Learning Activity Sheets

PRECALCULUS
Quarter 1 – Week 2

Parabolas
MARIA MARGARITA M. ORTEGA
Subject Teacher
 Lesson 2: Parabolas

Learning Outcomes of the Lesson

Upon completion of this lesson, you should be able to:
1. define a parabola; (STEM_PC11AG-Ia-5
2. determine the standard form of equation of a parabola. (STEM_PC11AG-Ib-1)

PRE-ASSESSMENT

Direction: Choose the letter that corresponds to your answer.

(Note: Use a separate sheet of paper in answering the activities.)

1. A conic that consists of all points equidistant from a fixed point called the focus F
and a fixed line l called the directrix, not containing F.
a. circle b. parabola c. hyperbola d. ellipse

2. The opening of a parabola given by the equation 𝑦𝑦 = −4𝑝𝑝𝑝𝑝.

a. upward b. downward c. to the right d. to the left

3. The coordinates of the vertex of the parabola that is represented by the equation
(𝑥𝑥 − 2)2 = 12(𝑦𝑦 − 1) is
a. (2, 4) b. (2, 1) c. (4, 2) d. (1, 2)

4. Referring to the equation of the parabola in problem 3, the coordinates of the focus
is
a. (2, 4) b. (2, 1) c. (4, 2) d. (1, 2)

5. What is the standard form of the equation of the parabola with vertex (2, −3) and
focus (4, −3)?
a. (𝑥𝑥 + 3)2 = 8(𝑦𝑦 − 2) c. (𝑥𝑥 − 3)2 = 8(𝑦𝑦 + 2)
b. (𝑦𝑦 + 3)2 = 8(𝑥𝑥 − 2) d. (𝑦𝑦 − 3)2 = 8(𝑥𝑥 − 2)

6. Find the focus of a parabola whose equation is 𝑦𝑦 2 = −8𝑥𝑥.

a. (0, −2) b. (0, 1) c. (−2, 0) d. (1, 0)

7. Find the directrix of the parabola given in problem 6.

a. 𝑥𝑥 = −2 b. 𝑦𝑦 = −2 c. 𝑥𝑥 = 2 d. 𝑦𝑦 = 2
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In problems 8 to 10, refer to the equation of the parabola 𝑥𝑥 = 8𝑦𝑦.

8. What is the coordinate of the focus?

a. (0, 0) b. (2, 0) c. (0, 2) d. (0, 4)

9. What is its vertex?

a. (0, 0) b. (0, 2) c. (2, 2) d. (0, 1)
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10. What is the length of the latus rectum?
a. 2 b. 4 c. 3 d. 1

Brief Introduction

A parabola is one of the conic sections. You have learned from the previous lesson
that it is formed when the plane intersects only one cone to form an unbounded curve.
We have already seen parabolas which open upward or downward, as graphs of
quadratic functions. Here, we will see parabolas opening to the left or right.

Definition and Equation of a Parabola

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Graphing a Parabola with a Vertex at the Origin

We can draw an approximate sketch of a parabola whose vertex is at the origin with
three pieces of information. We know that the vertex is located at (0, 0). Additional
information that we week is the direction in which the parabola opens and
approximately how wide or narrow to the draw the parabolic curve. The direction
toward which the parabola opens is found from the equation. An equation of the form
𝑦𝑦 2 = 4𝑝𝑝𝑝𝑝 opens either left or right. It opens right of 𝑝𝑝 > 0 and opens left if 𝑝𝑝 < 0. An
equation of the form 𝑥𝑥 2 = 4𝑝𝑝𝑝𝑝 opens either up or down. It opens up if 𝑝𝑝 > 0 and opens
down if 𝑝𝑝 < 0. How narrow or wide should you draw the parabolic curve? If we select a
few points that satisfy the equation, we can use those as graphing aids.
In Example 1, we found that the focus of that parabola is located at (2, 0). If we select
the 𝑥𝑥-coordinate of the focus 𝑥𝑥 = 2, and substitute that value into the equation of the
parabola 𝑦𝑦 2 = 8𝑥𝑥, we find the corresponding 𝑦𝑦 values to be 𝑦𝑦 = 4 and 𝑦𝑦 = −4. If we
plot the three points (0, 0), (2, −4), and (2, 4) and then connect the points with a
parabolic curve, we get the graph (left) shown below.
The line segment that passes through the focus (2, 0) that is parallel to the directrix
𝑥𝑥 = −2, and whose endpoints are on the parabola is called the latus rectum. The latus
rectum in this case has length 8. The latus rectum is a graphing aid that assists us in
determining how wide or narrow to draw the parabola.
In general, the points in a parabola of the form 𝑦𝑦 2 = 4𝑝𝑝𝑝𝑝 that lie above and below
the focus (p, 0 ) satisfy the equation 𝑦𝑦 2 = 4𝑝𝑝2 and are located at (𝑝𝑝, −2𝑝𝑝) and
(𝑝𝑝, 2𝑝𝑝). The latus rectum will have length 4|𝑝𝑝|. Similarly, a parabola of the form 𝑥𝑥 2 =
4𝑝𝑝𝑝𝑝 will have a horizontal latus rectum of length 4|𝑝𝑝|.

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Practice Exercise 1

Perform as indicated.

1. Find the focus and directrix of a parabola whose equation is 𝑦𝑦 2 = 16𝑥𝑥.

2. Find the focus, directrix, and length of the latus rectum of the parabola 𝑦𝑦 2 = −8𝑥𝑥,
and use these to graph the parabola.

3. Find the equation of a parabola whose focus is at the point (−5, 0) and whose
directrix is 𝑥𝑥 = 5.

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 In order to find the vertex of a parabola given a general second-degree equation,
first complete the square in order to identify (ℎ, 𝑘𝑘). Then determine whether the
parabola opens up, down, left, or right. Identify points that lie in the graph of the
parabola. Intercepts are often the easiest points to find, since they are the points where
one of the variables is set equal to zero.

Example 1

Example 2 Given the equation of the 𝑦𝑦 2 + 6𝑥𝑥 + 8𝑦𝑦 + 1 = 0, find the vertex, the focus,
and equation of the directrix, an equation of the axis, and the length of the latus rectum,
and draw a sketch of the graph.

Solution Rewrite the given equation as 𝑦𝑦 2 + 8𝑦𝑦 = −6𝑥𝑥 − 1

Completing the square of the terms involving 𝑦𝑦 on the left side of the equation by
adding16 on both sides, we obtain
𝒚𝒚𝟐𝟐 + 𝟖𝟖𝟖𝟖 + 𝟏𝟏𝟏𝟏 = −𝟔𝟔𝟔𝟔 + 𝟏𝟏𝟏𝟏

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 𝟓𝟓
(𝒚𝒚 + 𝟒𝟒)𝟐𝟐 = −𝟔𝟔(𝒙𝒙 − )
𝟐𝟐
Comparing this equation with (𝑦𝑦 − 𝑘𝑘)2 = 4𝑝𝑝(𝑥𝑥 − ℎ), we let
𝟓𝟓
𝒉𝒉 = −𝟒𝟒 𝒉𝒉 =
𝟐𝟐
and
𝟑𝟑
𝟒𝟒𝟒𝟒 = −𝟔𝟔 or 𝒑𝒑 = −
𝟐𝟐

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Therefore, the vertex is at (2 , −4); an equation of the axis is 𝑦𝑦 = −4; focus is at (1, −4);
and equation of the directrix is at 𝑥𝑥 = 4; and the length of the latus rectum is 6. A
Sketch of the graph is shown below.

Example 3 Find an equation of the parabola having as its directrix the line 𝑦𝑦 = 1 and
its focus the point 𝐹𝐹(−3, 7).

Solution Since the directrix is parallel to the 𝑥𝑥 axis, the axis will be parallel to the 𝑦𝑦
axis, and the equation will have the form (𝑥𝑥 − ℎ)2 = 4𝑝𝑝(𝑦𝑦 − 𝑘𝑘).
Since the vertex 𝑉𝑉 is halfway between the directrix and the focus, 𝑉𝑉has
coordinates (−3, 4). The directed distance from the vertex to the focus is 𝑝𝑝, and so

𝒑𝒑 = 𝟕𝟕 − 𝟒𝟒 = 𝟑𝟑

Therefore, an equation is

(𝒙𝒙 + 𝟑𝟑)𝟐𝟐 = 𝟏𝟏𝟏𝟏(𝒚𝒚 − 𝟒𝟒)

Squaring and simplifying, we have

𝒙𝒙𝟐𝟐 + 𝟔𝟔𝟔𝟔 − 𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟓𝟓𝟓𝟓 = 𝟎𝟎

A sketch of the graph of this parabola is shown at the right.

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Practice Exercise 2

Perform as indicated.

1. Given the equation of the parabola 𝑦𝑦 2 + 4𝑦𝑦 − 5𝑥𝑥 − 5 = 0, find the vertex, the focus,
an equation of the directrix, an equation of the axis, and the length of the latus
rectum. Draw a sketch of the graph.

2. Find an equation of the parabola having its vertex 𝑉𝑉 at (2, 4) and focus at 𝐹𝐹(−3, 4).
Draw a sketch of the graph.

ASSESSMENT

Please see attached sheet.

ANSWER KEY

𝑦𝑦 2 = −20𝑥𝑥 3.

(𝑦𝑦 − 4)2 = −20(𝑥𝑥 − 2) 2.

Axis: 𝑦𝑦 = −2
Length of latus rectum = 5

Directrix: 𝑥𝑥 = −
20
61
, −2)
Focus: (−
20
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(− , −2)
5
2.
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1. The focus is (4, 0) and the directrix is 𝑥𝑥 = −4.
Vertex: 1.

Practice Exercises 1 Practice Exercises 2

References

Stewart, James, et. al, Precalculus, Cengage Learning, 2016, pp. 788
Young, Cynthia y., Precalculus 2nd Edition, John Wiley & Sons Inc., 2014, pp. 911-919
Liethold, Louis, The Calculus with Analytic Geometry Third Edition, Harper and Low
Publishers Inc., 1976, pp. 579-583

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