Download PRECALCULUS-SLM-WEEK-3 ELLIPSE and more Study notes Mathematics in PDF only on Docsity!
Republic of the Philippines
DEPARTMENT OF EDUCATION
Region IX, Zamboanga Peninsula
Zamboanga del Sur National High School
Senior High School
Sta. Maria, Pagadian City
Name : __________________________________ Grade and Section : __________________
Subject: Precalculus Quarter : 1 Week: 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 𝑩
𝟐
Eccentricity
Circle 𝐵
2
− 4 𝑎𝑐<0 or A=C 𝑒 = 0
Parabola 𝐵
2
Ellipse 𝐵
2
Hyperbola 𝐵
2
2
2
We will collect the values of A, B, and C.
A= 1, B=0, and C=1. Solving for 𝐵
2
2
2
Note that B=0, A=C. Thus, the conic section is a circle.
2
2
We will collect the values of A, B, and C.
A= 4, B=0, and C=4. Solving for 𝐵
2
2
2
Note that B=0, A=C. Thus, the conic section is a circle.
2
2
We will collect the values of A, B, and C.
A= 4, B=0, and C=9. Solving for 𝐵
2
2
2
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 𝐵
2
− 4 𝑎𝑐 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.
Recalling these concepts are useful in teaching ellipse.
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.
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.
From the graph, the center is C (h,k) which is (0,0),
and the radius r is 3. Using the standard form
2
2
2
2
2
2
2
2
= 9 is the equation of the circle.
A=1 and C=1, then A=C.
first vertex first focus center radius
second focus second vertex
b.
(𝑥− 1 )
2
100
(𝑦+ 1 )
2
36
3600 [
2
2
= 1 ]
2
2
2
2
2
2
2
2
2
2
The general form is 36 𝑥
2
2
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 F
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.
Given are two points on the x-axis, F1( - c, 0) and F2(c, 0), the foci, both c units 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
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
[
2
2
2
]
4
2
2
2
Let F 1
and F 2
be two distinct points. The set of all points P, whose distances
from F 1
and from F 2
add up to a certain constant, is called an ellipse. The points
F
1
and F 2
are called the foci of the ellipse.
2
2
2
2
2
4
2
2
2
2
2
2
2
2
2
2
2
by letting 𝑏 = √(𝑎
2
2
), so a>b
2
2
2
2
When we let b=𝑎
2
2
, 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
𝑥
2
𝑎
2
𝑦
2
𝑏
2
= 1 ,where a>b. Let 𝑐 = √(𝑎
2
2
Figure 1.
(1) center : origin (0, 0)
(2) foci : F 1
(−c, 0) and F 2
(c, 0)
▪ Each focus is c units away from the center.
▪ For any point on the ellipse, the sum of its distances from the foci is 2a.
(3) vertices: V 1
(-1,0) and V 2
▪ 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: W 1
(0, −b) and W 2
(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 W 1
W
2
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 with equation
𝑥
2
25
𝑦
2
9
- Sketch the graph, and include these points.
Solution. With a
2
= 25 and b
2
= 9, we have a = 5, b = 3, and
2
2
foci: F 1
(−4, 0), F
2
(4, 0) vertices: V 1
(−5, 0), V
2
covertices: W 1
(0, −3), W
2
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
24
(𝑦− 5 )
2
49
2
2
Solution: (1) From 𝑎
2
= 49 and 𝑏
2
= 24 , we have a=7, b= √
6 ≈ 4. 9 , and
Solution: (1) From 𝑎
2
= 49 and 𝑏
2
= 24 , we have a=7, b=2√ 6 ≈ 4. 9 , and
2
2
) = 5. The ellipse is vertical.
Center: (-3,5)
Foci: F 1
(-3,0), F
2
Vertices: V 1
(-3,-2), V
2
Covertices: W 1
W
2
(2) We first change the given equation to standard form.
9(x
2
−14x) + 16(y
2
9(x
2
− 14x + 49) + 16(y
2
- 4y + 4) = 71 + 9(49) + 16(4)
9(x − 7)
2
2
(x − 7 )
2
(y + 2 )
2
We have 𝑎 = 8 and 𝑏 = 6. Thus, 𝑐 = √(𝑎
2
2
) = 2 √ 7 =5.3. The ellipse is horizontal.
Center: (7,-2)
Foci: F 1
F
2
Vertices: V 1
(-1,-2), V
2
Covertices: W 1
(7,-8), W
2
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 = √(𝑎
2
2
The equation is
(x + 3 )
2
(y + 2 )
2
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
61 units away from the center. From the length of the minor axis, 2b = 12 so b = 6. The standard
equation is
(x− 2 )
2
61
(y + 5 )
2
36
= 1. Each focus is 𝑐 = √
2
2
= 5 units away from (2, - 5), so their
coordinates are (-3,-5) and (7,-5).
Assessment: Circle the correct letter that corresponds the correct answer.
For items 1 – 3 Given:
𝑥
2
144
𝑦
2
169
- The coordinates of the vertices
a. (0, ±13) b. (0, ±5) c. (±12, 0) d. (0, 0)
- The coordinates of the foci
a. (0, ±13) b. (0, ±5) c. (±12, 0) d. (0, 0)
- The coordinates of the covertices
a. (0, ±13) b. (0, ±5) c. (±12, 0) d. (0, 0)
For items 4 – 6 Given: 36 𝑥
2
2
- The value of c, where 𝑐 = √𝑎
2
2
a. 4 b. 6 c. √ 20 d. 5
- The coordinates of the vertices
a. (- 4 , - 3), (8, - 3) b. (2, - 7), (2, 1) c. ( 2 ± 2 √ 5 , − 3 ) d. (2, - 9), (2, 3)
- The coordinates of the foci
a. (-4, - 3), (8, - 3) b. (2, - 7), (2, 1) c. (-2, - 3), (6, - 3) d. (2, - 9), (2, 3)
Match the equal equations, write the letter before each number.
2
2
− 126 𝑥 + 64 𝑦 = 71 A.
(𝑥− 4 )
2
4
(𝑦− 3 )
2
9
2
2
− 72 𝑥 − 24 𝑦 = − 144 B.
(𝑥− 1 )
2
4
(𝑦+ 2 )
2
16
(𝑥+ 3 )
2
7
(𝑦+ 1 )
2
16
= 1 C.
(𝑥− 7 )
2
64
(𝑦+ 2 )
2
36
(𝑥− 7 )
2
64
(𝑦+ 2 )
2
25
= 1 E. 25 𝑥
2
2
F. 16 𝑥
2
2
G. 25 𝑥
2
2
- The standard equation of the ellipse with covertices (-4, 8) and (10, 8), and focus at (3, 12)
a.
(𝑥− 3 )
2
64
(𝑦− 8 )
2
25
= 1 c.
(𝑥− 3 )
2
65
(𝑦− 8 )
2
49
b.
(𝑥− 3 )
2
49
(𝑦− 8 )
2
65
= 1 d.
(𝑥− 3 )
2
16
(𝑦− 8 )
2
4
