PRE CALCULUS
Name:
Grade & Section:
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Learning Module No. 2
What is this module all about?
The following topic/s are included in this module:
1. Parabola
What are you going to learn?
In this module, you are expected to meet the following competencies:
1. Define a parabola. STEM_PC11AGIa-5 (DepEd-MELC)
2. Determine the standard form of equation of a parabola. STEM_PC11AGIb-1 (DepEdMELC)
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DAY 1
PRETEST
Directions: Read and analyze the picture carefully. Label the parts of the parabola shown in the figure below.
directrix
Latus
Rectum
vertex
focus
axis of symmetry
Concept Notes
Definition of Parabola. A parabola is the set of
all points in a plane that are equidistant from a
fixed line (directrix) and a fixed point (focus) not
on the line. The midpoint between the focus and
the directrix is called the vertex, and the line
passing through the focus and the vertex is called
the axis of symmetry (focal axis) of the parabola.
The line through the focus perpendicular to the
axis and intercepted by the parabola is called
the latus rectum of a focal chord and the length
of the focal chord is called the focal width.
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LESSON 2.1 EXPLORING THE PARABOLA
Smooth, regular curves are common in the world around us. The orbit of the space shuttle around the
Earth is one kind of smooth curve; another is the shape of domed buildings such as the United States Capitol
and the St. Louis Gateway Arch. Television receiving dishes are a very common curved shape, as are the
curved rearview mirrors on your car. Mathematically, however, an interesting property of all the curves just
mentioned is that they are expressed as quadratic equations in two variables.
We learned from basic geometry that the graph of a quadratic function is an upward or downward
opening parabola. We have seen the role of the parabola in free-fall and projectile motion. When you kick a
soccer ball (or shoot an arrow, fire a missile, or throw a stone) it arcs up into the air and comes down again
following the path of a parabola.
Activity 1
Direction: Can you think of other examples of a parabolic curve in real life? Illustrate your answer.
Then, explain.
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LESSON 2.2 UNDERSTANDING THE GENERAL AND STANDARD EQUATIONS OF THE
PARABOLA
You could begin by defining conics in terms of the intersections of planes and cones, as the Greeks
did, or you could define them algebraically, in terms of the general second-degree equation.
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
c. (x – h)2 = 4p(y – k)
d. (y – k)2 = 4p(x – h)
a. x2 + Dx + Ey + F = 0
b. y2 + Dx + Ey + F = 0
e. x2 = 4py
f. y2 = 4px
Just like the circle, the equation of a parabola can also be expressed and written in standard and general form.
Activity 2.2
General form: 1.
2.
x2 + Dx + Ey + F = 0
y2 + Dx + Ey + F = 0 , where D , E and F are constants.
The standard form of the equation of a parabola with vertex at (h, k) is as follows:
1. (x – h)2 = 4p(y – k)
2. (y – k)2 = 4p(x – h)
,where p should not be zero.
The focus lies on the axis p units (directed distance) from the vertex. If the vertex is at the origin (0, 0) the
equation takes one of the following forms.
1. x2 = 4py
2. y2 = 4px
Directions: Identify the form of each equation by writing GF (general form) or SF (standard form). Write
your answer on the blank provided before each number.
GF
____ 1.
SF
____ 2.
____ 3.
SF
3
3x + y2 – 7y + 6 = 0
y2 = – 96x
56y = x2
SF
SF
____ 5.
____ 6.
GF
____ 4. (x – 7)2 = 24 (y +1)
y2 = -16 (x + 5)
x + x2 – y + 3 = 0
DAY 2
LESSON 2.3.1 EQUATIONS OF PARABOLAS (Standard Form to General Form)
Example: STANDARD TO GENERAL FORM
Transform the equation of a parabola given by (y + 5)2 = -2 (x – 7) to general form.
Step 1:
Square the binomial and expand.
(y + 5)2
= -2 (x – 7)
(y + 5) (y + 5) = -2 (x – 7)
y2 + 10y + 25
= -2x + 14
Step 2:
Place all the terms at the left side of the equation and equate to 0.
y2 + 10y + 25 + 2x – 14 = 0
Step 3:
Combine similar terms.
y2 + 10y + 2x + 25 – 14 = 0
y2 + 10y + 2x + 11 = 0
Therefore, the general form of (y + 5)2 = -2 (x – 7) is y2 + 10y + 2x + 11 = 0.
Activity 2.3.1
Directions: Transform each equation in standard form to general form. Show your complete solution.
1. (y + 3)2 = 6 (x – 2)
_________________________
2. (x – 4)2 = 12 (y + 5)
_________________________
3. (y – 3)2 = 6 (x + 1)
_________________________
4. (x + 3)2 = - 2 (y – 1)
_________________________
5. (x – 1)2 = - 8 (y + 2)
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LESSON 2.3.2 EQUATIONS OF PARABOLAS (General Form to Standard Form)
Example: GENERAL TO STANDARD FORM
Transform the equation of a parabola given by y2 – 4x – 12y + 28 = 0 to standard form.
Step 1:
Group the terms with the same variables.
y2 – 4x – 12y + 28 = 0
y2 – 12y – 4x + 28 = 0
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Step 2:
Move the remaining terms to the right side of the equation.
(y2 – 12y) = 4x – 28
Step 3:
Complete the square to make a perfect square trinomial. Balanced the equation.
(y2 – 12y) + ___
(y2 – 12y) + (
= 4x – 28 + ___
−𝟏𝟐 2
)
𝟐
−𝟏𝟐 2
)
𝟐
= 4x – 28 + (
y2 – 12y + 36
Step 4:
= 4x – 28 + 36
Simplify both sides of the equations.
y2 – 12y + 36
Step 5:
= 4x + 8
Express the trinomial as square of binomial to make the equation in vertex form (standard
form).
(y – 6)2 = 4 (x +2)
from
−𝟏𝟐
𝟐
Therefore, the standard form of y2 – 4x – 12y + 28 = 0 is (y – 6)2 = 4 (x +2).
Activity 2.3.2
Directions: Transform each equation in standard form to general form. Show your solutions.
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1. 3x + y2 – 6y + 6 = 0
_________________________
2. 6x + x2 – y + 3 = 0
_________________________
3. 3. y2 – 4x – 4 = 0
_________________________
4. 4. 8x + y2 + 6y + 25 = 0
_________________________
5. 5. 2x + x2 – y + 3 = 0
_________________________
DAY 3
LESSON 2.4.1 GRAPHS OF PARABOLAS
Transformation of Parabolas
Example 1: Sketch the graph of the parabola with vertex at the origin and focus at (2, 0). Identify its parts and
find the standard equation.
Solution: Since the parabola is at origin (0, 0) and its focus is (2, 0), we know that the parabola opens to the
right. The directrix, which has the same distance to the focus, has an equation x = – 2. The focal length p is 2
since focus is (p, 0), and the focal axis is x – axis or the horizontal axis. Thus, we use the equation y2 = 4px.
Substituting p = 2, we have the equation y2 = 4(2)x which gives us y2 = 8x (standard form).
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Activity 2.4.1
Drawing Activity
Sketch the graph of the parabola with the given vertex and focus. Identify its parts and find the standard
equation.
1. vertex at (0, 2) and focus at (0, 4).
2. vertex at (0, -2) and focus at (0, -4)
3. vertex at (3, 0) and focus at (5, 0)
4. vertex at (-3, 0) and focus at (-5, 0)
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DAY 4
LESSON 2.4.2 GRAPHS OF PARABOLAS
Example: Sketch the graph of the parabola with vertex at (2, 1) and focus at (2, 4). Identify its parts and find
the standard equation.
Solution: Since the vertex of the parabola is at (h, k) or (2, 1) and its focus is (2, 4), we know that the axis is
vertical since there is only change in the ordinate (y – coordinate).
We consider the equation (x – h)2 = 4p(y – k), where h = 2, k = 1 and p = 4 – 1 which is 3 and the directrix,
which has the same distance to the focus is 3 units below the vertex, which is y = k – p , y = – 2.
The focal length p is 3 since focus is (h, k + p). Therefore, (x – 2)2 = 4(3)(y – 1) or (x – 2)2 = 12(y – 1) is the
standard equation of the parabola.
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Activity 2.4.2
Sketch the graph of the parabola with the given vertex and focus. Identify its parts and find the standard
equation.
1. vertex at (3, 1) and focus at (3, -2)
2. vertex at (3, 1) and focus at (3, 4)
3. vertex at (-3, 1) and focus at (-6, 1)
4. vertex at (-3, 1) and focus at (1, 1)
Reminder: The quiz shall be accomplished at a given time only, either on a synchronous or asynchronous
session.
No part of this publication may be reproduced or transmitted in any form or by any means without
the prior written permission of the author.
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