5) ,vertices: (-1,0),(-1,10)
c. center: (-1,5) , vertices: (-6,5), (4,5)
d. center: (1,-5) , vertices: (-4,-5), (6,-5)

29. What are the vertices and asymptotes of the hyperbola 9y2 - 16x2 = 144?
4
a. vertices: (0,-4),(0,4) , asymptote: 𝑦 = ± 3 𝑥
3
b. vertices: (0,-4), (0,4) , asymptote: 𝑦 = ± 4 𝑥
4
c. vertices: (-4,0), (4,0) , asymptote: 𝑦 = ± 3 𝑥
3
d. vertices: (-4,0), (4,0) , asymptote: 𝑦 = ± 4 𝑥

30. Find the standard form of the equation of the hyperbola with the given
characteristics, vertices: (0,-6), (0,6) and foci (0,-7), (0,7).
2 𝑥2 2 𝑦2
a. 𝑦 − =1 c. 𝑥 − =1
36 49 36 13

2 𝑥2 2 𝑦2
b, 𝑦 − =1 d. 𝑥 − =49
36 13 36 13

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31. Write the equation of the ellipse that has its center at the origin with focus at (0,
4) and vertex at (0, 7).
2 𝑦2 2 𝑦2
a. 𝑥 + =1 c. 𝑥 + =-1
49 33 33 49
2 𝑦2 2 𝑦2
b. 𝑥 − =1 d. 𝑥 + =1
33 49 33 49

2 2
32. What is the graph of the hyperbola 9x − 9y = 81?

a. c,

b. d.

33. An arch 20 meters high has the form of parabola with vertical axis. The length of
horizontal beam placed across the arch 9 meters from the top is 60 meters. Find the
width of the arch at the bottom.
a. 44.72 meters b. 45.72 meters c. 89.44 meters d. 90.44 meters

34. A spotlight in a form of a paraboloid 9 inches deep has its focus 3 inches from
the vertex. Determine the radius of the opening of the spotlight.

a. 9.39 inches b. 10.39 inches c. 11. 39 inches d. 12.39 inches

35. A bridge is supported on an elliptical arch of height of 7 meters and width at the
base of 40 meters. A horizontal roadway is 2 meters above the center of the arch.
How far is it above the arch at 8 meters from the center?
a. 0.58 meters b. 1.58 meters c. 2.58 meters d. 3.58 meters

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LESSON Introduction of Conic
1 Sections and the Circle

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ illustrate the different types of conic sections: parabola, ellipse, circle,

hyperbola, and degenerate cases;
➢ determine the type of conic section defined by a given 2nd degree
equation in x and y;
➢ define a circle;
➢ determine the standard form of equation of circle;
➢ Graph a circle in a rectangular coordinate system;
➢ Derive and illustrate the equation of the circle;
➢ Find the center and the radius of the circle of an equation;
➢ Convert the general equation of circle to standard form and vice versa;

What’s In
Activity 1: Recall
You had learned in your previous mathematics in junior high school about
solving a quadratic equation by completing the squares. Let us recall your
knowledge about the subject using these following examples.
In order to find the roots of a certain quadratic equation, the following steps
will be used using completing the square method.

1. Rewrite the equation in the form x2 + bx = c.

2. Add to both sides the term needed to complete the square.
3. Factor the perfect square trinomial.
4. Solve the resulting equation by using the square root property.
1. Solve the equation x2 + 8x + 5 = 0 by completing the square.

Solution: x2 + 8x + 5 = 0

x2 + 8x = -5 Rewrite the equation in the form x2 + bx = c.

x2 + 8x + 16 = -5 + 16 Add the appropriate constant to complete the square.

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(x + 4)2 = 11 Factoring the perfect square trinomial

x+4=±√11 Solve using the square root method.

𝑥 = −4 ± √11
x1=0.68
x2= -7.38
2. Find the roots of x2 + 10x − 4 = 0 using completing the square method.

x2 + 10x = 4 Rewrite the equation in the form x2 + bx = c.

x2+10x+25=4+25 Add the appropriate constant to complete the square.

(x+5)2=29 Factoring the perfect square trinomial

(x+5)=±√29 Solve using the square root method.

𝑥 = −5 ± √29
X1=0.39
X2= -10.39
Completing the square method is useful in discussing conic sections specially
the equation of a circle.

What’s New

Geometric Figures or shapes are use in architectural designs. For this activity,
identify the following shapes as circle, parabola, ellipse, or hyperbola as shown in
the pictures being used in real-life. Write your answer on the space provided.

1) _ 2) 3) _ 4)

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 5) _ 6) 7) 8)

Photo source: file:///C:/Users/admin/Desktop/conic-sections.pdf and https://bit.ly/3iLqvXz

Does the activity ignite your interest to study more about geometric shapes
particularly different conic sections like your answers in the activity? Can you name
other architectural designs not in the pictures that used the idea of geometrical
shapes? Does the shape matters on the durability, functionality and artistic designs?
Studying this module will help you appreciate nature and man’s creation that
would help daily life activities.

What is It
We present the conic sections, a particular class of curves which sometimes
appear in nature and which have applications in other fields. In this lesson, we first
illustrate how each of these curves is obtained from the intersection of a plane and a
cone, and then discuss the first of their kind, circles. The other conic sections will be
covered in the next lessons.
Conic sections (or conics), is a curved formed by a plane passing through a
double-napped circular cone. One of the first shapes we learned, a circle, is a conic.
When you throw a ball, the trajectory it takes is a parabola. The orbit taken by each
planet around the sun is an ellipse. Properties of hyperbolas have been used in the
design of certain telescopes and navigation systems. We will discuss circles in this
lesson, leaving parabolas, ellipses, and hyperbolas for subsequent lessons.
• Circle (Figure 1.1) – is a special case of ellipse in which the plane is
perpendicular to the axis of the cone.
• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form
a bounded curve
• Parabola (Figure 1.2) – the plane is parallel to a generator line of the cone
• Hyperbola (Figure 1.3) – the intersection is an unbounded curve and the
plane is not parallel to a generator line of the cone and the plane intersects
both halves of the cone.

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 •

Figure 1.1 Figure 1.2 Figure 1.3

We can draw these conic sections on a rectangular coordinate plane and find
their equations. To be able to do this, we will present equivalent definitions of these
conic sections in subsequent sections, and use these to find the equations.
There are other ways for a plane and the cones to intersect, to form what are
referred to as degenerate conics: a point, one line, and two lines. See Figures 1.4, 1.5,
and 1.6.

Figure 1.4 Figure 1.5 Figure 1.6

The graph of the second degree equation of the form 𝐴𝑥2 + 𝐵𝑥 + 𝐶𝑦2 + 𝐷𝑥 +
𝐸𝑦 + 𝐹 = 0 is determine by the values of 𝐵2 − 4𝑎𝑐.
Table 1
Graphs of Quadratic Equations
Conic Section Value of 𝐵2 − 4𝑎𝑐 Eccentricity
Circle 𝐵2 − 4𝑎𝑐<0 or A=C 𝑒=0
Parabola 𝐵2 − 4𝑎𝑐 = 0 𝑒=1
Ellipse 𝐵2 − 4𝑎𝑐 < 0, 𝐵 G 0 𝑜𝑟 𝐴 G 𝐶 0<𝑒<1
Hyperbola 𝐵2 − 4𝑎𝑐 > 0 𝑒>1

Example1.1 Determine the type conic section that each general equation will
produce.

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 1. 2𝑥2 + 4𝑥𝑦 + 3𝑦2 + 12𝑦 − 1 = 0 3. 3𝑥2 + 3𝑦2 + 18𝑥 − 16𝑦 + 31 = 0
2. 3𝑥2 − 2𝑦2 + 6𝑥 + 10𝑦 − 16 = 0 4. 𝑥2 + 4𝑥𝑦 − 4𝑥 + 𝑦2 − 12𝑦 = 0

Solutions: We will collect all the values of A, B, C in each equation. Then solve for
the value of 𝐵2 − 4𝑎𝑐. Interpret the result based on table 1.
1. 2𝑥2 + 4𝑥𝑦 + 3𝑦2 + 12𝑦 − 1 = 0
𝐴 = 2, 𝐵 = 4, 𝐶 = 3
𝐵2 − 4𝑎𝑐 = 42 − 4(2)(3) = 16 − 24 = −8 < 0
Note that 𝐵 G 0 and 𝐴 G 𝐶. Thus, the conic section is an ellipse.
2. 3𝑥2 − 2𝑦2 + 6𝑥 + 10𝑦 − 16 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = −2
𝐵2 − 4𝑎𝑐 = 02 − 4(3)(−2) = 0 + 24 = 24 > 0
Thus, the conic section is hyperbola.
3. 3𝑥2 + 3𝑦2 + 18𝑥 − 16𝑦 + 31 = 0
𝐴 = 3, 𝐵 = 0, 𝐶 = 3
𝐵2 − 4𝑎𝑐 = 02 − 4(3)(3) = 0 − 36 = −36 < 0
Note that 𝐵 = 0 and 𝐴 = 𝐶. Thus, the conic section is a circle.
4. 4𝑥2 + 4𝑥𝑦 − 4𝑥 + 𝑦2 − 12𝑦 = 0
𝐴 = 4, 𝐵 = 4, 𝐶 = 1
𝐵2 − 4𝑎𝑐 = 42 − 4(4)(1) = 16 − 16 = 0
The conic section is a parabola.

Definition and Equation of a Circle

A circle may also be considered a special kind of ellipse (for the special case
when the tilted plane is horizontal). As we get to know more about a circle, we will
also be able to distinguish more between these two conics.
See Figure 1.7, with the point 𝐶(3,1)shown. From the figure, the distance of
𝐴(−2,1) from C is AC = 5. By the distance formula, the distance of 𝐵(6,5) from C is
BC = √(6 − 3)2 + (5 − 1)2 = 5. There are other points P such that PC = 5. The
collection of all such points which are 5 units away from C, forms a circle.

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 Figure 1.7 Figure 1.8

Let C be a given point. The set of all points P having the same distance
from C is called a circle. The point C is called the center of the circle,
and the common distance its radius.

The term radius is both used to refer to a segment from the center C to a point
P on the circle, and the length of this segment.
See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0.
A point P(x, y) is on the circle if and only if PC = r. For any such point then, its
coordinates should satisfy the following.

𝑃𝐶 = 𝑟

√(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟

(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
This is the standard equation of the circle with center C(h, k) and radius r. If
the center is the origin, then h = 0 and k = 0. The standard equation is then
x2 + y2 = r2.

Example 1.2. Graph the circle 𝑥2 + 𝑦2 = 9.

The given equation is in standard form with center at the origin C(0,0) and
radius. We can rewrite the equation into this form 𝑥2 + 𝑦2 = 32.following the standard
equation x2 + y2 = r2. Thus, r=4. To be able to graph the circle, we take all the points
that are 4 units from the center (0,0) to all directions along the plane. See Figure 1.9
below.

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 Figure 1.9

Example1.2. In each item,

give the standard equation
of the circle satisfying the
given conditions.
(1) center at the origin, radius 4
(2) center (−4, 3), radius √7
(3) circle in Figure 1.7
(4) circle A in Figure 1.10
(5) circle B in Figure 1.10
(6) center (5, 6), tangent to the y-axis Figure 1.10
(7) center (5, −6), tangent to the x-axis
(8) It has a diameter with endpoints A(−1, 4) and B(4, 2).
Solution:
(1) Since the center of the circle is the origin, then h=0 and k=0, the standard
equation of the circle given radius (r=4) is 𝑥2 + 𝑦2 = 16.
(2) Since the center of the circle is not the origin, we will use the standard
equation (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
2
(𝑥 − (−4))2 + (𝑦 − 3) = (√7)
(𝑥 + 4)2 + (𝑦 − 3)2 = 7
(3) The center is (3, 1) and the radius is 5
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(𝑥 − 3)2 + (𝑦 − 1) = (5)2
(𝑥 − 3)2 + (𝑦 − 1)2 = 25
(4) By inspection, the center is (- 2, -1) and the radius is 4.
The equation is (x + 2)2 + (y + 1)2 = 16.
(5) Similarly by inspection, we have (x − 3)2 + (y − 2)2 = 9.
(6) The center is 5 units away from the y-axis, so the radius is
r= 5 (you can make a sketch to see why). The equation is

15
 (x− 5)2 + (y + 6)2 = 25.
(7) Similarly, since the center is 6 units away from the x-axis,
the equation is (x − 5)2 + (y + 6)2 = 36. 4+2 3
(8) The center C is the midpoint of A and B: C= −1+4 , The
= ( , 3).
2 2 2

29
radius is then r=AC=√ (−1 − 3) + (4 − 3)2 = √ . The circle has an
2 4

3 29
equation (𝑥 − )2 + (𝑦 − 3) =
2
.
2 4

After expanding the standard equation, say for example the standard form in
3 29
example 1.8, (𝑥 − )2 + (𝑦 − 3)2 = , can be written as 𝑥2 + 𝑦2 − 3𝑥 − 6𝑦 + 4 =
2 4

0, an equation of the circle in general form. If the equation of a circle is given in the
general form

Ax2 + By2 + Cx + Dy + E = 0, A ≠ 0,

x2 + y2 + Cx + Dy + E = 0,
we can determine the standard form by completing the square in both variables.
Steps below show the knowledge we had in our previous activity about completing
the square.

3 29
2 4
9 29
−3𝑥 + + 𝑌2 − 6𝑦 + 9 =
4 4
29 9
𝑥2 + 𝑌2 − 3𝑥 − 6𝑦 = − −9
4 4
𝑥2 + 𝑌2 − 3𝑥 − 6𝑦 = −4

𝑥2 + 𝑌2 − 3𝑥 − 6𝑦 + 4 = 0

In completing the square like the expression (𝑥2 + 14𝑥), means determining
the term to be added that will produce a perfect polynomial square. Since the
coefficient of x2 is already 1, we take half the coefficient of x and square it, and
we get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To complete the
16
square in, say, 3x2 + 18x, we factor the coefficient of x2 from the expression:
3(x2 + 6x), then add 9 inside. When completing a

17
square in an equation, any extra term introduced on one side should also be
added to the other side.

Example 1.3. Identify the center and radius of the circle with the given equation in
each item. Sketch its graph, and indicate the center.

(1) 𝑥2 − 6𝑥 + 𝑦2=7

(2) 𝑥2 + 𝑦2 − 14𝑥 + 2𝑦 = −14

(3) 16𝑥2 + 16𝑦2 + 96𝑥 − 40𝑦 = 315

Solution. The first step is to rewrite each equation in standard form by completing the
square in x and in y. From the standard equation, we can determine the center and
radius.

(1) 𝑥2 − 6𝑥 + 𝑦2 = 7 (Given)

𝑥2 − 6𝑥 + 9 + 𝑦2 = 7 + 9 (Adding 9 both sides)

𝑥2 − 6𝑥 + 9 + 𝑦2 = 16 (Simplify right side of the equation)

(𝑥 − 3)2 + 𝑦2 = 16 (Factoring into perfect square binomial)

(𝑥 − 3)2 + 𝑦2 = 42

Center (3,0), r=4, see Figure 1.11

(2) 𝑥2 + 𝑦2 − 14𝑥 + 2𝑦 = −14 (Given)

𝑥2 − 14𝑥 + 𝑦2 + 2𝑦 = −14 (Rearrange by terms)

𝑥2 − 14𝑥+49 + 𝑦2 + 2𝑦 + 1 = −14 + 49 + 1 (Adding 49 & 1 both sides)

𝑥2 − 14𝑥+49 + 𝑦2 + 2𝑦 + 1 = 36 (Simplify right side of the equation)

(𝑥 − 7)2 + (𝑦 + 1)2 = 36 (Factoring into perfect square binomial)

(𝑥 − 7)2 + (𝑦 + 1)2 = 62

Center (7,-1), r=6, see Figure 1.12

(3) 16𝑥2 + 16𝑦2 + 96𝑥 − 40𝑦 = 315 (Given)

16𝑥2 + 96𝑥 + 16𝑦2 − 40𝑦 = 315 (Rearrange by terms)

5
16(𝑥2 + 6𝑥) + 16(𝑦2 − 𝑦) = 315 Applying common monomial factor
2

18
 5 25
( ) 25 ( )
2 2
16(𝑥 + 6𝑥 + 9) + 16 (𝑦 − 𝑦+ ) = 315 + 16 9 + 16( ) (Adding 16 9 +
2 16 16
25
16( ) both sides)
16
5 2
16(𝑥 + 3)2 + 16(𝑦 − ) = 484 (Simplify & factoring into perfect
4
square binomial)
5 1
1 · {16(𝑥 + 3)2 + 16(𝑦 − )2} = 484 · (Multiplying 1 both sides)
16 4 16 16

5 121
(𝑥 + 3)2 + (𝑦 − 4)2 = 4
(Simplify)

5 11 2
(𝑥 + 3)2 + (𝑦 − )2 = ( )
4 2

Center (-3, 5), r=11 or 5.5, see Fig. 1.13

4 2

Figure 1.11 Figure 1.12 Figure 1.13

In the standard equation (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2 both the two squared terms
on the left side have coefficient 1. This is the reason why the preceding example, we
multiplied 1 at the last equation.
16

What’s More
Activity 1.1: Let Me try!

Let us find out if you really understood the discussed concept by answering these
follow-up exercises.
1. Determine the type of conic section that each general
equation will produce. Show your solution.

a. 𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 + 1 = 0 b. 𝑥2 + 𝑦2 + 8𝑥 − 4𝑦 − 2 = 0

19
c. 5𝑥2 + 5𝑦2 − 9𝑥 − 14𝑦 + 26 = 0 d. 9𝑥2 + 16𝑦2 − 54𝑥 + 4𝑥𝑦 − 64𝑦 + 1 = 0

20
2. In each item, give the standard equation of the circle satisfying the
given conditions.

a. center at the origin, contains (0, 3) b. center (1, 5), diameter 8

c. circle A in Figure 1.10 d. circle B in Figure 1.10

e. circle C in Figure 1.10 f. center (-2,-3), tangent to the y- axis

g. center ( -2,-3), tangent to the x- h. contains the

axis points ( -2,0) and (8, 0),
radius 5.

21
 Figure1.10
3. Identify the center and radius of the circle with the given equation in each item.
Sketch its graph, and indicate the center.
(a) x2 + y2 + 8y = 33

b.4x2 + 4y2 − 16x + 40y + 67 = 0

c. 4x2 + 12x + 4y2 + 16y − 11 = 0

22
d. x2 + y2 - 6x + 4y +4 = 0

e. x2 + y2 – 4x - 8y + 20 = 0

23
f. What is the equation of a circle having a diameter with endpoints at A(4,5) and
B (-2,3). Sketch the graph.

g. What is the general equation of a circle whose center is at M(5,-3) and whose
radius is 4 units. Sketch the graph.

What I Have Learned

Let me check your knowledge by filling the blanks with a correct symbols/letter or
terms in order to complete the statement/s.
There are four types of conic sections. When the plane is perpendicular to the
axis of the cone and intersects each generator, a/an (1). is formed.
However, when the plane is tilted slightly so that it intersects each generator, but only
intersects one nappe of the cone, a/an (2)_ is formed. When the
plane is tilted further so that it is parallel to one and only one generator and intersects
only one nappe of the cone, a /an(3) _ is formed. A
hyperbola is generated when plane intersects both nappes.

24
 Given a general equation of the conic sections, we can determine what type
of conics by collecting the values (4)_ , (5) and (6) . Then solve the value of
𝐵2 − 4𝑎𝑐 and interpret the result based on the table of the graphs of quadratic
equations.
The first type of conic section is circle. It is defined as a set of all points in a
plane equidistant from a fixed point called (7) _ of the circle and the constant
equal distance is called (8)_ _. The standard form of the equation of a circle is
((𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟 with the center: (9)
2 2 2 , and radius: (10) _. However, when
the circle has a center at origin: C(0,0), the standard equation would be(11) .
This equation of the circle Ax2 + By2 + Cx + Dy + E = 0, is called (12) .
This equation can be converted into standard form using completing of (13)
_. To graph the equation of the circle into thecoordinate plane, use
the center represented by (14) . After locating the center, use the value of the
(15)_ to move in all directions and then connectthe dots
to form a circle.

What I Can Do

Performance Task: Let’s do this!

Materials: Grid paper, Philippine new coins (1, 5 and 10 peso coins), ruler, and pen.
Procedures:
A. Center (0,0)
1. Use the grid paper below and draw 3 sets of Cartesian coordinate plane. Use
1 unit in labelling the x- and y- axes.
2. Locate the center (0,0) of each 3 sets of Cartesian plane by putting a visible
dots.
3. Place each coin (designated set) at the center or on the dot in each Cartesian
plane. Using pen, draw a circle by tracing the circumference (edge) of the
each coin.
4. After you draw, remove the coins. From the center (dot), draw a line to any
point of the circle.
5. Get your ruler, measure in centimeters the line (radius) you created in each
Cartesian plane and record the values.
6. Solve the equation of each circle using the obtained value of the radius.
7. Compare the equation obtained and make an observation note of the activity.

25
 Set A (1 peso coin)

Set B (5 peso coin)

Set C (10 peso coin)

26
Observation Note:

B. Center(h,k)

1. Use the grid paper below and draw 3 sets of Cartesian coordinate plane. Use
1 unit in labelling the x- and y- axes.
2. Locate the center (2,3) of each 3 sets of Cartesian plane by putting a visible
dots.
3. Place each coin (designated set) at the center or on the dot in each Cartesian
plane. Using pen, draw a circle by tracing the circumference (edge) of the
each coin.
4. After you draw, remove the coins. From the center (dot), draw a line to any
point of the circle.
5. Get your ruler, measure in centimeters the line (radius) you created in each
Cartesian plane and record the values.
6. Solve the equation of each circle using the obtained value of the radius.
7. Compare the equation obtained and make an observation note of the activity.

27
 Set A (1 peso coin)

Set B (5 peso coin)

Set C (10 peso coin)

28
Observation Note:

Online connect! For additional knowledge and information about the topics please
visit the link indicated below.

1. https://www.youtube.com/watch?v=HO2zAU3Eppo

2. https://www.youtube.com/watch?v=auD46ZWZxQo

3. https://www.youtube.com/watch?v=JUvo3GrgWHk

4. shorturl.at/bKU67

29
LESSON
The Parabola
2

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ define a parabola;
➢ determine the standard form of equation of parabola
➢ graph a parabola in a Cartesian coordinate system;
➢ describe and discuss the parts of parabola;
➢ convert the general equation of parabola to standard form and vice
versa.

What’s In

Activity 2.1: Recall

Let us recall previous lessons in quadratic function. Write the correct answer
of the following questions below.

What is the standard form of a quadratic function? _

What is Vertex form of a quadratic function? _
What do you call the graph of quadratic function?
Recalling this concepts are useful in studying the new lesson in this module
as you go along the parts.

30
 What’s New
A parabola is one of the conic sections. You have learned from theprevious
lesson that it is formed when the plane intersects only one cone to form an
unbounded curve. The same thing with circle, you will learn more about the
opening of the graph, equation in standard form and general form.
Let us discover some important parts of the graph of a parabola.

Figure 2.1
Follow-up Activity! Study figure 2.1 and fill in the blank to complete the statement.
Knowing that the graph of quadratic function is a parabola and you already had
the idea on its part. But, there are new parts to be introduced in teaching parabola as
one of the conics. A parabola is the set of all points in a plane equidistant from a fixed
point and a fixed line. The fixed point is called _ andthe fixed line is called the _
_.The of the parabola is the midpoint
of the perpendicular segment from the focus to the directrix, while the line that passes
through it and the focus is called the _. The line segmentthrough the
focus perpendicular to the axis of symmetry is called the _ _
whose length is 4ac.

What is It

Parabola is the set of all points in a plane equidistant from a fixed point and
a fixed line.

31
 This part presents how to convert general form of a parabola to its standard
form and vice versa. Table 2.1 presents the general and standard equations of the
parabola with vertex at origin and at (h,k).

Table 2.1
General and Standard Equations of the Parabola
Vertex General Form Standard Form
𝑦2+ Dx+ F=0 𝑦2 = 4𝑐𝑥
𝑦2 = −4𝑐𝑥
(0,0)
𝑦2+ Dx+ F=0 𝑥2 = 4𝑐𝑦
𝑥2 = −4𝑐𝑦
𝑦2+ Dx+ Ey + F=0 (𝑦 − 𝑘)2 = 4𝑐(𝑥 − ℎ)
(𝑦 − 𝑘)2 = −4𝑐(𝑥 − ℎ)
(h,k)
𝑥2+ Dx+ Ey + F=0 (𝑥 − ℎ)2 = 4𝑐(𝑦 − 𝑘)
(𝑥 − ℎ)2 = −4𝑐(𝑦 − 𝑘)

Example 2.1Convert the general equations to standard form:

a. 𝑦2 + 12𝑥 + 2𝑦 + 25 = 0 b. 2𝑥2 − 12𝑥 − 𝑦 + 16 = 0
Solution: a. 𝑦2 + 12𝑥 + 2𝑦 + 25 = 0
𝑦2 + 2𝑦 = −12𝑥 − 25
𝑦2 + 2𝑦 + 1 = −12𝑥 − 25 + 1
(𝑦 + 1)2 = −12𝑥 − 24
(𝑦 + 1)2 = −12(𝑥 + 2)
b. 2𝑥2 − 12𝑥 − 𝑦 + 16 = 0
2𝑥2 − 12𝑥 = 𝑦 − 16
2(𝑥2 − 6𝑥) = 𝑦 − 16
2(𝑥2 − 6𝑥 + 9) = 𝑦 − 16 + 2(9)
2(𝑥 − 3)2 = 𝑦 − 16 + 18
2(𝑥 − 3)2 = 𝑦 + 2
𝑦+2
(𝑥 − 3)2 =
2
Example 2.2 Convert the general equations to standard form:

32
 a. (𝑦 − 3)2 = 7(𝑥 − 8) b. (𝑥 + 2)2 = −8(𝑦 + 5)
Solution: a. (𝑦 − 3)2 = 7(𝑥 − 8)
𝑦2 − 6𝑦 + 9 = 7𝑥 − 56
𝑦2 − 6𝑦 + 9 − 7𝑥 + 56 = 0
𝑦2 − 7𝑥 − 6𝑦 + 65 = 0
b. (𝑥 + 2)2 = −8(𝑦 + 5)
𝑥2 + 4𝑥 + 4 = −8𝑦 − 40
𝑥2 + 4𝑥 + 4 + 8𝑦 + 40 = 0
𝑥2 + 4𝑥 + 8𝑦 + 44 = 0
Consider the point F(0, 2) and the line ℓ having equation y = 2, as shown in
Figure 1.14. What are the distances of A(4, 2) from F and from ℓ? (The latter is taken
as the distance of A from Aℓ, the point on ℓ closest to A). How about the distances of
B(−8, 8) from F and from ℓ (from Bℓ)?

AF = 4 and AAℓ = 4
BF=√(−8 − 0)2 + (8 − 2) = 10 and
2 BBℓ = 10
There are other points P such that PF = PPℓ (where Pℓ is the closest point on
line ℓ). The collection of all such points forms a shape called a parabola.

Let F be a given point, and ℓ a given line not containing F. The set
of all points P such that its distances from F and from ℓ are the
same, is called a parabola. The point F is its focus and the line ℓ its
directrix.

Figure 1.14

33
 Figure 1.15

Consider a parabola with focus F(0, c) and directrix ℓ having equation y = c. See
Figure 1.26. The focus and directrix are c units above and below, respectively, the
origin. Let P(x, y) be a point on the parabola so PF = PPℓ, where Pℓ is the point on ℓ
closest to P. The point P has to be on the same side of the directrix as the focus (if P
was below, it would be closer to ℓ than it is from F).

PF = PPℓ
√𝑥2 + (𝑦 − 𝑐)2= y − (−c) = y + c

x2 + y2 − 2cy + c2 = y2 + 2cy + c2

x2 = 4cy
The vertex V is the point midway between the focus and the directrix. This
equation, x2 = 4cy, is then the standard equation of a parabola opening upward with
vertex V (0, 0).

Suppose the focus is F(0,-c) and the directrix is y = c. In this case, a point P
on the resulting parabola would be below the directrix (just like the focus). Instead of
opening upward, it will open downward. Consequently, PF = √x2 + (y + c)2 and
PPℓ = 𝑐 − 𝑦(you may draw a version of Figure 1.15 for this case). Computations
similar to the one done above will lead to the equation x2 = −4cy.

We collect here the features of the graph of a parabola with standard equation
x2 = 4cy or x2 = −4cy, where c > 0.
F

Figure 1.16

34
 Figure 1.17

(1) vertex : origin V (0, 0)

If the parabola opens upward, the vertex is the lowest
point. If the parabola opens downward, the vertex is the
highest point.
(2) directrix : the line y = −c or y = c
• The directrix is c units below or above the vertex.
(3) focus: F(0, c) or F(0, −c)
• The focus is c units above or below the vertex.
Any point on the parabola has the same distance from
the focus as it has from the directrix.
(4) axis of symmetry : x = 0 (the y-axis)
This line divides the parabola into two parts which are
mirror images of each other.

Example 2.3. Determine the focus and directrix of the parabola with the given
equation. Sketch the graph, and indicate the focus, directrix, vertex, and axis of
symmetry.

(1) x2 = 12y (2) x2 = −6y

Solution.

(1) The vertex is V (0, 0) and the parabola opens upward. From
4c = 12, c = 3. The focus, c = 3 units above the vertex, is F(0,
3). The directrix, 3 units below the vertex, is y = −3. The axis
of symmetry is x = 0.

35
 Figure 1.18
(2) The vertex is V (0,0) and the parabola opens downward. From 4c=6,
c=3. The focus, c=3 units below the vertex, is F (0,- 3). The directrix, 3 units above the
2 2 2 2
vertex, is y=3. The axis of symmetry is x=0.
2

Figure 1.19

Example 2.4. What is the standard equation of the parabola in Figure 1.14?

Solution: From the figure, we deduce that c=2. The equation is thus 𝑥2 = 8𝑦

In all four cases below, we assume that c > 0. The vertex is V (h, k), and it lies
between the focus F and the directrix ℓ. The focus F is c units away from the vertex
V, and the directrix is c units away from the vertex. Recall that, for any point on the
parabola, its distance from the focus is the same as its distance from the directrix.

36
 (x − h) = 4c(y − k) (y − k) = 4c(x − h)

(x − h) = −4c(y − k) (y − k) = −4c(x −h)

directrix ℓ: horizontal directrix ℓ: vertical

axis of symmetry: x=h, vertical axis of symmetry: y=k, horizontal

Note the following observations:

• The equations are in terms of x h and y k: the vertex coordinates are subtracted
from the corresponding variable. Thus, replacing both h and k with
0 would yield the case where the vertex is the origin. For instance, this
replacement applied to (x h)2 = 4c(y k) (parabola opening upward) would yield
x2 = 4cy, the first standard equation we encountered (parabola opening
upward, vertex at the origin).
• If the x-part is squared, the parabola is “vertical”; if the y-part is squared, the
parabola is “horizontal.” In a horizontal parabola, the focus is on the left or right
of the vertex, and the directrix is vertical.
• If the coefficient of the linear (non-squared) part is positive, the parabola opens
upward or to the right; if negative, downward or to the left.

Example 2.4. Figure 1.20 shows the graph of parabola, with only its focus and vertex
indicated. Find its standard equation. What are its directrix and its axis of symmetry?

Solution. The vertex is V (5, −4) and the focus is F(3, −4). From these, we deduce the
following: h = 5, k = −4, c = 2 (the distance of the focus from the vertex). Sincethe
parabola opens to the left, we use the template (y − k)2 =−4c(x − h). Our equation is
(y + 4)2 = −8(x − 5).

Its directrix is c = 2 units to the right of V , which is x = 7. Its

axis is the horizontal line through V : y = −4.

37
 Figure 1.20

The standard equation (𝑦 + 4)2 = −8(𝑥 − 5)from the preceding example

can be rewritten as y2 + 8x + 8y 24 = 0, an equation of the parabola in general
form.
If the equation is given in the general form Ax2 + Cx + Dy + E = 0 (A and C
are nonzero) or By2 + Cx+ Dy + E = 0 (B and C are nonzero), we can determine
the standard form by completing the square in both variables.

Example 2.5. Determine the vertex, focus, directrix, and axis of symmetry of the
parabola with the given equation. Sketch the parabola, and include these points and
lines.
(1) y2 − 5x + 12y = −16 (2) 5x2 + 30x + 24y = 51
Solution.

(1) We complete the square on y, and move x to the other side.

y2 + 12y = 5x − 16
y2 + 12y + 36 = 5x − 16 + 36 = 5x + 20
(y + 6)2 = 5(x + 4)
The parabola opens to the right. It has vertex V (−4, −6). From 4c = 5, we get c = 5 =
1.25. The focus is c = 1.25 units to the right of V: F(−2.75, −6). The (vertical) directrix

38
is c = 1.25 units to the left of V: x = −5.25. The (horizontal) axis is through V: y = −6.

(2) We complete the square on x, and move y to the other side.

5𝑥2 + 30𝑥 = −24𝑦 + 51

5(𝑥2 + 6𝑥 + 9) = −24𝑦 + 51 + 5(9)
5(𝑥 + 3)2 = −24𝑦 + 96
5(𝑥 + 3)2 = −24(𝑦 − 4)
−24
(𝑥 + 3)2 = (𝑦 − 4)
5
In the last line, we divided by 5 for the squared part not to have any
coefficient. The parabola opens downward. It has vertex V(-3,4). From 4c = 24, we
5
get c = 6 = 1.2. The focus is c = 1.2 units below V:F(−3, 2.8). The (horizontal)
mdirectrix is c = 1.2 units above V: y = 5.2. The (vertical) axis is through V : x = −3.

Example 2.6 A parabola has focus F(7, 9) and directrix y = 3. Find its standard
equation.
Solution. The directrix is horizontal, and the focus is above it. The parabola then
opens upward and its standard equation has the form (x - h)2 = 4c(y-k). Since the
distance from the focus to the directrix is 2c = 9-3 = 6, then c = 3. Thus, the vertex is
V (7, 6), the point 3 units below F. The standard equation is then (x − 7) = 12(y − 6).

What’s More
Activity 2.1: Let Me try!
Let us find out if you really understood the discussed concept by answering
these follow-up exercises.

39
1. Convert the following general form to standard form of a parabola.

a. 4𝑥2 − 24𝑥 − 40𝑦 − 4 = 0 c. 𝑦2 − 6𝑥 + 2𝑦 − 17 = 0

b. 𝑥2 − 2𝑥 − 6𝑦 + 25 = 0 d. 2𝑦2 − 𝑥 − 12𝑦 + 18 = 0

2. Convert the following standard form to general form of a parabola.

a. (𝑦 + 1)2 = −2(𝑥 + 2) c. (𝑥 − 3)2 = 12(𝑦 − 4)

3(𝑥−3)
b. (𝑥 − 1)2 = 8(𝑦 − 3) d. (𝑦 − 3)2 = −
2

40
3. Determine the vertex, focus, directrix, and axis of symmetry of the parabola with
the given equation. Sketch the parabola, and include these points and lines

a. y2 = 20x

b. 3x2 = −12y

c. . x2 − 6x − 2y + 9 = 0

41
d. 3y2 + 8x + 24y + 40 = 0

4. A parabola has focus F( 11, 8) and directrix x = -17. Find its standard equation

5. Find the equation of a parabola with vertex at the origin whose properties are
given below.
a. length of latus rectum is 10 and parabola opens downward
b. equation of directrix is y=8
c. focus at (0,4)
d. diretcrix is x=7

42
6. Determine the standard equation of the parabola in Figure below given only its
vertex and directrix. Then determine its focus and axis of symmetry

7. Determine the standard equation of the parabola in the figure given only its focus
and vertex. Determine its directrix and axis of symmetry.

8. Find an equation of the parabola with vertex at (-1,4) and y=5 as the line of
directrix. Draw the general appearance of this graph.

43
 What I Have Learned

Let me check your knowledge by filling the blanks with a correct symbols/letter
or terms in order to complete the statement/s.
1. A parabola is the set of all points in the plane that are equidistant from a fixed
point called the _ and fixed line called the of the parabola.
2. The graph of the equation 𝑥2 = 4𝑐𝑦 is a parabola with focus F( , ) and directrix

y=_ _. So the graph of 𝑥2 = 12𝑦 is a parabola with focus F( , ) and directrix

y=_ .
3. The graph of the equation 𝑦2 = 4𝑐𝑥 is a parabola with focus F( , ) and directrix
x= . So the graph of 𝑦2 = 12𝑥 is a parabola with focus F( , ) and directrix
x= .
4. Label the focus, directrix and vertices on the graphs given for the parabolas
below.
a. 𝑥2 = 12𝑦 of 𝑦2 = 12𝑥

What I Can Do

Performance Task: Let’s do this!

Materials: Flashlight, ball and ring
1. Get a flashlight and perform the task in the picture. A flashlight is held to
form a lighted area on the ground, as shown in the figure. Is it possible to angle

44
the flashlight in such a way that the boundary of the lighted area is a parabola?
Explain your answer.
Answer:

2. In a basketball game, it is crucial to be able to execute a throw which creates a

parabola that can deliver the ball through a hoop with ease. Shooting at a 90 degree
is optimal yet impractical as in order to shoot to this degree, one must be directly under
the rim. This is where a parabola is used to complete the objective of shooting the ball
to acquire points.

Go to the basketball court (if available or accessible, if not innovate) and perform the
following.

a. Standing and facing on one of the posts, move your feet backward 3 times and then
perform shooting the ball. Repeat the process of moving your feet 7 times, 10 times,
12, times 15 times and 20 times and then shoot the ball.

2. While doing the activity of shooting the ball in different distances, what can yousay
on the following?

a. Does shooting the ball create parabolic arc? Explain

b. Do you think the parabolic arcs formed are of the same measurement? Explain

45
c. Are the parabolic arcs formed in shooting the ball are dependent on the distance
of a person throwing the ball? Or are they related? Explain

d. Do you think that a player or famous player studied the shooting style in order to
get the perfect shoot?

Online connect! For additional knowledge and information about the topics please
visit the links/url indicated below.

shorturl.at/eikV1
shorturl.at/cfRT5
https://www.youtube.com/watch?v=ZJf9shWlMz0
shorturl.at/HKSU6

46
 LESSON The Ellipse
3

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ define an ellipse;
➢ determine the standard form of equation of an ellipse;
➢ graph an ellipse in a Cartesian coordinate system;
➢ discuss the parts of an ellipse;
➢ convert the general equation of an ellipse to standard form and vice
versa.

What’s In

Activity 2.1: Recall

Let us recall previous lessons in the graphs of quadratic equations. We will

prove that a given equation without graphing will result to a certain conic section.
Now, let us try circle.

Table 1
Graphs of Quadratic Equations
Conic Section Value of 𝐵2 − 4𝑎𝑐 Eccentricity
Circle 𝐵2 − 4𝑎𝑐<0 or A=C 𝑒=0
Parabola 𝐵2 − 4𝑎𝑐 = 0 𝑒=1
Ellipse 𝐵2 − 4𝑎𝑐 < 0, 𝐵 G 0 𝑜𝑟 𝐴 G 𝐶 0<𝑒<1
Hyperbola 𝐵2 − 4𝑎𝑐 > 0 𝑒>1

1. 𝑥2 + 𝑦2 + 6𝑥 − 16 = 0.
We will collect the values of A, B, and C.
A= 1, B=0, and C=1. Solving for 𝐵2 − 4𝑎𝑐.
𝐵2 − 4𝑎𝑐=02 − 4(1)(1) = 0 − 4 = −4 < 0

47
 Note that B=0, A=C. Thus, the conic section is a circle.

2. 4𝑥2 + 4𝑦2 − 4𝑥 − 12𝑦 + 1 = 0

We will collect the values of A, B, and C.
A= 4, B=0, and C=4. Solving for 𝐵2 − 4𝑎𝑐.
𝐵2 − 4𝑎𝑐=02 − 4(4)(4) = 0 − 64 = −64 < 0
Note that B=0, A=C. Thus, the conic section is a circle.

3. 4𝑥2 + 9𝑦2 − 48𝑥 + 72𝑦 + 144 = 0

We will collect the values of A, B, and C.
A= 4, B=0, and C=9. Solving for 𝐵2 − 4𝑎𝑐.
𝐵2 − 4𝑎𝑐=02 − 4(4)(9) = 0 − 144 = −144 < 0
Note that B=0, A±C. Thus, the conic section is not a circle but an ellipse.
Comparing the three equations, 1 and 2 are both circles, while 3 is an
ellipse. Although the values of B=0 for the three equations, only 1 and 2 have
same values of A and C, while equation 3 has different values of A and C. For
a circle, the values of A and C should be the same, while in ellipse should be
different. Circles and ellipse are quite related in terms of their graphs, but do not
be confused in determining the two by evaluating their A, B, and C values and
solving 𝐵2 − 4𝑎𝑐 in their standard form in order to be précised in sketching the
graph.

Let us show the graph of a circle to prove that A and C are of the same
values.

From the graph, the center is C (h,k) which

is (0,0), and the radius r is 3. Using the
standard form

(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2 ,

(𝑥 − 0)2 + (𝑦 − 0)2 = 32

𝑥2 + 𝑦2 = 9 is the equation of the circle.

A=1 and C=1, then A=C.

Recalling these concepts are useful in teaching ellipse.

48
 What’s New

Unlike circle and parabola, an ellipse is one of the conic sections that most
students have not encountered formally before. Its shape is a bounded curve which
looks like a flattened circle. The orbits of the planets in our solar system around the
sun happen to be elliptical in shape. Also, just like parabolas, ellipses have reflective
properties that have been used in the construction of certain structures. These
applications and more will be encountered in this lesson.
Name the parts of the two figures below using the terms found in the box.

first vertex first focus center radius

second focus second vertex

Can you tell the difference between the graphs? Let us leave the question
unanswered and do some discussions and activities in order for you to understand
better the topic.

What is It

An Ellipse is a set of all points in a plane, the sum whose distances from two
fixed points is constant. The fixed points are called foci.
This section presents how to convert general form of ellipse to its standard
form and vice versa.
Example 3.1. Convert the following general equation to standard form.
a. 9𝑥2 + 8𝑦2 = 288 b. 3𝑥2 + 4𝑦2 + 24𝑥 − 16𝑦 + 52 = 0

49
Solution:
a. 9𝑥2 + 8𝑦2 = 288
9𝑥2 8𝑦2 288
+ =
288 288 288
𝑥2 𝑦 2
+ =1
32 36
2 𝑦2
The standard form is 𝑥 + =1
32 36

b. 3𝑥2 + 4𝑦2 + 24𝑥 − 16𝑦 + 52 = 0

(3𝑥2 + 24𝑥) + (4𝑦2 − 16𝑦) + 52 = 0 Regroup the terms
3(𝑥2 + 8𝑥) + 4(𝑦2 − 4𝑦) = −52 Apply common Factor
3(𝑥2 + 8𝑥 + 16) + 4(𝑦2 − 4𝑦 + 4) = −52 + 3(16) + 4(4) Apply completing the square
3(𝑥2 + 8𝑥 + 16) + 4(𝑦2 − 4𝑦 + 4) = −52 + 48 + 16 Simplify right side
3(𝑥2 + 8𝑥 + 16) + 4(𝑦2 − 4𝑦 + 4) = 12 Simplify right side
3(𝑥 + 4)2 + 4(𝑦 − 2)2 = 12 Factoring completely
3(𝑥+4)2 4(𝑦−2)2 12
+ = Divide both sides by 12
12 12 12
(𝑥+4)2 (𝑦−2)2
+ =1 Simplify
4 3
(𝑥+4)2 (𝑦−2)2
The standard form is + = 1.
4 3
Example 3.2. Convert the following standard form to general form:
(𝑦−2)2 (𝑥−3)2 (𝑥−1)2 (𝑦+1)2
a. + =1 b. + =1
25 9 100 36
Solution:

(𝑦−2)2 (𝑥−3)2
a. + = 1
25 9
(𝑦 − 2)2 (𝑥 − 3)2
225 [ + = 1]
25 9

9(𝑦 − 2)2 + 25(𝑥 − 3)2 = 225

9(𝑦2 − 4𝑦 + 4) + 25(𝑥2 − 6𝑥 + 9 = 225
9𝑦2 − 36𝑦 + 36 + 25𝑥2 − 150𝑥 + 225 = 225
9𝑦2 − 36𝑦 + 36 + 25𝑥2 − 150𝑥 + 225 − 225 = 0
25𝑥2 + 9𝑦2 − 150𝑥 − 36𝑦 + 36 = 0
50
The general form is 25𝑥2 + 9𝑦2 − 150𝑥 − 36𝑦 + 36 = 0

51
 (𝑥−1)2 (𝑦+1)2
b. . + = 1
100 36

(𝑥 − 1)2 (𝑦 + 1)2
3600 [ + = 1]
100 36

36(𝑥 − 1)2 + 100(𝑦 + 1)2 = 3600

36(𝑥2 − 2𝑥 + 1) + 100(𝑦2 + 2𝑦 + 1) = 3600
36𝑥2 − 72𝑥 + 36 + 100𝑦2 + 200𝑦 + 100 = 3600
36𝑥2 − 72𝑥 + 36 + 100𝑦2 + 200𝑦 + 100 − 3600 = 0
36𝑥2 + 100𝑦2 − 72𝑥 + 200𝑦 − 3464 = 0
The general form is 36𝑥2 + 100𝑦2 − 72𝑥 + 200𝑦 − 3464 = 0.

Consider the points F1(−3, 0) and F2(3, 0), as shown in Figure 1.22. What
is the sum of the distances of A(4, 2.4) from F1 and from F2? How about the sum of
the distances of B(and C(0, −4)) from F1 and from F2?
AF1 + AF2 = 7.4 + 2.6 = 10
BF1 + BF2 = 3.8 + 6.2 = 10
CF1 + CF2 = 5 + 5 = 10
There are other points P such that PF1 + PF2 = 10. The collection of all such points
forms a shape called an ellipse.

Figure 1.22 Figure 1.23

52
 Let F1 and F2 be two distinct points. The set of all points P, whose
distances from F1 and from F2 add up to a certain constant, is called
an ellipse. The points F1 and F2 are called the foci of the ellipse.

Given are two points on the x-axis, F1( -c, 0) and F2(c, 0), the foci, both cunits
away from their center (0, 0). See Figure 1.23. Let P(x, y) be a point on the ellipse. Let
the common sum of the distances be 2a (the coefficient 2 will make computations
simpler). Thus, we have PF1 + PF2 = 2a
𝑃𝐹1 = 2𝑎 − 𝑃𝐹2
√(𝑥 + 𝑐)2 + 𝑦2 = 2𝑎 − √(𝑥 − 𝑐)2 + 𝑦2
𝑥2 + 2𝑐𝑥 + 𝑐2 + 𝑦2 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦2 + 𝑥2 − 2𝑐𝑥 + 𝑐2 + 𝑦2
𝑎√(𝑥 − 𝑐)2 + 𝑦2 = 𝑎2 − 𝑐𝑥
𝑎2[𝑥2 − 2𝑐𝑥 + 𝑐2 + 𝑦2] = 𝑎4 − 2𝑎2𝑐𝑥 + 𝑐2𝑥2
(𝑎2 − 𝑐2)𝑥2 + 𝑎2𝑦2 = 𝑎4 − 𝑎2𝑐2 = 𝑎2(𝑎2 − 𝑐2)

𝑏2𝑥2 + 𝑎2𝑦2 = 𝑎2𝑏2 by letting 𝑏 = √(𝑎2 − 𝑐2), so a>b

𝑥2 𝑦 2
+ =1
𝑎2 𝑏2

When we let b=𝑎2 − 𝑐2, we assumed a > c. To see why this is true, look at
∆PF1F2 in Figure 1.23. By the Triangle Inequality, PF1 + PF2 > F1F2, which implies
2a > 2c, so a > c.
We collect here the features of the graph of an ellipse with standard equation
𝑥2 𝑦2
+ = 1,where a>b. Let 2 2 .
𝑎2 𝑏2 𝑐 = √(𝑎 −𝑏 )

Figure 1.24

(1) center : origin (0, 0)

(2) foci : F1(−c, 0) and F2(c, 0)
▪ Each focus is c units away from the center.
53
 ▪For any point on the ellipse, the sum of its distances from the
foci is 2a.
(3) vertices: V1(-1,0) and V2 (1,0)
▪ The vertices are points on the ellipse, collinear with the center
and foci.
▪ If y = 0, then x = ±a. Each vertex is a unit away from the center.
The segment V1V2 is called the major axis. Its length is 2a. It
divides the ellipse into two congruent parts.
(4) covertices: W1(0, −b) and W2(0, b)
▪ The segment through the center, perpendicular to the major
axis, is the minor axis. It meets the ellipse at the covertices. It
divides the ellipse into two congruent parts.
▪ If x = 0, then y = ±b. Each covertex is b units away from the
center.
▪ The minor axis W1W2 is 2b units long. Since a > b, the major
axis is longer than the minor axis.

Example 3.3. Give the coordinates of the foci, vertices, and covertices of the ellipse
𝑥2 𝑦2
with equation + = 1. Sketch the graph, and include these points.
25 9

Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and

𝑐 = √(𝑎2 − 𝑏2) = 4.
foci: F1(−4, 0), F2(4, 0) vertices: V1(−5, 0), V2(5, 0)
covertices: W1(0, −3), W2(0, 3)

Example 3.4. Find the (standard) equation of the ellipse whose foci are F1( 3, 0)
and F2(3, 0), such that for any point on it, the sum of its distances from the foci is 10.
See Figure 1.22.
Solution. We have 2a = 10 and c = 3, so a = 5 and 𝑏 = √(𝑎2 − 𝑐2) = 4. The
2 𝑦2
equation is 𝑥 + =1
25 16
The ellipses we have considered so far are “horizontal” and have the origin as
their centers. Some ellipses have their foci aligned vertically, and some have centers

54
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.

In all four cases below, a > b and The foci F1 and F2 are 𝑐 = √(𝑎2 − 𝑏2)
units away from the center. The vertices V1 and V2 are a units away from the center,
the major axis has length 2a, the covertices W 1 and W2 are b units away from the
center, and the minor axis has length 2b. Recall that, for any point on the ellipse, the
sum of its distances from the foci is 2a.

Center Corresponding Graphs

(0, 0)

𝑥2 𝑦2 𝑥2 𝑦2
+ = 1, 𝑎 > 𝑏 + = 1, 𝑏 > 𝑎
𝑎2 𝑏2 𝑏2 𝑎2

(h, k)

(𝑥−ℎ)2 (𝑦−𝑘)2 (𝑥−ℎ)2 (𝑦−𝑘)2

+ = 1,b>a
𝑎2
+ 𝑏2
= 1,a>b 𝑏 2 𝑎2

major axis: horizontal major axis: vertical

minor axis: vertical minor axis: horizontal

55
 In the standard equation, if the x-part has the bigger denominator, the ellipse is
horizontal. If the y-part has the bigger denominator, the ellipse is vertical
Example 3.5. Give the coordinates of the center, foci, vertices, and covertices of the
ellipse with the given equation. Sketch the graph, and include these points.
(𝑥+3)2 (𝑦−5)2
1. 24
+ 49
= 1
2. 9𝑥2 + 16𝑦2 − 126𝑥 + 64𝑦 = 71
Solution: (1) From 𝑎2 = 49 and 𝑏2 = 24, we have a=7, b=2√6 ≈ 4.9, and
Solution: (1) From 𝑎2 = 49 and 𝑏2 = 24, we have a=7, b=2√6 ≈ 4.9, and

𝑐 = √(𝑎2 − 𝑏2) = 5. The ellipse is vertical.

Center: (-3,5)
Foci: F1(-3,0), F2(-3,10)
Vertices: V1 (-3,-2), V2(-3,12)
Covertices: W1(-3,-2√6 , 5) ≈ (−7.9,5)
W2(-3+2√6 , 5) ≈ (1.9,5)

(2) We first change the given equation to standard form.

9(x2 − 14x) + 16(y2 + 4y) = 71

9(x2 − 14x + 49) + 16(y2 + 4y + 4) = 71 + 9(49) + 16(4)
9(x − 7)2 + 16(y + 2)2 = 576
(x − 7)2 (y + 2)2
+ =1
64 36

We have 𝑎 = 8 and 𝑏 = 6. Thus, 𝑐 = √(𝑎2 − 𝑏2) = 2√7=5.3. The ellipse is horizontal.

Center: (7,-2)
Foci: F1(7-2√7 ,-2) ) ≈ 1.7, −2)
F2 (7+2√7 ,-2) ) ≈ 12.3, −2)

56
Vertices: V1 (-1,-2), V2(15,-2)
Covertices: W1(7,-8), W2(7,4)

Example 3.6. The foci of an ellipse are (-3,-6) and ( -3, 2). For any point on the
ellipse, the sum of its distances from the foci is 14. Find the standard equation of the
ellipse.
Solution. The midpoint (−3, −2) of the foci is the center of the ellipse. The ellipse
is vertical (because the foci are vertically aligned) and c=4. From the given sum, 2a=14 so
a=7. Also, 𝑏 = √(𝑎2 − 𝑐2) = √33. The equation is
(x + 3)2 (y + 2)2
+ =1
33 49

Example 3.7. An ellipse has vertices (2-√61,− 5) and (2+√61,− 5) and its minor axis
is 12 units long. Find the standard equation and its foci.
Solution: The midpoint (2, −5) of the vertices is the center of the ellipse, which is
horizontal. Each vertex is 𝑎 = √61 units away from the center. From the length of the
(X−2)2 (y + 5)2
minor axis, 2b = 12 so b = 6. The standard equation is + = 1. Each
61 36
focus is 𝑐 = √(𝑎2 − 𝑏2) = 5 units away from (2, -5), so their coordinates are (-3,-5)
and (7,-5).

What’s More
Activity 3.1: Let Me try!

Let us find out if you really understood the discussed concept by answering
these follow-up exercises.

57
1. Convert the following general form to standard form of an ellipse.

a. 16𝑥2 + 4𝑦2 − 32𝑥 + 16𝑦 − 32 = 0 c. 9𝑥2 + 4𝑦2 − 72𝑥 − 24𝑦 + 144 = 0

b. 4𝑥2 + 9𝑦2 + 48𝑥 + 72𝑦 + 144 = 0 d. 49𝑥2 + 9𝑦2=441

2. Convert the following standard form to general form of an ellipse .

2 2 2 2
(𝑦−3) (𝑥−4) (𝑥−1) (𝑦−1)
a. + =1 c. + =1
9 4 4 2

(𝑥−2)2 (𝑦−3)2 (𝑦+3)

2
(𝑥−2)
2
b + =1 d. . 4
+ 1 =1
20 36

For numbers 3-6, give the coordinates of the center, foci, vertices, and covertices of
the ellipse with the given equation. Sketch the graph, and include these points.

58
3. .

4.

5. . Give the coordinates of the foci, vertices, and covertices of the ellipse with
𝑥2 𝑦2
equation + = 1. Then sketch the graph and include these points.
169 144

59
6. 25𝑥2 + 9𝑦2 = 225

For numbers 7-10, answer as directed.

7. Find the standard equation of the ellipse whose foci are F1(0,-8)
and F2(0, 8), such that for any point on it, the sum of its distances
from the foci is 34.

60
8. 8. Find an equation of an ellipse with center at (0,0), one focus at
(3,0), and a vertex (-4,0). Sketch the graph.

9. The covertices of an ellipse are (5,6) and (5,8). For any point on the ellipse, the
sum of its distances from the foci is 12. Find the standard equation of the ellipse.

10. An ellipse has foci ( and ( , and its major axis is 10 units
long. Find its standard equation and its vertices.

61
 What I Have Learned
Let me check your knowledge by filling the blanks with a correct
symbols/letter or terms in order to complete the statement/s.
1. An ellipse is the set of all points in the plane for which the of the
distances from two fixed points F1 and F2 is constant. The points F1 and F2 are called
the _ of the ellipse.
𝑥2 𝑦2
2. The graph of equation + 𝑏2
= 1 with a>b is an ellipse with vertices ( , )
𝑎2
𝑥2 𝑦2
and ( , ) and foci (±𝑐, 0),where c = _ . So the graph of + = 1 is
52 42

an ellipse with vertices (_,_) and (_, ) and foci (_, ) and ( , ).
𝑥2 𝑦2
3. The graph of the equation + = 1 with a>b>0 is an ellipse with
𝑏2 𝑎2
vertices ( , and ( , ) and foci (0,±𝑐)where c= . So the graph
𝑥2 𝑦2
of + = 1 is an ellipse with vertices ( , ) and ( , ) and foci
42 52
( , ) and ( , ).
4. Label the vertices and foci on the graphs given for the ellipses:

𝑥2 𝑦2 𝑥2 𝑦2
a. + b. + =1
52 42 42 52

62
 What I Can Do

DISCOVERY∎ DISCUSSION∎ WRITING

Materials: bond paper, cylindrical bottle, compass
1. A flashlight shines on a wall, as shown in the figure. What is the shape of the
boundary of the lighted area? Explain your answer.

2. Get a piece of bond paper and wrapped around a cylindrical bottle, and then use a
compass to draw a circle on the paper, as shown in the figure. When the paper is
laid flat, is the shape drawn on the paper an ellipse? Explain your findings.

Online connect! For additional knowledge and information about the topics please
visit the links/url indicated below.

shorturl.at/cknx6
shorturl.at/pGTZ1
shorturl.at/agAO6

63
 LESSON The Hyperbola
4

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ Determine hyperbola and its properties.

➢ determine the standard form of equation of hyperbola;
➢ sketch hyperbola given various equations;
➢ discuss the parts of hyperbola;

What’s In

Just like ellipse, a hyperbola is one of the conic sections that most students
have not encountered formally before. 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.

Hyperbolas can be used in so-called “trilateration”, or positioning problems. It is

possible to locate the place from which a sound, such as gunfire, emanates. Long
Range Aid to Navigation (LORAN for short) system, of ship or aircraft utilizes
hyperbolas.

Definition and Equation of a Hyperbola

Given two distinct points F1 and F2 in the plane and a fixed distance d, a
hyperbola is the set of all points (x,y) in the plane such that the absolute value of
the difference of each of the distances from F1 and F2 to (x,y) is d. The points F1 and
F2 are called the foci of the hyperbola.

64
 In the figure above: the distance of F1 to (x1,y1) - distance of F2 to (x1,y1) = d
And the distance of F2 to (x2,y2) - distance of F2 to (x2,y2) = d.
Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall
assume that the center is (0, 0), the vertices are (a, 0) and (−a, 0) and the foci are (c,
0) and (−c, 0). We label the endpoints of the conjugate axis (0, b) and (0, −b). (Although
b does not enter into our derivation, we will have to justify this choice as you shall see
later.) As before, we assume a, b, and c are all positive numbers. Schematically we
have

Since (a, 0) is on the hyperbola, it must satisfy the conditions of hyperbola. That
is, the distance from (−c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal
the fixed distance d. Since all these points lie on the x-axis, we get

distance from (−c, 0) to (a, 0) − distance from (c, 0) to (a, 0) = d

(a + c) − (c − a) = d

2a = d

In other words, the fixed distance d from the definition of the hyperbola is actually
the length of the transverse axis! (Where have we seen that type of coincidence
before?) Now consider a point (x, y) on the hyperbola. Applying the definition, we get

distance from (−c, 0) to (x, y) − distance from (c, 0) to (x, y) =2a

− = 2a

− = 2a

65
 = 2a + (x − c )2 + y 2 , Transpose the second radical to the right
2 2

 (x + c )2 + y 2  =  2a +  , Square both sides of the equation

   

x 2 + 2cx + c 2 + y 2 = 4a2 + 4a + x 2 − 2cx + c 2 + y 2, Cancel like terms

4cx = 4a2 + 4a x 2 − 2cx + c 2 + y 2 , Combine like terms and divide the Eq. by 4

cx − a2 = a (x − c )2 + y 2 , Transpose a2 to the left side of the Eq.

(cx − a ) 2 2 =  a 2
(x − c )2 + y 2  ,Square again both sides of the equation


( )
c 2 x 2 − 2a2cx + a4 = a2 x 2 − 2cx + c 2 + y 2 , Distribute a2 to the quantity

c 2 x 2 − a2 x 2 − a2 y 2 = a2c 2 − a 4 , Factor out the common term

(c 2 ) ( )
− a2 x 2 − a2 y 2 = a2 c 2 − a2 , Regroup the terms
b2 x 2 − a2 y 2 = a2b2,Replace (c 2 − a2 ) with b 2 since b2 = c 2 − a2

x 2 y2
2
− 2 = 1, Divide both sides of the Eq. by a 2b 2
a b

What’s New
Activity 1
You’ll need: Patty paper Ruler Sharpie Compass Colored paper Tape or glue
stick As you do each of the following, be careful not to smudge your work. For any
step that includes the use of a Sharpie, wait about 30 seconds after marking before
you do any folding.
1. Using a Sharpie, draw a point near the center of your wax paper.
2. Label this point 𝑂.
3. Using a pencil and a compass, draw a circle of radius 4 cm†, using the
point you drew in step 1 as the center.
4. Using a Sharpie, draw another point somewhere outside your circle. Where
66
you put the point will affect the final result. Don’t put your point too close to the edge
of the paper or too close to the circle, or it will be difficult to do the rest of the activity.
Try to arrange it so that everyone in your group has their point at different distances
and in different positions from their circle.
5. Label this point 𝐹.
6. Fold the paper so that the point 𝐹 lands on the circle (or the circle lands on
𝐹).

67
 7. Crease the paper.
8. “Slide” the point along the circle just a little bit so that a different place on the
circle is over the point 𝐹.
9. Crease again.
10. Keep sliding, folding and creasing the paper so that different places on
the circle land on the point. The closer together your creases are, the more
refined your shape will be.
11. Keep doing this until you have gone all the way around the circle.
12. Unfold your paper. Do you see a definitive shape?
13. Carefully darken the outline of your shape with a Sharpie. You’ve just
drawn a hyperbola!
14. Tape or glue the edges of your patty paper to a piece of colored paper.
15. Write “The Hyperbola” and your name at the top of the colored paper.

What is It

We collect here the features of the graph of a hyperbola with standard equation.

(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

68
 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 V1V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = b x and y = − b x, the lines ℓ1 and ℓ2 in the figure
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 𝑦2
equation, replace 1 by 0 and in the resulting equation 𝑥
2
− = 0, solve
𝑎 𝑏2
for y.

• To help us sketch the asymptotes, we point out that the asymptotes ℓ 1 and ℓ2
are the extended diagonals of the auxiliary rectangle drawn in Figure 1.45.
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
V1V2.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.

69
70
Activity 2

71
72
73
 Example 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.) − =1 2. ) 4x2 − 5y2 + 32x + 30y = 1
25 9

Solution: 1.) From a2 = 25, and b2 = 9, we have a = 5, b = 3, and

c = a2 + b2 = √34 , 𝑐 ≈ 5.8. The h2 yperbola is vertical. To determine
( ) (x − 7)2
y+2 =0
the asym5ptotes, we write 25 − 9 , which is equivalent to

y+2= (x − 7).
3 We can then solve for y.

center: C(7, -2) foci: F1(7,−2 − 34)  (7,−7.8) and F2 (7,−2 + 34)  (7,3.8)
5 41 5 29
y = x − and y = − x +
asymptotes: 3 3 3 3

The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.

74
Solution 2.) We first change the given equation to standard form.
( )
4 x 2 + 8x − 5(y 2 − 6y) = 1
4(x 2 + 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
5 4 =1

We have a =  2.2 and b = 2. Thus, c = = 3. The hyperbola is horizontal.

(x + 4)2 (y − 3)2
To determine the asymptotes, we write − = 0 which is equivalent
2 5 4
( )
to y - 3 =  x + 4 , and solve for y.
5
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)

2 2 8
asymptotes: y = x+ + 3 and y = − x− +3
5 5 5

The conjugate axis drawn has its endpoints b = 2 units above and below
the center.

75
 What’s More

Let us find out if you really understood the concept about hyperbola by
answering these exercises.

A. Reduce eachof the following equations of hyperbolas in

standard form.
1.) 4x 2 − y 2 + 2y − 5 = 0
2.) 36x2 − 81y2 + 24x + 328 = 0
3.) 49x2 − 25y2 + 98x + 200y + 874 = 0
4.) 28x2 − 64y2 − 28x − 128y − 505 = 0

B. Determine the foci, vertices, and asymptotes of the hyperbola with

2
x2 y
equation - = 1. Sketch the graph, and include these points and
16 33

lines, the transverse and conjugate axes, and the auxillary rectangle.

For each equation of the hyperbola, find the center, foci, vertices, endpoints of
conjugate axis. Determine the equation of the asymptotes and sketch the graph.

1.)
(y + 6)2 −
(x − 4)2 =1 2.) 9x2 + 126x − 16y2 − 96y + 153 = 0
25 39

What I Have Learned

Let me check your knowledge by filling the blanks with a correct

symbols/letter or terms in order to complete the statement/s.
1. A hyperbola is the set of all points in the plane for which the of the
distances from two fixed points F1 and F2 is constant. The points F1 and F2 are
76
called the of the hyperbola.

𝑥2 𝑥2
2. The graph of the equation − = 1 with a>0,b>0 is a hyperbola with vertices
𝑎2 𝑎2
( , _) and ( , ) and foci(±𝑐, 0) where c= _. So the graph of

77
𝑥2 𝑦2 ) and ( _, ) and foci ( _, _)
− = 1 is a hyperbola with vertices ( ,
42 32
and ( _, _).

𝑦2 𝑥2
3. The graph of the equation − = 1 with a>0,b>0 is a hyperbola with
𝑎2 𝑎2
vertices ( , ) and ( _, _) and foci (0,±𝑐), where c= . So the
𝑦2 𝑥2
graph of − = 1 is a hyperbola with vertices ( , ) and ( _, ) and foci
42 32
( , _) and ( ,_ ).

4 Label the vertices, foci, and asymptotes on the graphs given for the hyperbola.
𝑥2 𝑦2 𝑦2 𝑥2
a. − =1 b. − =1
42 32 42 32

What I can Do

DISCOVERY∎ DISCUSSION∎ WRITING

1. Light from a Lamp. The light from a lamp forms a lighted area on a wall, as shown in
the figure. Why is the boundary of this lighted area a hyperbola? How can one hold a
flashlight so that its beam forms a hyperbola on the ground?

78
Online connect! For additional knowledge and information about the topics please visit
the links/url indicated below.

Hyperbola (Part 1) - Conic Sections Class 11 CBSE

https://www.youtube.com/watch?v=WEyYaIWIUp0

Hyperbola (Part 2) - Conic Sections Class 11 CBSE

https://www.youtube.com/watch?v=Ni0gjU8-Pn4

79
 Equation and Important
Characteristics of the
LESSON 5
Different Types of
Conic Sections

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ recognize the equation and important characteristics of the different

types of conic sections discuss the parts of hyperbola;

What’s In

In this lesson, we will identify the conic section from a given equation. We will
analyze the properties of the identified conic section. We will also look at problems
that use the properties of the different conic sections. This will allow us to synthesize
what has been covered so far.

Identifying the Conic Section by Inspection

The equation of a circle may be written in standard form
Ax2 + Ay2 + Cx + Dy + E = 0,

that is, the coefficients of x2 and y2 are the same. However, it does not
follow that if the coefficients of x2 and y2 are the same, the graph is a circle.

80
 For a circle with equation (x − h) + (y − k ) = r 2 , we have r 2  0. This is
2 2

not the case for the standard equations of (A) and (B).

In (A), becase the sum of two squares can only be 0 if and only if each square
1 3
is 0, it follows that x − = 0 and y + = 0. The graph is thus the single point
2 2
1 3
 ,− .
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.

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 + By2 + Cx + Dy + E = 0.

Some terms may vanish, depending on the kind of conic section.

(1) Circle: both x2 and y2 appear, and their coefficients are the same

Ax2 + Ay2 + Cx + Dy + E = 0

Example: 18x2 + 18y2 − 24x + 48y − 5 = 0

Degenerate cases: a point, and the empty set
(2) Parabola: exactly one of x2 or y2 appears
Ax2 + Cx + Dy + E = 0 (D ƒ= 0, opens upward or downward)
By2 + Cx + Dy + E = 0 (C ƒ= 0, opens to the right or left)
Examples: 3x2 - 12x + 2y + 26 = 0 (opens downward)
– 2y2 + 3x + 12y − 15 = 0 (opens to the right)

(3) Ellipse: both x2 and y2 appear, and their coefficients A and B have the
same sign and are unequal
Examples: 2x2 + 5y2 + 8x − 10y − 7 = 0 (horizontal major axis) 4x2
+y2 − 16x − 6y + 21 = 0 (vertical major axis)
81
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 y2 appear, and their coefficients A and B have
different signs
Examples: 5x2 − 3y2 − 20x − 18y − 22 = 0 (horizontal transverse axis)
– 4x2 + y2 + 24x + 4y − 36 = 0 (vertical transverse axis)
Degenerate case: two intersecting lines.

What’s New

Activity 1

Verify a Model
Alex says he canrepresent the enclosedregion as
(x + 6 )2 + 4
Julie says she canrepresent the enclosedregion as
x2 + 12x + 40.
Which student do you agree with? Justify your choice.
You can draw on the picture to justify your thinking.

82
 What is It

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.
(x − 2)
2 (y + 1)2
(1) 4x + 9y − 16x + 18y + 25 = 0 
2 2
+ =0
32 22

 one point (2,-1)

(x − 2)2 (y + 1)
2

(2) 4x2 + 9y2 − 16x + 18y + 61 = 0  + = −1

32 22

 empty set
(x − 2)2 (y + 1)
2

(3) 4x2 − 9y2 − 16x − 18y + 7 = 0  + =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 thedegenerate 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).

83
Activity 2

Classify the graph of the equation as circle, a parabola, an ellipse, or a hyperbola.

Show your proof.

1.) 9x2 + 4y2 − 18x + 16y − 119 = 0

2.) x2 + y2 − 4x − 6y − 23 = 0

3.) 16x2 − 9y2 + 32x + 54y − 209 = 0

4.) x 2 + 4x − 8y + 20 = 0
5.) y 2 + 12x + 4y + 28 = 0

6.) 4x2 + 25y2 + 16x + 250y + 541 = 0

7.) x 2 + y2 + 2x − 6y = 0

8.) y 2 − x2 + 2x − 6y − 8 = 0

9.) 3x2 + 3y2 − 6x + 6y + 5 = 0

10.) 2x2 − 4y2 + 4x + 8y − 3 = 0

What’s More

Activity 3

Mark the point at the intersection of circle 1 and line 1. Mark both points that are on
line 2 and circle 2. Continue this process, marking both points on line 3 and circle 3,
84
and so on. Then connect the points with a smooth curve.

85
Activity 4

An ellipse is the set of points such that the sum of distance from two fixed points is
constant. The two fixed points are called foci.

Circles in the figure are named by letters, circle 1 is circle c, circle 2

is circle d, and so on and so forth

86
 5

Circles in the figure are named by letters, circle 1 is circle c, circle 2

is circle d, and so on and so forth.

87
 What I have Learned

Activity 6

88
 What I can Do
A Activity 7

Online connect! For additional knowledge and information about the topics
please visit the links/url indicated below.

➢ Determining What Type of Conic Section from General

Form
https://www.youtube.com/watch?v=auD46ZWZxQo
➢ Conic Sections - Circles, Ellipses, Parabolas, Hyperbola -
HowTo Graph & Write In Standard Form
https://www.youtube.com/watch?v=PLrgwD9TleU

89
 Solving Situational
LESSON
Problems Involving
6
Conic Sections

What I Need to Know

Upon completion of this lesson, you should be able to:

➢ solve situational problems involving conic sections

What’s In

There are four conics in the conics sections- Parabolas, Circles, Ellipses and
Hyperbolas. We see them everyday, but we just don't notice them. They appear
everywhere in the world and can be man-made or natural. The applications of conics
can be seen everyday all around us. Conics are found in architecture, physics,
astronomy and navigation. If you get lost, you can use a GPS and it will tell you where
you are (a point) and it will lead you to your destination (another point). Bridges,
buildings and statues use conics as support systems. Conics are also used to describe
the orbits of planets, moons and satellites in our universe.

What’s New

Activity 1

Mathematics is fun, isn’t it? To develop your artistic minds and also to apply your
brilliant ideas about the different conic sections you will be task to create or draw a
figure resembling the different types of conic sections. It’s like a tessellation but not a
tessellation because it can have gaps or overlapping which is not true for tessellation.
Use A-4 size bond paper for your drawing. Add colors on your drawing.

90
Use the sample below as your guide.

What Is It

Since conics have so many real-life applications, it is essential that we

know how to solve problems involving conic sections. Below are the
different conic sections that you have learned in the previous lessons
with their corresponding properties and equations that will help you in
solving situational problems involving conics.

91
\
Example of Situational Problems Involving Conics

1. A street with two lanes, each 10 ft wide, goes through a semicircular tunnel with
radius 12 ft. How high is the tunnel at the edge of each lane? Round off to 2
decimal places.

Solution: We draw a coordinate system with origin at the middle of the highway,

as shown. Because of the given radius, the tunnel’s boundary is on the

circle x2 + y2 =122. Point P is the point on the arc just above the edge

of a lane, so its x-coordinate is 10. We need its y-coordinate. We then

solve 102 + y2 = 122 for y > 0, giving us y = 2 11  6.63ft.

2. A satellite dish has a shape called a paraboloid, where each cross-section

92
 is a parabola. Since radio signals (parallel to the axis) will bounce off the
surface of the dish to the focus, the receiver should be placed at the focus. How
far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5 ft
deep at the vertex?

Solution: The second figure above shows a cross-section of the satellite disc drawn
on a rectangular coordinate system, with the vertex at the origin. From the problem,
we deduce that (6, 4.5) is a point on the parabola. We need the distance of the focus
from the vertex, i.e., the value xof2 c= in4cy
x 2 = 4cy.
62 = 4c(4.5)
62
c= =2
4 • 4.5

Thus, the receiver should be 2 ft away from the vertex.

3. A tunnel has the shape of a semi-ellipse that is 15 ft high at the center, and 36
ft across at the base. At most how high should a passing truck be, if it is 12 ft wide,
for it to be able to fit through the tunnel? Round off your answer to two decimal
places.

Solution: Refer to the figure above. If we draw the semi-ellipse on a

rectangular coordinate system, with its center at the origin, an
x 2 y2
equation of the ellipse which contains it, is + =1
182 152

93
To maximize its height, the corners of the truck, as shown in the figure, would have
to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n) be the corner
of the truck in the first quadrant, where n > 0, is the (maximum) height of the truck.
Since this point is on the ellipse, it should
2
fit the equation. Thus, we have
2
6 + n
=1
182 152

n = 10  14.14ft

4. An explosion was heard by two stations 1200 m apart, located at F 1(−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 can 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.

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

x2 y2
+ = 1.
115600 244400

94
 What’s More

Activity 2 Solve the following problems, show all your solutions.

1. A piece of a broken plate was dug up in an archaeological site. It was put

on top of a grid with the arc of the plate passing through A( 7, 0), B(1, 4)
and C(7, 2). Find its center, and the standard equation of the circle
describing the boundary of the plate.

2. The cable of a suspension bridge hangs in the shape of a parabola. The

towers supporting the cable are 400 ft apart and 150 ft high. If the cable, at
its lowest, is 30 ft above the bridge at its midpoint, how high is the cable 50 ft
away (horizontally) from either tower?

3. The orbit of a planet has the shape of an ellipse, and on one of the foci isthe
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.

4. LORAN navigational transmitters A and B are located at (-130 , 0) and (130, 0)

respectively. A receiver P on a fishing boat somewhere in the first quadrant
listens to the pair (A, B) of the transmissions and computes the difference of
the distance from boat A and B as 240 miles. Find the equation of thehyperbola
on which P is located.
5. A tunnel through a mountain for a four lane highway is to have
an elliptical opening. The total width of the highway (not the
opening) is to be 16m, and the height at the edge of the road
must be sufficient for a truck 4m high to clear if the highest
point of the opening is to be 5m approximately. How wide must
the opening be?

6. At a water fountain, water attains a maximum height of 4m at

horizontal distance of 0.5m from its origin. If the path of water
is a parabola, find the height of water at a horizontal distance
of 0.75m from the point of origin

95
 7.
What I have Learned

8.
Solving Situational Problems Involving Conic Sections is possible by applying
the strategies below and by using the equations and characteristics of
the different conic sections previously learned

Problem Solving Strategies:

A. Collect information
1. write down what is/are given
2. write down what is asked for

B. Visualize
1. If possible draw a sketch
2. Label the sketch

C. Analysis
1. Write down any related formulas or principles
2. Write out the sequence of what you must do

D. Solve numerically(Solution)
1. Substitute numbers from the problem into your formula
2. Evaluate the resulting equation to get an answer usually numerical.
3. Don’t forget units (minutes, meter, miles, etc.)

E. Verification/Checking
Check your answer for reasonableness.

96
 What I Can

Activity 3 Answer the following multi-step problems involving

conics. Show your complete solutions.

1.) Cross section of a nuclear cooling tower is in the shape of a

= 1. The tower is 150m tall and

302 442

2.) The maximum and minimum distances of the Earth from the
sun respectively are 152 106km and 94.5 106km. The sun is

Online connect! For additional knowledge and information about the topics
please visit the links/url indicated below.

8.6 Conic Sections - Word Problems

https://www.youtube.com/watch?v=1eVzYUEi93o

Solve a word problem involving parabolas

https://www.youtube.com/watch?v=oXKkgIRnfEU

How to solve a word problem involving ellipsis

https://www.youtube.com/watch?v=KTiuwijOch4

Solving Applied Problems Involving Hyperbolas

https://www.youtube.com/watch?v=upsEcpmuPKA

97
Summary
Circle is the set of all points in a plane that are equidistant from a given point in
plane, called center. Any segment whose endpoints are the center and a point on
the circle is radius of the circle.

Standard form of the equation of a circle

(x − h)2 + (y − k)2 = r2, where (h,k) is the center and r is the radius.
General Form of the equation of the circle
x2 + y2 + Dx + Ey + F = 0
Parabola can be defined as the set of all points in a plane that are the same
distance from a given point called focus and a given line called directrix.
Equations of Parabola:
General Form Standard Form
𝑦2+ Dx+ F=0 𝑦2 = 4𝑐𝑥
𝑦2 = −4𝑐𝑥
𝑦2+ Dx+ F=0 𝑥2 = 4𝑐𝑦
𝑥2 = −4𝑐𝑦
𝑦2+ Dx+ Ey + F=0 (𝑦 − 𝑘)2 = 4𝑐(𝑥 − ℎ)
(𝑦 − 𝑘)2 = −4𝑐(𝑥 − ℎ)
𝑥2+ Dx+ Ey + F=0 (𝑥 − ℎ)2 = 4𝑐(𝑦 − 𝑘)
(𝑥 − ℎ)2 = −4𝑐(𝑦 − 𝑘)

The vertex of the parabola lies halfway between the focus and the directrix, and the
axis of symmetry is the line that runs through the focus perpendicular to the
directrix.
An ellipse is the set of all points in the plane the sum of whose distances from two
fixed points F1 and F2 is a constant. These two fixed points are the foci (plural of
focus) of the ellipse.

98
Equation of Ellipse

General form Standard Form

𝐴𝑥2 + 𝐶𝑦2 + 𝐹 = 0, 𝐴 < 𝐶 𝑥2 𝑦 2
+ = 1, 𝑎 > 𝑏
𝑎2 𝑏2
𝐴𝑥2 + 𝐶𝑦2 + 𝐹 = 0, 𝐴 > 𝐶 𝑦 2 𝑥2
+ = 1, 𝑎 > 𝑏
𝑎2 𝑏2
𝐴𝑥2 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0, 𝐴 < 𝐶 (𝑥 − ℎ)2 (𝑦 − 𝑘)
+ = 1, 𝑎 > 𝑏
𝑎2 2 𝑏2 2
𝐴𝑥2 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0, 𝐴 > 𝐶 (𝑦 − 𝑘) (𝑥 − ℎ)
+ = 1, 𝑎 > 𝑏
𝑎2 𝑏2

A hyperbola is the set of all points in the plane, the difference of whose distances
from two fixed points F1 and F2 is a constant. These two fixed points are the foci of
the hyperbola.

General form Standard Form

𝐴𝑥2 − 𝐶𝑦2 + 𝐹 = 0 𝑥2 𝑦 2
− =1
𝑎2 𝑏2
𝐶𝑦2 − 𝐴𝑥2 + 𝐹 = 0 𝑦 2 𝑥2
− =1
𝑎2 𝑏2
𝐴𝑥2 − 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 (𝑥 − ℎ)2 (𝑦 − 𝑘)2
− =1
𝑎2 2 𝑏2 2
𝐶𝑦2 − 𝐴𝑥2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 (𝑦 − 𝑘) (𝑥 − ℎ)
− =1
𝑎2 𝑏2

center : origin (0, 0)

foci : F1(−c, 0) and F2(c, 0)

• Each focus is c units away from the center.

Vertices: V1(−a, 0) and V2(a, 0)

The vertices are points on the hyperbola, collinear with

the center and foci.

Asymptotes: y = b x and y = − b x, the lines ℓ1 and ℓ2 in the figure

The asymptotes of the hyperbola are two lines passing through the center
which serve as a guide in graphing the hyperbola:

99
 Assessment (Post-test)
Multiple Choice: Encircle the letter of the best answer.

1. It is the intersection of a plane and a double-napped cone.

a. locus b. conic section c. circle d. radius

2. A collection of points satisfying a geometric property can also be

referred to as a _ of points.
a. locus b. conic section c. circle d. radius

3. A _ is the set of all points (x, y) in a plane that are

equidistant from a fixed point, called the center.
a. locus b. conic section c. circle d. radius

4. A/n _ is the set of all points in the plane that are

equidistant from a fixed point F(called the focus) and a fixed line l
(called the directrix).
a. circle b. ellipse c. parabola d. hyperbola

5. A/n is the set of all points in the plane the sum of

whose distances from two fixed points F1 and F2 is a constant.
a. circle b. ellipse c. parabola d. hyperbola

6. A/n _ is the set of all points in the plane, the difference

of whose distances from two fixed points F1 and F2 is a constant.
a. circle b. ellipse c. parabola d.
hyperbola

7. 6𝑥2 + 12𝑥 + 6𝑦2 − 8𝑦 = 100 is an example of a

a. Circle b. parabola c. ellipse d. hyperbola

8. 4𝑥2 + 12𝑥 + 𝑦2 − 8𝑦 = 64 is an example of a

a. Circle b. parabola c. ellipse d. hyperbola

9. 6𝑦 = 3𝑥2 − 15 is an example of a
a. Circle b. parabola c. ellipse d. hyperbola

100
10. 2𝑥2 + 87𝑥 − 25 = 𝑦2 + 4𝑦 is an example of a
a. Circle b. parabola c. ellipse d. hyperbola
11. 6𝑥2 − 15𝑥 = 3𝑦2 + 4𝑦 + 13 is an example of a
a. Circle b. parabola c. ellipse d. hyperbola

12. A hyperbola is the only conic that has

a. asymptotes b. focus c. vertex d. a minor axis

𝑥2 𝑦2
13. − = 1 has a major axis of length
4 25

a. 2 b. 3 c. 4 d. 5

(𝑥+3)2 (𝑦−2)2
Use − = 1 to answer the next 4 questions.
9 4

14. The center is at _?

a. (-3, 2) b. (-3, -2) c. (3, 2) d. (3, -2)

15. The vertices are ?

a. (0, 2), (-6, 2) b. (-3, 5), (-3, -1) c. (-1, 2), (-5, 2) d. (-3, 4), (-3, 0)

16. The foci are at _?

𝑎. (−3 ± √7, 2) 𝑏. (−3 ± √13, 2) c. (−3, 2 ± √13) d. (−3, 2 ± √7)

17. The length of the major axis is ?

a. 3b. 6 c. 9 d. 12

18. The standard form of 9𝑥2 − 4𝑦2 = 36 is

𝑥2 𝑦2 2 𝑦2 2 𝑦2 2 𝑦2
𝑎. − =1 b. 𝑥 − =1 c. 𝑥 − = 1 d. 𝑥 − =1
4 9 36 9 25 4 5 20

19. The earth’s orbit is an ellipse with the sun at one of the foci. If the farthest
distance of the sun from the earth is 105.5 million km and the nearest
distance of the sun from the earth is 78.25 million km, find the eccentricity
of the ellipse.
a. 0.15 b. 0.25 c. 0.35 d. 0.45

𝑥2 𝑦2
20. − = 1 will have a minor and major axis with length (In that order)
9 16

a. 3, 4 b. 6, 8 c. 9, 16 d. 18, 32
101
21. Which of the following shows the correct graph of the circle?

2 2
a. x +y =4

2 2
b. y =x + 16

2 2
c. x +y = 16

2 2
d. x +y =1

22. Which graph represents the equation 𝑥2 − 10𝑥 + 𝑦2 = −9?

a. c.

d.

23. What is the focus and vertex of the parabola (𝑦 − 2)2 + 16(𝑥 − 3) = 0?
a. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−3,14)
b. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−3, −18)
c. Vertex: 𝑉(−3, −2) , Focus: 𝐹(−7, −2)
d. Vertex: 𝑉(3,2) , Focus: 𝐹(−1,2)

102
24. Find the equation of the parabola with vertex at (5, 4) and focus at (-3, 4).

2 2
a (y-4) = - 32( x - 5 ) c. (y+4) = - 32( x - 5 )
2 2
b (y-4) = 32( x - 5 ) d. (y-4) = 8( x - 5 )
2 (𝑦+9)2
25. Find the center and foci of the ellipse (𝑥+5) + .
5 9

a. center: (5,9) , foci: (5,7),(5,11)

b. center (-5,-9) , foci: (-5,-11), (-5,-7)
c. center: (-5,-9) , foci: (-7,-9), (-3,-9)
d. center: (5,9) , foci: (3,-9), (7,-9)

26. Find the center and vertices of the ellipse 4x2 + 9y2 - 24x + 72y + 144 = 0

a. center: (-4,3) , vertices: (-7,3), (-1,3)

b. center: (-3,4) , vertices: (-5,4), (-1,4)

c. center: (3,-4) , vertices: (1,-4), (5,-4)

d. center: (3,-4) , vertices: (0,-4), (6,-4)

27. Which of the following shows the correct graphical representation of the ellipse
𝑦2
𝑥2 + =1?
16 4

a. c.

b. d.

28. Find the center and vertices of the hyperbola 11x2 - 25y2 + 22x + 250y - 889 = 0.
a. center: (1,-5) , vertices: (1,-10), (1,0)
b. center: (-1,5) ,vertices: (-1,0),(-1,10)
c. center: (-1,5) , vertices: (-6,5), (4,5)
d. center: (1,-5) , vertices: (-4,-5), (6,-5)

103
29. What are the vertices and asymptotes of the hyperbola 9y2 - 16x2 = 144?
4
a. vertices: (0,-4),(0,4) , asymptote: 𝑦 = ± 3 𝑥

3
b. vertices: (0,-4), (0,4) , asymptote: 𝑦 = ± 4 𝑥

4
c. vertices: (-4,0), (4,0) , asymptote: 𝑦 = ± 3 𝑥

3
d. vertices: (-4,0), (4,0) , asymptote: 𝑦 = ± 4 𝑥

30. Find the standard form of the equation of the hyperbola with the given
characteristics, vertices: (0,-6), (0,6) and foci (0,-7), (0,7).
2 𝑥2 2 𝑦2
a. 𝑦 − =1 c. 𝑥 − =1
36 49 36 13

2 𝑥2 2 𝑦2
b, 𝑦 − =1 d. 𝑥 − =49
36 13 36 13

31. Write the equation of the ellipse that has its center at the origin with focus at (0,
4) and vertex at (0, 7).
2 𝑦2 2 𝑦2
f. 𝑥 + =1 c. 𝑥 + =-1
49 33 33 49
2 𝑦2 2 𝑦2
g. 𝑥 − =1 d. 𝑥 + =1
33 49 33 49

2 2
32. What is the graph of the hyperbola 9x − 9y = 81?

a. c,

b. d.

104
 33. An arch 20 meters high has the form of parabola with vertical axis. The length
of horizontal beam placed across the arch 9 meters from the top is 60 meters. Find
the width of the arch at the bottom.

a. 44.72 meters b. 45.72 meters c. 89.44 meters d. 90.44 meters

34. A spotlight in a form of a paraboloid 9 inches deep has its focus 3 inches from
the vertex. Determine the radius of the opening of the spotlight.

a. 9.39 inches b. 10.39 inches c. 11. 39 inches d. 12.39 inches

35. A bridge is supported on an elliptical arch of height of 7 meters and width at the
base of 40 meters. A horizontal roadway is 2 meters above the center of the arch.
How far is it above the arch at 8 meters from the center?

a. 0.58 meters b. 1.58 meters c. 2.58 meters d. 3.58 meters

105
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Camilon, M.G.,et.al. 2017. Precalculus for Senior High School. Quezon City:
Educational Resources Publication.

Carl Stitz, Ph.D. , Jeff Zeager, Ph.D.,

July 4, 2013 Lakeland Community College Lorain County Community College
Precalculus Corrected Edition

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November 30, 2012

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Analytic Geometry (Worktext) 2003 edition

Khan, Sal. 2001. Intro to Conic Sections. Accessed July 8, 2020.

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9
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2010. http://mathispower4u.wordpress.com/. July 22. Accessed 9 2020, July.

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2017. http://bit.ly/ProfDaveSubscribe. November 2017. Accessed July 9, 2020.

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sections.pdf .

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Pinoybix Engineering Copyright © 2020

Hyperbola (Part 1) - Conic Sections Class 11 CBSE

https://www.youtube.com/watch?v=WEyYaIWIUp0
November 20, 2019

Hyperbola (Part 2) - Conic Sections Class 11 CBSE

https://www.youtube.com/watch?v=Ni0gjU8-Pn4
November 22, 2019

Jeff Eicher
8.6 Conic Sections - Word Problems
https://www.youtube.com/watch?v=1eVzYUEi93o
March 9, 2020

Study Force
Solve a word problem involving parabolas
https://www.youtube.com/watch?v=oXKkgIRnfEU
28 Aug 2018

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