Defining Hyperbola and Determining
Its Standard Equation Form
I
Lesson
After going through this module, you are expected to: 1.) Define a hyperbola; 2.) Determine the standard form and general form of the equation of a hyperbola; 3.) Identify parts of a hyperbola given specific conditions; 4.) Graph a hyperbola in a rectangular coordinate system; and 5.) Recognize the importance of hyperbola into real-life situation.
Learning Task 1: Draw the line of the given linear equation in a rectangular coordinate system.
1. y = 4x
2. y = 2x + 5
4.
5.
3. y = -2x - 5
Guide Questions:
1. How will you determine the slope of the given linear equation?
2. What is the first thing that you need to do in drawing a line pertaining to
linear equation?
3. How will you draw a line pertaining to linear equation?
D
Learning Task 2: Derive the formula of the standard form of the equation of a hyperbola which is oriented horizontally. Apply the distance formula by applying PF 1
- PF2 = 2a and take note of c2 =a2 + b2.
Guide Questions:
1. What can you say about a hyperbola?
2. Where is the center located?
3. How did you derive a the standard form of the equation of a hyperbola
which is oriented horizontally?
PIVOT 4A CALABARZON
28
WEEK
4
A hyperbola is a curve that occurs at the intersection of a cone and a
plane. It is also the set of all points in the plane such that the difference of their
distances from two fixed points (foci) is constant. In deriving the standard form of
the equation of a hyperbola, distance formula will be utilized. Based on the illustration,
. It implies that the difference of the distance of PF1 and PF2
is equal to the length of a transverse axis.
Orientation
Vertex at the Origin
Vertex at (h, k)
Horizontal
Vertical
Note: We can easily determine the orientation by looking which of x and y is positive. If y is positive, then the orientation is vertical and if x is positive then it is
horizontal.
Parts
Definition
Center
C
Vertices
(V1 & V2)
Endpoints of Conjugate Axis
(W1 & W2)
Foci (F1 & F2)
This can be at the origin or (h, k). This is middle inner most part
of an ellipse.
Each vertex is a units from the center. Moreover, the transverse
axis has length 2a.
Each endpoint of conjugate axis is b units from the center. Moreover, the length of conjugate axis is 2b.
Asymptotes
Each focus is c units from the center and collinear to the vertices.
These are the two lines that intersect at the center.
Patterns in Finding Parts of an Ellipse
Parts
ORIENTATION – HORIZONTAL
Vertex at the Origin
Vertex at (h, k)
Center
C: (0, 0)
C: (h, k)
Vertices
V1:(a, 0) & V2: (-a, 0)
V1:(h+a, k) & V2: (h-a, k)
Endpoints of Conjugate Axis
W1:(0, b) & W2: (0, -b)
W1:(h, k+b) & W2: (0, k-b)
Foci
F1:(c, 0) & F2: (-c, 0)
F1:(h+c, k) & F2: (h-c, k)
Equation of Asymptotes
29
PIVOT 4A CALABARZON
ORIENTATION – VERTICAL
Parts
Vertex at the Origin
Vertex at (h, k)
Center
C: (0, 0)
C: (h, k)
Vertices
V1:(0, a) & V2: (0, -a)
V1:(h, k+a) & V2: (h, k-a)
Endpoints of Conjugate Axis
W1:(b, 0) & W2: (-b, 0)
W1:(h+b, k) & W2: (h-b, k)
Foci
F1:(0, c) & F2: (0, -c)
F1:(h, k+c) & F2: (h, k-c)
Equation of Asymptotes
Note: In hyperbola, we use c2 = a2 + b2
Illustrative Examples
I. Center at the Origin
Direction: Identify the parts of the given specific conditions, standard form and general
form of the equation of an ellipse.
Solution: First, determine the orientation of the equation of a hyperbola. Since x is
positive therefore the orientation is horizontal. Second, identify the values of a, b and c.
To find c, use the formula
Since the orientation is horizontal, center is at the origin and the values of
a, b and c are 5,
and 6 consecutively therefore by substituting this to the
given pattern we can identify the parts of a hyperbola.
Center Vertices
Endpoints of
Conjugate Axis
Foci
Equation
of the
Asymptotes
(0, 0)
Standard Form of the Equation of an Ellipse:
General Form of the Equation of of an Ellipse:
PIVOT 4A CALABARZON
30
2. Find the parts of a hyperbola oriented horizontally whose center at the origin,
length of transverse axis is 6 and length of conjugate axis is
.
Solution: Since the length of transverse axis (2a) is 6 therefore a = 3. Likewise, the length of the conjugate axis (2b) is
therefor
.
To find c, use the formula
Since the orientation is horizontal, center is at the origin and the values of
a, b and c are 3,
, and 7 consecutively therefore by substituting this to
the given pattern we can identify the parts of a hyperbola.
Center Vertices
Endpoints of
Conjugate Axis
Foci
Equation
of the
Asymptotes
(0, 0)
Standard Form of the Equation of an Ellipse:
General Form of the Equation of an Ellipse: 40x2 - 9y2 – 360= 0
II. Center at (h, k)
Direction: Identify the parts of the given specific conditions, standard form and
general form of the equation of a hyperbola.
1.
Solution: First, determine the orientation of the equation of an ellipse. Since y
is positive therefore the orientation is vertical. Second, identify the values of a, b
and c.
To find c, use the formula
Since the orientation is vertical, center is at the (-1, 5) and the values of a, b
and c are 10, 5, and
consecutively therefore by substituting this to
the given pattern we can identify the parts of a hyperbola.
31
PIVOT 4A CALABARZON
Center
Vertices
Endpoints of Conjugate Axis
Foci
(-1, 5)
V1:(h, k+a)
W1:(h+b, k)
F1:(h, k+c)
V1:(-1, 5+10)
W1:(-1+5, 5)
V1:(-1, 15)
W1:(4, 5)
V2: (h, k-a)
W2: (h-b, k)
V2: (-1, 5-10)
W2: (-1-5, 5)
V2: (-1, -5)
W2: (-6, 5)
Equation of the
Asymptotes
F2: (h, k-c)
y = 5+2(x+1)
y = 5+2x+2
y = 2x + 7
y = 5-2(x+1)
y = 5-2x-2
Standard Form of the Equation of an Ellipse:
General Form of the Equation of of an Ellipse:
2. Find the parts of a hyperbola whose transverse axis has length of
foci 12 units left and right of the center (0, 2).
and
Solution: Since the length of the transverse axis (2a) is
therefore
. Likewise, the foci have length 12 units left and right therefore c = 12.
Since the foci are counted left and right of the center therefore the orientation
is horizontal.
To find b, use the formula
Since the orientation is horizontal, center is at the (0, 2) and the values
of a, b and c are
, 9, and 12 consecutively therefore by substituting this to
the given pattern we can identify the parts of a hyperbola.
PIVOT 4A CALABARZON
32
Center
Vertices
Endpoints of Conjugate Axis
Foci
(0, 2)
V1:(h+a, k)
W1:(h, k+b)
F1:(h+c, k)
W1:(0, 2+9)
F1:(0+12, 2)
W1:(0, 11)
F1:(12, 2)
W2: (0, k-b)
F2: (h-c, k)
W2: (0, 2-9)
F2: (0-12, 2)
W2: (0, -7)
F2: (-12, 2)
V2: (h - a, k)
Equation of the
Asymptotes
Standard Form of the Equation of an Ellipse:
General Form of the Equation of an Ellipse:
E
Learning Task 3: Identify the length of transverse axis and conjugate axis, and
the orientation of the given standard form of the equation of a hyperbola.
1.
2.
3.
Learning Task 4: Identify parts, standard form and general form of the equation
of a hyperbola given its specific condition.
1. Find the parts of a hyperbola whose center is located at (-6, -3), one of its
foci is located at (-11, -3) and one of the endpoints of the conjugate axis is
located at (-6, 0).
2. The center of a hyperbola is at (-4,1), one of the vertices is at (-4,3) and the
length of the conjugate axis is 6 units.
33
PIVOT 4A CALABARZON
Learning Task 5: To find the answer, find the parts of a hyperbola given the
standard general form of its equation. Write the corresponding letter on the box
above the correct answer in the decoder.
Decode the Answer
It is an hour-glass shape, which means that it has two hyperbolas one on
each side.
1.
Orientation:
Letter
Parts
P
Center
Answer
K
Vertices
B
Endpoints of
Conjugate Axis
T
Foci
N
Equation of
Asymptotes
Graph:
2. 121y2 – 4x2 - 484 = 0
Orientation:
Letter
Parts
O
Center
Answer
R
Vertices
E
Endpoints of
Conjugate Axis
W
Foci
C
Equation of
Asymptotes
Graph:
(3, 1) &
(-1, 1)
(0, 0)
(5, 1) &
(-3, 1)
PIVOT 4A CALABARZON
(11, 0) &
(-11, 0)
&
(0, 0)
(1, 1)
(11, 0) &
(-11, 0)
&
34
(0, 0)
(0, 2) &
(0, -2)
(0, 2) &
(0, -2)
(5, 1) &
(-3, 1)
A
Learning task 6: After going through the different activities in this module, I am
sure that you learned a lot. I want you to share with me your thoughts about our
lesson, by completing the following statements:
I learned that ____________________________________
___________________________________________________.
I want to learn more about ______________________
___________________________________________________.
I can apply what I learned in _____________________
__________________________________________________.
Learning Task 7: Designing a Vase
I am sure that you are now equipped with skills and knowledge about hyperbola
that could help you in designing a hyperbolic object.
I want you to create your own design a hyperbolic vase of the table. You will be
sketching this in a rectangular coordinate system with 1 centimeter distance from
each number. You will submit a sketch of the design with corresponding equation
and parts of a hyperbola. Your output will be graded using the given rubric.
Category
Excellent
Very Satisfactory
Satisfactory
Needs
Improvement
Content- Accuracy
(20)
100% of the solutions
are correct. (20)
80 -99% of the solutions are correct (17)
60-79% of the solutions are correct (14)
Below 60% of the
solutions are correct
(11)
Presentation of
Output
Output is exceptionally
attractive in terms of
design, layout and neatness (15)
Output is attractive in
terms of design, layout and neatness (13)
Output is acceptably
attractive though it
may be a bit messy
(11)
Output is distractingly messy and not
attractive (9)
Complete understanding
of the mathematical
concepts is evident in
the presentation. (15
Substantial understanding of the mathematical concepts is
applied. (13)
Partial understanding
of the mathematical
concepts is applied.
(11)
Limited understanding of the mathematical concepts is applied. (9)
(15)
Mathematical Content/ Reasoning
(15
Source: Haydee Mojica’s Module in Pre-Calculus—Division of Cavite Province
35
PIVOT 4A CALABARZON
WEEKS
5-6
Solving Problems Involving Iconic Functions
Lesson
I
After going through this module, you are expected to: 1.)Transform general
to standard form of the equation; 2.) Recognize the types of conics and its orientation given the standard and general form of the equation; 3.) Identify the type of
degenerate cases given the standard and general form of the equation; 4.) Solve
problems involving conic sections; 5.) Appreciate the significant value of conics
into real-life situation.
Learning Task 1: Given the standard form of the equation. Identify the type of
conics being presented.
1.
2.
3.
4.
5.
Guide Questions:
1. What is your basis in identifying the types of conics given the standard form
of the equation?
2. Based on our previous discussion, can you determine the orientation of given type of conics?
D
Learning Task 2: Transform the following general form of the equation into
standard form and identify the type of conics, and the values of A and B on each
example.
1.
2.
3.
4.
11x2 + 7y2 + 22x – 126y + 501 = 0
2x2 + 2y2 – 20x – 4y – 20 = 0
2y2 – 15x2 – 40y – 90x + 35 = 0
y2 – 8y – 12x – 68 = 0
Guide Questions:
1. How did you transform the general form to standard form of the given
equation?
2. Did you see any pattern to easily recognize the type of conics given the
general form of the equation?
3. What do you think will be the role of identifying the values of A and B given
the general form of the equation in recognizing the type conics?
PIVOT 4A CALABARZON
36
Recognize conic sections and degenerate cases
The relationship between coefficients A and B are significant in identifying
the type of conics and its characteristics.
If A = B ≠ 0, the equation represents a circle.
If A ≠ B and AB > 0, the equation represents an ellipse.
The orientation of an ellipse is horizontal when A < B and vertical when A > B.
If A = 0 or B = 0, but not both, the equation represents a parabola.
The parabola opens a.) upward when B = 0 and D <0, b.) downward when B =
0 and D>0, c.) to the right when A = 0 and C <0 and d.) to the left when A =
0 and C>0
If AB < 0, the equation represents a hyperbola.
The orientation of a hyperbola is horizontal when A> 0 and vertical when B > 0.
In recognizing type of conics representing a standard form of the equation, it is
essential to familiarize yourself with the following:
Circle: (x – h)2 + (y – k)2 = r2
Ellipse:
: if a 2 >b2 then the orientation if horizontal and
if b2 >a2 then the orientation is vertical.
Parabola: If (x-h)2 = 4p(y-k) then the parabola opens upward. If (x-h)2 = -4p
(y-k) then the parabola opens downward. If (y-k)2 = 4p(x -h) then the parabola opens to the right. If (y-k)2 = -4p(x -h) then the parabola opens to
the left.
Hyperbola:
tal.
If then the orientation of hyperbola is horizonIf then the orientation of hyperbola is vertical.
In determining the types of degenerate cases, it is essential to familiarize yourself with the following:
If the left side in the standard form of the equation of a circle and ellipse are
equal to 0 then it is a point while if it has negative value, then it is an empty set.
If the left side in the standard form of the equation of a hyperbola is equal to 0
then it forms two intersecting lines. (Source: Learners Materials in Pre-Calculus)
Illustrative Examples
I. Direction: Identify the type of conics and its orientation given the standard
form of the equation.
Standard Form of the Equation
Type of Conics
Orientation
Ellipse
Vertical since 121>15
Hyperbola
(x-17)2 = -80(y+12)
Parabola
37
Vertical since y comes
first before x
Downward since it follows
the formula (x-h)2 = -4p(yk)
PIVOT 4A CALABARZON
II Direction: Identify the the values of A and B, type of conics and its orientation
given the general form of the equation.
General Form of the Equation
1. 5x2 + 55x−12y + 450 = 0
2.-11x2 + 20y2 + 18x – 15y + 25 = 0
3. 5x2 + 5y2 + 20x – 30y + 100 = 0
4.
121x2
+
16y2
+ 4x −128y + 292 = 0
A
B
Type of
Conics
5
0
Parabola
-11
20
Hyperbola
5
5
Circle
None
121
16
Ellipse
Vertical since
A>B
Orientation
Upward since
A>0, B=0 and
D<0
Vertical since
B>A
III. Direction: Transform the given general form of the equation and identify the
types of degenerate cases.
1. x2 + 5x + y2 −y + 7 = 0
Solution:
x2 + 5x + y2 −y + 7 = 0
(x2 + 5x +
)+ (y2 −y +
= -7 +
+
(x + )2 + (y - )2 =
Since the left side of the standard form of a circle has a negative value therefore it
is an empty set.
2. 36x2+360x-64y2−512y-124 = 0
Solution:
36x2+360x-64y2−512y-124 = 0
36(x2 + 10x + 25) - 64(y2 + 8y + 16) = 124 + 900 – 1024
36(x+5)2 – 64(y+4)2 = 0
Since the left side of the standard firm of a hyperbola has a 0 value therefore it
Situational Problems Involving Conic Sections
In solving situational problems involving conic sections, it is essential to follow
these steps:
1. Identify the given.
2. Determine what is being asked.
3. Show solution
Illustrative Examples
1. Romans built their amphitheaters always in an ovoid shape, or in simple words, an ellipse. One of the amphitheaters in Rome is the Colosseum, known for being the largest of
all the others, as well as the fact that it has been a part of Roman history.
The Colosseum, from its exterior walls, is 189 meters long and 156 meters wide. It has a
central arena (with the same elliptical shape) which is 87 meters long and 55 meters wide.
Given that the Colosseum (and the arena) are both horizontally aligned, if the Colosseum’s
center is found on the origin of the Cartesian plane (0,0), find the distance between the
amphitheater’s foci and their same-side central arena’s vertices (i.e. distance between the
focus with a positive x and the C.A. vertex with a positive x, and the same for negative).
PIVOT 4A CALABARZON
38
31
Identify the given values.
Major axis of the Colosseum = 189 meters;
= 94.5 m; Vertices = (94.5, 0) and (-94.5, 0)
Minor axis of the Colosseum = 156 meters;
= 78 m; Co-Vertices = (0, 78) and (0, -78)
Major axis of the central arena = 87 meters;
= 43.5 m; Vertices = (43.5, 0) and (-43.5, 0)
Minor axis of the central arena =
= 27.5 m; Co-Vertices = (0, 27.5) and (0, -27.5)
55 meters;
Determine what is being asked
Find the distance between the amphitheater’s foci and their same-side central arena’s vertices (i.e. distance between the focus with a positive x and the C.A. vertex
with a positive x, and the same for negative).
Solution
The formula for the foci is given as
. To find the foci of the Colosseum, we need to substitute the given values to the formula.
Now, we know that the foci of the ellipse are situated at (52.35, 0) and (52.35, 0) because the center of the ellipse is at the origin. From this we can use
the distance formula between the foci and the vertices of the central arena, or we
can simply subtract the two x values.
Final Answer
Therefore, the distance between the foci of the Colosseum and its central
arena’s vertices is 8.85 m.
2. A student from Tanza National Comprehensive High School got curious about
the circular logo just outside their school. He speculates that the logo has a diameter of 2 meters. He tells his math teacher that the equation of the circle can
be
if 1 meter is treated as 1 unit. His teacher says that
his constant is wrong. What should the constant be to make his answer right?
39
PIVOT 4A CALABARZON
Given:
The diameter of the circle is 2 meters.
The equation of the circle can be expressed as for
some constant k.
What is being asked?
What should the constant be to make his answer right?
Solution:
We can use the student’s equation and compress it into the standard form of the
equation, also with the use of completing the square.
Because 1 meter is
then
must be
treated as 1 unit, and the radius of the circle is 1 meter,
equal to , or basically, 1.
Substituting back to the student’s equation, it should then be:
Final Answer:
Therefore, the constant should be (positive) 2.
3. As nighttime approached, a woman opened the lampshade in her room, and immediately noticed that the lamp cast vertical hyperbolic shadows on the wall. As
she had nothing else to do, she decided to do some math. She found out that the
center of the hyperbola was 3 units away from the vertices, and the foci were 2
units away from the vertices. If the center is situated at the origin, find the general
form of the equation of these hyperbolic shadows, along with the lengths of its
transverse and conjugate axes.
Given:
Center: (0,0); Vertices: (0,3) and (0, -3); Foci: (0,5) and (0, -5)
What is being asked?
If the center is situated at the origin, find the general form of the equation of these
hyperbolic shadows, along with the lengths of its transverse and conjugate axes.
PIVOT 4A CALABARZON
40
Solution:
The standard form of the equation of a vertical hyperbola with center at the
origin is
and
. Since we already have
(distance from vertices to center)
(distance from foci to center), we only need
(distance from co-vertices to
center). To find this, we can use the formula
and manipulate it to get
a formula for finding b.
We can then substitute this to the equation.
The transverse axis in a hyperbola is the segment connecting the two vertices, therefore, the formula for this is 2a . Easily, we get its length by substituting,
and we get 6 units.
On the contrary, the conjugate axis is the segment connecting the covertices, which means the formula for this is 2b. We then get 8 units as an answer.
Final Answer
The general
form
of
the
equation
of
the
hyperbolic
shadow
is
with 6 units as length of transverse axis and 8 units as
length of conjugate axis.
E
Learning Task 3: Identify the type of conics or degenerate cases of the following
general form of the equation. Determine also the orientation if the equation represents a type of conics.
A. Circle
E. Point
1.
2.
3.
4.
5.
B. Parabola
F. Empty Set
C. Ellipse
D. Hyperbola
G. Two Intersecting Lines
x2 + y2 −18y−19 = 0
x2 −10x−48y + 244 = 0
−5x2 +60x+7y2 +84y +72 = 0
−144x2 −1152x + 25y2 −150y – 5679 = 0
9x2−72x−16y2−128y−256 = 0
41
PIVOT 4A CALABARZON
Learning Task 4: Identify the type of conics or degenerate cases of the following
standard form of the equation. Determine also the orientation if the equation represents a type of conics.
A – Circle
B – Parabola
C. Ellipse
D. Hyperbola
E. Point
F. Empty Set
G. Two Intersecting Lines
1. (x – 3)2 + (y + 6)2 = 0
2. (y – 5)2 = 18(x – 3)2
4.
5.
3.
Learning Task 5: Solve each situational problem involving conic sections.
1. A mathematician was shopping for furniture when he stumbles across the clocks area
and finds different sizes of clocks. He then thinks what he’s finding in the clock that he
would buy. He tells the shopkeeper that he would like a clock with a diameter that is irrational. The shopkeeper then gave him 3 clocks as a challenge. The first clock had this
equation:
; the second had the ends of its diameter on the
points (3,6) and (9,8); the last clock had an area of 144 square units. Which clock(s) would
fit the mathematician’s request?
2. An engineer is working on making a diagram for the cooling tower he was going to construct in the future. Cooling towers can be seen as towers with inward-curving sides, or
simply, hyperbola-looking sides. His daughter asks him how high the center of the hyperbolic sides of the tower was
, and he answers that the center can be situated at (0,15) in the Cartesian plane. The engineer then says that the asymptotes of the hyperbola are the following: , and the area of its auxiliary rectangle is 168 square units. The
daughter then finds the foci of the hyperbola. At what points on the Cartesian plane are
these foci found on?
3. Suspension bridges are known to be normal around the world, especially in developed
countries. These bridges usually have towers of some sort and cables, both of which have
great importance in the integral stability of the bridge. A STEM Student makes some plans
for his project on a suspension bridge model. He first tries to draw his model. The two towers are 12 meters high and 20 meters apart (10 meters away each from the origin in the
center), and the cable he drew between and from the top of the two towers are of course,
parabolic. The lowest point of the cable is 3 meters above the origin. He shows this model
to his teacher, and the teacher asks him, “What is the vertical distance from the road to
the cable at 5 meters away from the center?” What is the answer to the teacher’s question?
4. A certain celestial body is orbiting around a star somewhere in space. An astronomer
tries to graph out the elliptical orbit of the celestial body using dimensional analysis and
scaling. In his drawing, the celestial body is in the origin, and he finds out that the equation of the elliptical orbit is
for some value of k. If his
findings are true, find the location of the star (the center of the ellipse) in the Cartesian
plane.
PIVOT 4A CALABARZON
42
A
Learning Task 6: After going through the different activities in this module, I am
sure that you learned a lot. I want you to share with me your thoughts about our
lesson, by completing the following statements:
I have learned that ________________________________________
__________________________________________________________.
I realized that _____________________________________________
___________________________________________________________.
Learning Task 7: PHOTO CONIC SECTIONS: You will be working individually
about the application of conic sections into real-life situation by capturing photos
of different conic sections that you can see within the town. Also, you need to construct own situational problems that have in connection with each picture. The
solution of each problem is also needed.
GOAL: You are task to capture photos of different conic sections that you can see
within your town and construct own problems involving each picture.
ROLE: The students will act as photographer and problem maker.
AUDIENCE: The audience will be the Municipal Mayor.
SITUATION: The Municipal Mayor of your town, would like to know the ravishing
and flabbergasting establishments at your town. One of his consultants, gave him
an advice that the best pictures that they can present are those pictures that have
in connection with conic sections. However, he is very problematic because the
Provincial Governor asked him to present pictures so that they can be able to see
the captions and the histories of each pictures.
PRODUCT: The photographer will give stunning presentations of pictures in a
unique way with captions, problems and solutions.
STANDARDS:
43
PIVOT 4A CALABARZON