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Mathematics C30
Module 1
Lesson 5
Mathematics C30
Trigonometric Equations
Introduction to the Conics
299
Lesson 5
Mathematics C30
300
Lesson 5
Introduction
In an earlier lesson, you were shown how to find values of x, if an equation like sin x  1 ,
were given. With the calculator in degree mode, using the inverse and sine functions, or
the sin 1 function, the solution x  90  was found. Even though one solution to the
equation has been found, you may recall that there are infinitely many other solutions.
This lesson continues the study of solving equations. The use of trigonometric identities
enables one to solve more complicated equations.
The latter part of this lesson begins the study of conics with a study of circles and
parabolas. The study of ellipses and hyperbolas will begin the next lesson in Module 2.
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Lesson 5
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Lesson 5
Objectives
After completing this lesson you will be able to
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
solve a trigonometric equation by finding a particular
solution and by finding the general solution.

convert the equation of a circle from the general form to
the standard form, and vice versa.

sketch the graph of a circle.

convert the equation of a parabola from the general
form to the standard form and vice versa.

sketch the graph of a parabola.
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Lesson 5
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Lesson 5
5.1 Solving Equations – General and Particular
Solutions
y
2
2
1
150°
30°
1
30°
x
1
is given, one solution for  is the principal angle,   30  or
2
5

radians ,
radians. Another solution is the principal angle,   180   30   150  or
6
6
since sine is positive in the second quadrant also.
If an equation like sin  
All the principal angles satisfying the equation are called particular solutions to the
equation. Thus, particular solutions are found from 0 to 2  or 0° to 360°.
All other angles which are coterminal with the particular solutions are also solutions to
1
the equation, sin x  . The general solution to the equation is the infinite set of all
2
angles which are coterminal with the particular solutions. The general solution may be
written in a compact way.

x  30   n 360  and x  150   n 360 , n  I
or

x

5
 n 2   and x 
 n 2 , n  I
6
6
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Lesson 5
Example 1
Find all the particular and general solutions to 2 sin    3 .
Solution:
Rewrite the equation

sin   
3
2
Sine is negative in the third and fourth quadrants.
y
x
– 3
2
2
– 3
Recognize that the reference angles are both 60° by recalling the 30   60   90  triangle.
2
3
60°
1
Therefore, the particular solutions are:
 180   60   240  and 360   60   300 
or


 4
 5

and 2   
.
3
3
3
3
The general solution is:

  240   n 360  and   300   n 360  , or  
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4
5
 n 2   and  
 n 2  , n  I .
3
3
Lesson 5
Example 2
Find the particular solutions to cos 2  
1
.
2
Solution:
Solve for cos  .
1
cos   
Note the  value when you square root
both sides.
2
1
1

There are now two equations to solve, cos  

Recognize that the reference angles are 45° by recalling the way the sides are labelled
in the 45   45   90  triangle.
2
2
and cos   
.
2
1
45°
1
Each solution to the equation will have a reference angle of 45°.
y
Therefore, the particular solutions are:

for cos  
2
1
2
,   45  and 315°.
1
45° 1
x
45°
–1
2
y

for cos   
1
2
,   135  and 225°.
1
–1

In radians, the solutions are
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2
45°
–1
x
45°
2
 3 5 7
,
,
,
.
4 4
4
4
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Lesson 5
Example 3
Use a calculator to solve 2 + cos x = 2.0731 for all x in [ 0 , 2 ) .
Solution:
Rewrite the equation.



2  cos x  2 .0731
cos x  0 .0731
With the calculator in radian mode, the solution is 1.4976 radians.
This solution is in the first quadrant.
The other solution, in the fourth quadrant, is 2  1.4976  4.7856 .
Note that since  1 .4976 is not in [0 , 2 ) , it is not a solution. The positive angle of
measure 4.7856 is coterminal with  1 .4976 and is in [0 , 2 ) .
Exercise 5.1
1.
Find all the solutions 0    360  for each equation.
a.
csc    2
b.
cot   1
c.
sin   
d.
sec   
1
2
e.
csc   1
f.
tan   0
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Lesson 5
2.
3.
4.
5.
Find the general solutions, in radians, for each equation.
3
2
a.
cos  
b.
sin   1
c.
tan   1
d.
2 sin   1  0
e.
cot   1  0
Find all the solutions 0    2 for each equation.
3
4
a.
sin 2  
b.
cos 2   1
c.
tan
d.
3 cot 2   1  0
2
3
Find all the solutions in degrees, for each equation.
a.
cos x   0 .6406
b.
tan x  85
c.
sin
d.
cos x 
2
x  0.5869
2
5
Solve sin x  0 .6789 for
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
3
 x  .
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4
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Lesson 5
5.2 Solving Equations of the Form
sin (ax + b) = c.
The technique for solving these equations is similar to that of the previous section and will
simply be illustrated by the examples.
1.
Replace ax  b with A.
2.
Solve for A.
3.
If the answer is to be a general solution, find A as a general
solution by adding 2 n  now.
4.
Replace A with ax  b and solve for x.
Example 1
Solve sin
1
1
x   , for all possible values of x. Then find the solution, 0  x  2 .
2
2
Solution:
1
2
Step 1: Use the solution to the simpler equation, sin A   .
1
2
7
A
6
Step 2:
sin A  
(Remember sin is negative in quadrants 3 and 4.)
and
7
 2 n , n I
6
Step 3:
A
Step 4: Substitute
1
x for A.
2
1


x  A.
 Let
2


1
7
x
 2 n , n I
2
6
A
11
6
A
11 
 2 n , n I
6
1
11 
x
 2 n , n  I
2
6
Multiply each expression by 2 to isolate the x.
x
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7
 4 n , n  I
3
x
310
11 
 4 n , n  I
3
Lesson 5
Find the solution, 0  x  2 .

Test various values of n  I to find 0  x  2 .

If n = 0, x 

If n = 1 or  1 , x clearly falls out of the range from 0 to 2  .
7
11
and
. These are not in [0 , 2 ) .
3
3
Therefore, there are no solutions for Example 1 in [ 0 , 2 ).
Example 2
This example will be used as a model for Example 3.
Solve tan x  1 .
Solution:

Since tangent is negative in the second and fourth quadrants, the two particular
3
7
solutions are the principal angles, x   , and x   .
4
4
y
1
1
–1

The reference angle is
–1
x

.
4
The general solution to the equation is:

x
3
7
  n 2 , and x    n 2 , n  I .
4
4
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Lesson 5
Example 3
Solve tan 2 x    1 .
Solution:
Use the general solution to Example 2 to solve tan 2 x    1 .
(Let A  2 x   .)
tan A  1
has the solution
 tan 2 x    1
3
  n 2  ,
4
7
A    n 2  , n  I
4
A
3
  n 2   ,
4
7
2 x      n 2 , n  I
4
2x   
has the solution
Solve for x.
3
  n 2 
4
3
2 x      n 2  
4
7
2 x    n 2 
4
7
x    n , n  I
8
2x   
7
  n 2  
4
7
2 x      n 2  
4
11
2x 
  n 2  
4
11
x
  n , n  I
8
2x   
or
Therefore, the general solution to tan 2 x    1 is:

x
7
11
  n , n  I and x 
  n , n  I
8
8
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Lesson 5
Exercise 5.2
1.
2.
a.
1
Find the general solutions to cos x   .
2
b.
1
1

Find the general solution to cos  x      .
2
2

a.
Find the general solution to sin 3 x    
b.
Find all solutions in 0, 2  .
3.

1
Find the general solution to cos  x    1 .
3
4
4.
Find all x values for which sec x is undefined.
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3
.
2
Lesson 5
5.3 Using Algebraic Techniques and
Trigonometric Identities to Solve Equations
Some trigonometric equations must be reduced to a simpler form before they can be solved
easily. The following examples illustrate some of the techniques that can be used.
Example 1 (Factoring)
Solve 3 sin 2   4 sin   1  0 for the particular solutions.
Solution:

This is a quadratic equation and can first be solved for sin  by factoring.
3 sin   1 sin   1  0
3 sin   1  0
sin   1  0
1
sin   
3
sin   1
recall
if a b  0
then a  0
or b  0

set each factor to 0 and isolate
the trig function
Using a calculator, the solution to sin   
1
is  0 .3398 radians.
3
y
0.3398
0.3398
x
The particular solutions are:
   0.3398  3.4814 , and
 2  0.3398  5.9434 .
The solution to sin   1 is  
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.
2
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Lesson 5
Some trigonometric equations must first be changed to quadratic form before they can be
solved. This is done by applying one of the trigonometric identities.
Example 2 (Factoring)
Solve sin 2   cos 2   sin  .
Solution:


Before the quadratic formula can be applied, the trigonometric ratios must be the
same.
The equations must contain all sines or all cosines, but not several different functions.
Substitute for cos 2 
sin 2   cos 2   sin   0


sin 2   1  sin 2   sin   0
Simplify
sin 2   1  sin 2   sin   0
2 sin 2   sin   1  0
2 sin   1sin   1  0
Factor
Solve for  as in the previous example.

2 sin   1  0  sin   

sin   1  0  sin   1
1
2
Alternatively, the quadratic formula may be used to solve 2 sin 2   sin   1  0 .
1  1  4 2  1  1  9 1  3


2 2 
4
4
1
sin   1, 
2
sin  



 n 2  , n  I .
2
1
7
If sin    , then     n 2  and
2
6
11

  n 2 , n  I .
6
If sin   1 , then  
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Lesson 5
Example 3 (Squaring Both Sides)
Solve the equation sin   cos   1 .
Solution:
Square both sides of the equation.
sin
  cos   1 2
2
sin 2   2 sin  cos   cos 2   1
sin
2

  cos 2   2 sin  cos   1
1  2 sin  cos   1
2 sin  cos   0
sin 2   0
Determine the general solution.

Double-angle identity
Let A  2
sin A  0
A  n
2  n 

  n, n  I
2
In the interval [ 0 , 2 ) , the particular solutions would be   0 ,

3
,  , and
, n  I.
2
2
Each solution must be checked in the original equation since squaring both
3
sides may introduce values which are not solutions. The values  and  do
2

not satisfy the equation; therefore,   0 ,
is the solution.
2
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Lesson 5
Example 4 (Using an Addition or Subtraction Identity)
Solve sin 5 x cos 3 x  sin 3 x cos 5 x  1 .
Solution:
Apply the subtraction identity sin  A  B   sin A cos B  sin B cos A .

Substitute A = 5x and B = 3x.
sin 5 x  3 x   1
sin 2 x  1

 n 2  
2

x   n , n  I
4
2x 
Determine the general solution.
Exercise 5.3
1.
Find all the particular solutions in radians for each equation in the interval 0, 2  .
a.
2 cos x  1tan
b.
sin  cos   0
c.
2 cos 2   cos   0
d.
1  sin   2 sin
e.
3 sin 2   5 sin   2  0
f.
tan
g.
cos 2 x  3 sin x  1
h.
cos x  sin x
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x  1  0
2

x  sec x  1  0
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Lesson 5
2.
Find the general solution in degrees.
2
2
a.
cos 3 x cos x  sin 3 x sin x 
b.
sin  x  180  cos 90   sin 90  cos  x  180  
c.
tan 3 x  tan x
 1
1  tan 3 x tan x
d.
sin   tan 
e.
cos 2   2 cos 2   0
3
2
5.4 Conic Sections – The Circle
Conic Section
Par abola
Cir cle
H yper bola
E llipse
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Lesson 5
The conic sections are the curves generated by the intersections of a plane with one or two
sections of a cone.
y
For a plane parallel to the top or bottom, a circle is produced.
x
y
The closed curve produced by the intersection of a single section with
an inclined plane is an ellipse.
x
y
The curve produced by a vertical plane intersecting both sections is
a hyperbola.
x
y
The curve produced by the intersection of a single section with a
plane,
parallel to a side of the cone, is a parabola.
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x
Lesson 5
Because of this simple geometric description, the conic sections were studied by the
Greeks long before their application to planetary orbits was known. Apollonius wrote the
classic ancient work on the subject entitled On Conics. Kepler was the first to notice that
planetary orbits were ellipses, and Newton was then able to derive the shape of orbits
mathematically using calculus, under the assumption that gravitational force goes as the
inverse square of distance. Depending on the energy of the orbiting body, orbit shapes,
which are any of the four types of conic sections, are possible.
In this course, the equations of the four conic sections will be studied, and the properties of
the graph of the conic will be determined from the equation.
Since the algebraic technique of completing the square is important to the work on conics,
a brief review is provided first.
Completing the Square
A complete or perfect square is a product of two equal factors.

 x  b 2 ,  x  b 2 , or 2 a  5 b 2 , etc.
Some quadratic expressions can be expressed as a perfect square of a binomial.


x 2  2 bx  b 2   x  b 
x 2  2 bx  b 2   x  b 
2
2
These equations can be checked by multiplying the right hand side by the FOIL method.
An expression that has one of the above forms can be written as a perfect square.
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Lesson 5
Example 1
Write each quadratic expression as a perfect square of a binomial.
x2  6x  9
u 2  2u  1
16 c 2  24 ac  9 a 2
a.
b.
c.
Solution:
a.
b.
c.
x 2  6 x  9  x 2  2 3  x   3    x  3 
2
u 2  2 u  1  u  1 
2
2
16 c 2  24 ac  9 a 2  4 c   2 4 c 3 a   3 a 
2
 4 c  3 a 
2
2
It will be necessary, in the study of circles, to change a quadratic expression in such a way
that a portion of it may be written as a perfect square.
Example 2
Given x 2  6 x  5 , add the correct amount to each side of the equation so that the
left side may be expressed as a perfect square.
Solution:

This equation is in the form ax 2  bx  c , where a  1 .
2
b
Determine   .
2
Add 9 to both sides.
Write as a perfect square.
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1 
 2 6   9


x2  6x  9  5  9
 x  3 2
 14
321
Lesson 5
Example 3
Complete the square, for each variable separately, in the expression
x 2  2 x  y 2  10 y  9 .
Solution:
Add 1 to the x terms and 25 to the y terms.
x
2
 

 2 x  1  y 2  10 y  25  9  1  25
Write the expression with each variable as a perfect square.
 x  1  2   y  5 2
 35
Note that the same amount has to be added to the right side of the equation to
maintain equality.
2
b
In general, if   is added to the left and right side of x 2  bx  K , then the left
2
side can be written as a perfect square.
2
2
2
2
b
b
x 2  bx     K   
2
2
b

b
x    K  
2

2
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Lesson 5
Example 4
Complete the square in x 2  7 x  11 .
Solution:
2

2
49
1 
7 
 2 7    2   4


 
Add
49
to both sides.
4
2
7 
7 
x 2  7 x     11   
2
2
2
2
7
49

 x    11 
2
4

2
7
44 49


x   
2
4
4

2
7
93

x   
2
4

Completing the Square when the Coefficient of x 2 is not 1.
To complete the square in equations where the coefficient of x 2 is not 1, and a  0 , such as
in ax 2  bx  K , you must first factor the a and then complete the square of the remaining
factor.
b 

a x 2  x   K
a 

2
2
 2 b
 b  
 b 

a x  x     K  a 

a
 2 a  
 2a 

2

 1  b 
 b 
Inside the brackets, “the square of one-half the coefficient of x” is     or   .
 2a 
 2  a 
2
2
 b 
Since the new term on the left side is actually a   , this is what must be added
 2a 
to the right side to maintain equality.
2
 b 
A common error is to add only   to the right side.
 2a 
Mathematics C30
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Lesson 5
Example 5
Complete the square, for each variable separately, in 2 x 2  5 x  3 y 2  9 y  1 .
Solution:
Combine common terms.
2 x
Factor
5 

2 x 2  x   3 y 2  3 y  1
2 

.
Complete each square.
2
 

 5x  3 y2  9 y  1


2
2
2
2


5
5 
3 
5
3
2  x 2  x      3  y 2  3 y      1  2   3 
2
 4  
 2  
4 
2


2
2
5
3
25 27


2 x    3 y    1 

4
2
8
4


2
2
5
3
87


2 x    3 y   
4
2
8


Example 6
Complete the square for the variable x in the following expressions.
a.
y  x2  9x 1
b.
y  3x2  7x  5
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Lesson 5
Solution:
a.


y  x2  9x 1
2
2

9 
9
y  x 2  9 x      1   
 2  
2

2
9
Note that   is added and subtracted on the same side of the equation. It could
2
have just as well been added to the y on the left side of the equation.
2
9
81

y  x   1 
2
4

2
9
77

y  x   
2
4


b.
Factor the 3.
Complete the square.

y  3x2  7x  5
7 

y  3 x 2  x   5
3 

2
2

7
7  
7 
y  3  x 2  x      5  3 
3
 6  
6

2
7 
Note that the new term added is not   but is
6
be subtracted.
2
7 
3   , so the same amount must
6
2
7
11

y  3 x   
6  12

The Exercises at the end of this section contain several completing the square problems,
which you may do before going on to conics.
Mathematics C30
325
Lesson 5
The Circle
Geometrically, a circle is defined to be a curve created by the intersection of a horizontal
plane with a vertical cone.
Another definition of a circle is that it is the set of points in a plane that are a fixed
distance from a fixed point. The fixed distance is called the radius of the circle and the
fixed point is called the center of the circle.
Using this definition, the equation of a circle can be developed.
A Circle whose Center is at ( h, k) and whose Radius is r
y
P(x, y)
r
C(h, k)
x
Let P  x, y  be any point on a circle, whose center C is the point (h, k), and whose radius is
r.

The distance from C to P equals r.
Use the distance formula.
x  h 2   y  k 2
r
Square both sides.
 x  h  2   y  k 2
 r2
Remember that the distance formula can determine the distance between
two points  x, y  and  x1 , y1  on the coordinate plane using the formula,
 y  y1 2  x  x1 2 .
The Standard Equation of a circle with radius r and center (h, k) is
 x  h  2   y  k 2  r 2 ,
where:
Mathematics C30
• r is the radius, and
• h, k  is the center.
326
Lesson 5
Example 7
Find the standard equation of the circle whose center is (4, 1) and radius is 5.
Solution:



r=5
h=4
k=1
Write the standard form.
 x  h  2   y  k 2
 r2
Substitute the values for r, h, and k.
 x  4 2   y  1 2
 x  4  2   y  1 2
 52
 25
Example 8
5 and center is 7,  5  .
Find the standard equation of the circle whose radius is
Solution:



r 5
h 7
k  5
Write the standard form.
Substitute the values for r, h, and k.
Mathematics C30
 x  h 2   y  k 2
 r2
 x  7 2   y   5 2  
 x  7 2   y  5 2  5
327
5

2
Lesson 5
Example 9
A circle whose center is at 5,  2  passes through the point (2, 2). Determine its
equation and sketch the graph of the circle.
Solution:

The radius must be found first by calculating the distance from the center to the given
point on the circle.
Use the distance formula.
r=
5  2 2   2  2 2
r  25
r 5
The equation of the circle is  x  5    y  2   25 .
2
2
Sketch the graph.
y
x
Use a compass and sketch the circle by putting the point at the
center 5,  2  with a radius of 5.
Mathematics C30
328
Lesson 5
The General Form of the Equation of the Circle
The standard form of a circle can be expanded.
x
2
 

x  y  2 hx  2 yk  h  k  r
2

 0
 2 xh  h 2  y 2  2 yk  k 2  r 2
2
2
2
2
Let the constants be:



C  2 h
D  2 k
E  h2  k2  r2
Therefore, x 2  y 2  Cx  Dy  E  0 .
The General Equation of a Circle
x 2  y 2  Cx  Dy  E  0
where C, D, and E, are constants.
The recognizing features are:
• The x 2 and y 2 have coefficients 1.
• x 2 and y 2 are added.
Example 10
Write the general form of the circle with center (0, 1) and radius 6.
Solution:
Write the standard form.
Expand
 x  0 2   y  1  2
 36
x 2  y 2  2 y  1  36  0
x 2  y 2  2 y  35  0

Note that the general form does not show the x term, 0  x .
Mathematics C30
329
Lesson 5
Completing the Square to Determine the Center and Radius
Example 11
Find the center and radius of the circle whose general equation is
x 2  6 x  y 2  4 y  12 .
Solution:
Group the common variables.
Complete the squares inside
the brackets.
Write in standard form.



x
x 2  6 x  y 2  4 y  12

x  6 x  9   y  4 y  4   12  9  4
2
2
 
 6 x  y 2  4 y  12
2
 x  3 2   y  2 2  25
 x  h  2   y  k 2  r 2
h 3
k  2
r 5
The center h, k  is 3,  2  .
The radius, r, is 5.
Example 12
Change the given general form of a circle to standard form and state the center and
radius of the circle.
2 x 2  8 x  2 y 2  12 y  48
Mathematics C30
330
Lesson 5
Solution:

Since the coefficients of x 2 and y 2 are 2, the entire equation may be divided by 2 to get
a coefficient of 1 for x 2 and y 2 .
Divide the expression by 2.
Group the common variables.
Complete the squares.
Write in standard form
x
x 2  4 x  y 2  6 y  24
 

x  4 x  4   y  6 y  9   24  4  9
2
 4 x  y 2  6 y  24
2
2
 x  2 2   y  3 2
 37
37 and center  2, 3  .
This is a circle with radius
Exercise 5.4
1.
2.
For what value of m is each a perfect square?
a.
x2  6x  m
b.
x2 
2
xm
3
c.
x2 
1
xm
2
d.
2 x2  5x  m
e.
1 2
x  2x  m
2




Complete the square in the variable x.
a.
3

y   x2 
2

b.
y  2 x 2  4 x  11
c.
y  3x2  x  2
Mathematics C30

x  7

331
Lesson 5
3.
4.
5.
6.
Complete the square in each variable separately.
a.
x2  x  y2  y  4
b.
x2  2x  y2  8 y  0
c.
2 x 2  4 x  3 y 2  12 y  1
d.
x2  y2  2x  6 y  7  0
e.
5x2  4 x  3  2 y2  4 y  8  0
f.
x2  2 y2  4 y  0
Write the standard form of the equation of each circle.
a.
center (2, 0), radius 2
b.
center 3,  4  , radius 5
c.
center (0, 0), passing through (5, 4)
d.
center 3,  3  , passing through (0, 1)
e.
x2  y2  9  0
f.
x2  y2  4 x  2 y  4  0
State why each is not an equation of a circle.
a.
y  2x  3
b.
x2  y2  2x  6 y 1  0
c.
x2  3 y2  2x  2 y  5  0
Determine if x 2  y 2  2 x  2 y  100  0 is the equation of a circle. Explain your
answer.
Mathematics C30
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Lesson 5
5.5 The Parabola
The geometric description of a parabola is the curve which is formed when a plane
intersects a cone and the plane is parallel to a slant side of the cone.
Another definition of a parabola, is the one which can be used to find the equation of the
curve.
Definition: The parabola is the set of all points equidistant from a fixed point F and a
fixed line not containing F.
In the diagram below, FP  MP for every point P on the curve.
The fixed point is called the focus of the parabola and the fixed line is called the directrix.
The axis of symmetry is the line through the focus perpendicular to the directrix. The
portion of the parabola on one side of the axis is the mirror image of the portion on the
other side of the axis.
The vertex is the point at which the locus intersects the axis of symmetry. The vertex is
always midway between the focus and the directrix.
axis of
symmet r y
arm
P ( x, y)
arm
F
focu s
ver t ex
M

direct rix
FP  PM for every point P.
By use of the distance formula, if the focus and directrix are known, then the equation of
the parabola can be found.
Mathematics C30
333
Lesson 5
Example 1
Find the equation of the parabola whose focus is the point (4, 5) and whose directrix
is the straight line y  3 .
Solution:
y
5
4
3
2
1
–1
–2
F(4, 5)
P( x, y)
x
1 2 3 4
M (x, –3)




y = –3
Dir ect r ix
Let P(x, y) be any point on the parabola.
Join PF.
Draw PM perpendicular to the directrix y  3 .
The coordinates of M are  x,  3  .
Use the definition of a parabola and the distance formula to determine the equation.
PF  PM
 x  4 2   y  5 2   x  x 2   y  3 2
 x  4 2   y  5 2   x  x 2   y  3 2
 x  4 2   y  5 2   y  3 2
 x  4 2   y  3 2   y  5 2
 x  4 2  y 2  6 y  9  y 2  10 y  25
 x  4 2  16 y  16
 x  4 2  16  y  1 
If this equation is solved for y you get y 
Square both sides.
1 2 1
x  x  2 , which is the familiar quadratic
16
2
equation.
Mathematics C30
334
Lesson 5
From the definition of a parabola, the following points are important to note.

The arms of the parabola extend indefinitely away from the directrix. The parabola
does not cross the directrix but is entirely on the same side of the directrix that the
focus is on.

The vertex is midway between the focus and the directrix.

The axis of symmetry passes through the vertex and the focus and is perpendicular to
the directrix.

From any point on the parabola the distance to the focus equals the perpendicular
distance to the directrix.
Example 2
From the information given about a parabola, determine the remaining information
and then, make a sketch of the parabola including the focus and directrix.
a
b
c
d
Focus
(2, 0)
0,  3 
 2, 2 
Directrix
x  2
y3
Vertex
x=4
(0, 0)
(4, 2)
Axis of Symmetry
Solution:
y
a.

The axis of symmetry is y = 0, or the x-axis.

The vertex is (0, 0), which is the same distance
to the focus as to the directrix.
y= 0
(2, 0)
x
x = –2
Mathematics C30
335
Lesson 5
y
b.

The axis of symmetry is the y-axis or x  0 .

The vertex is (0, 0), on the axes of symmetry
and midway between the focus and directrix.
y= 3
x
(0, –3)
x= 0
y
c.

The axis of symmetry is y = 0, or the x-axis.

Since the vertex is 4 units away from the
directrix, it must also be 4 units from the focus.
This puts the focus at  4 , 0  on the axis of
symmetry.
y= 0
x
x= 4
d.



y
The distance between the focus and vertex is
4 + 2 = 6.
Both points lie on the line y = 2 so the axis of
symmetry is y = 2.
The directrix is x = 10 since it must be
perpendicular to the axis of symmetry and 6
units away from the vertex.
f(–2, 2)
v(4, 2)
y= 2
x
x = 10
Mathematics C30
336
Lesson 5
The Standard Equation of a Parabola
Suppose that the vertex of a parabola is at some point (h, k), and that the directrix is a
distance p units away from the vertex. (Note that p is defined to be a distance and
distance is always a positive quantity.) The focus is then also a distance p units away
from the vertex. If a parabola with arms upward is considered first, the following diagram
and coordinates are obtained.
y
P (x, y )
F
p
p
V (h, k )
y= k –p
M
x


The coordinates of M are  x, k  p  .
The coordinates of F are h , k  p .
The equation of the above parabola can now be determined using the definition that a
parabola is a set of points equidistant from a fixed line and a point not on the line, just as
in Example 1. This can be done using the distance formula.
PF  PM
 x  h 2   y  k  p 2   x  x 2   y  k  p 2
 x  h 2   y  k   p 2   y  k   p 2
 x  h 2   y  k 2  2 p  y  k   p 2   y  k 2  2 p  y  k   p 2
 x  h 2  4 p  y  k 
Mathematics C30
337
Lesson 5
Similarly, if the arms of the parabola are downward, the equation obtained would be
 x  h 2
 4 p y  k  .
y
x
M
p
V (h, k )
p
F


P (x, y )
The coordinates of M are  x, k  p  .
The coordinates of F are h , k  p .
PF  PM
 x  h 2   y  k  p 2   x  x 2   y  k  p 2
 x  h 2   y  k   p 2   y  k   p 2
 x  h 2   y  k 2  2 p y  k   p 2   y  k 2  2 p y  k   p 2
 x  h 2  4 p y  k 
Mathematics C30
338
Lesson 5
A horizontal parabola as shown in the next diagram will have the equation
 y  k 2  4 p x  h  . If the arms were to the left, the equation would be
 y  k 2
 4 p x  h  .
y
x
M (h – p, k )
P (x, y )
2
( y – k ) = 4p (x – h )
p
p
(h, k )
F (h + p, k )
x= h –p
Standard form of the Equation of a Parabola




 x  h 2  4 p  y  k 
 x  h 2  4 p y  k 
 y  k 2  4 p  x  h 
 y  k 2  4 p x  h 
arms up
arms down
arms right
arms left
The vertex of each parabola is (h, k) and p is the distance
from the vertex to the directrix or the focus.
The standard equations of the vertical and horizontal parabolas must be remembered.
When finding the equation of a parabola you need not always use the distance formula.
You may use the standard equation and simply fill in the values for h, k, p.
Mathematics C30
339
Lesson 5
Example 3
Find the standard form of the equation of a parabola for which the following
information is given.
a.
b.
c.
vertex (0, 0); directrix x + 1 = 0
focus (2, 2); vertex (1, 2)
focus  1, 0  ; directrix y  2
Solution:
a.
vertex (0, 0); directrix x  1  0
y
Sketch the parabola.
V (0 , 0 )
x
x = –1

The arms must be to the right.
Determine the equation.

k 0

h 0

p 1
Substitute the values.
 y  k 2
 4 p x  h 
V 0, 0  is the vertex.
The distance from the vertex to the directrix.
 y  k 2
 y  0 2
 4 p x  h 
 4 1  x  0 
y2  4 x
Mathematics C30
340
Lesson 5
b.
focus (2, 2); vertex (1, 2)
y
Sketch the parabola.

Remember that the directrix and focus are on
opposite sides of the vertex.
F (2 , 2 )
V (1 , 2 )
x

The focus is one unit away from the vertex so the directrix is the y-axis, one unit away
from the vertex.

Therefore, p = 1.

From V(1, 2) we get h = 1, k = 2.

The arms open to the right.
Write the equation.
Substitute the values.
Simplify.
c.
 y  k 2  4 p x  h 
 y  2 2  4 1  x  1 
 y  2 2  4  x  1 
focus  1, 0  ; directrix y  2
y
Sketch the parabola.
F(–1, 0)





The given information shows that the arms of the
parabola must be up.
The vertex must be V  1,  1 .
p=1
h  1
k  1
Write the equation.
Substitute the values.
Simplify.
Mathematics C30
V
x
y = –2
 x  h 2  4 p y  k 
 x  1 2  4 1  y  1 
 x  1 2  4  y  1 
341
Lesson 5
The General Equation of a Parabola
Any quadratic equation of the form y  ax 2  bx  c, or x  ay 2  by  c with a  0 is called
a general equation of a parabola.
These quadratic equations are converted to the standard form, which shows the vertex
and the distance p from the vertex to the focus. This is done by completing the square of
the second degree variable.
Example 4
The equation of a parabola is given by y 
of the vertex and the value of p.
1 2
x  2 x  4 . Determine the coordinates
8
Solution:

Use the completing the square method to convert the equation to the standard form.
y
1 2
x  2x  4
8
Multiply by 8.
Complete the square in the
expression which contains
the quadratic variable.
8 y  x 2  16 x  32
Factor 8 from the left term.
8 y  32   x  8 

8 y  x

8 y  x 2  16 x  32
2

 16 x  64  32  64
8 y   x  8   32
2
2
8 y  4   x  8 
 x  8 2
2
 8 y  4 
This equation is now in the form,  x  h   4 p y  k  .
2



The vertex is V 8,  4  .
The arms are up.
Since 4p = 8, p = 2.
Mathematics C30
342
Lesson 5
Sketch the graph of the parabola.
y
(8, –2)
x
2
(8, –4)
p= 2
Dir ect r ix
y = –6
Example 5
Write the standard form of the general parabola equation, y 2  8 x  4 y  4  0 , and
sketch the parabola.
Solution:
y 2  4 y  8 x  4
Complete the square in the y variable.
y 2  4 y  4  8 x  4  4
 y  2 2  8 x  8
 y  2 2  8  x  1 
This is in the form  y  k   4 p x  h  .
2



The vertex h , k   1,  2  .
Since  4 p  8 , p  2 .
The arms are to the left.
Mathematics C30
343
Lesson 5
Sketch the parabola.
y
2
F(–1, –2)
x
2
V(1, –2)
x= 3
Information about Parabolas
form of equation
vertex
focus
directrix
direction of opening
axis of symmetry
 x  h 2
 4 p y  k 
(h, k)
h , k  p
yk p
up (+4p)
down  4 p 
xh
 y  k 2
 4 p x  h 
(h, k)
h  p, k 
xh p
right (+4p)
left  4 p 
yk
Application
A parabaloid is a three-dimensional object whose shape is obtained by rotating a parabola
about its axis. Some examples of this are automobile headlamps, television antennas, and
search lights. All of these make use of a reflecting property of the parabola.
F
When light rays come in parallel to the axis of the parabola, they are all reflected to the
focus. Conversely, when there is a light source at the focus, the light rays will reflect off
the surface parallel to the axis.
Mathematics C30
344
Lesson 5
Example 6
Suppose that a parabolic TV antenna has a diameter of 3 m and a depth of 0.6 m.
How far away from the vertex should the device catching the radio waves be placed?
Solution:
Sketch a cross section of the antennae on the x-y axis with the vertex at the origin and the
arms up.
y
(1.5, 0.6)
0.6 m
x
1.5 m

The information given indicates that one point on the parabola is (1.5, 0.6) and the
vertex is at (0, 0).



The standard equation of such a parabola is  x  h   4 p y  k  .
h 0
k 0
2
Write the equation.
Substitute the values.
 x  0 2
 4 p y  0 
x  4 py
2

Solve for p by substituting (1.5, 0.6) to get p = 0.9375 m.

The focus is at (0, 0.9375), and the device should be placed on the axis of the parabola,
0.9375 m from the vertex.
Mathematics C30
345
Lesson 5
Exercise 5.5
1.
Sketch the parabola from the given information. Label the focus, vertex, directrix
and axis of symmetry in each case.
a.
b.
c.
2.
Use the definition of a parabola and the distance formula to find the standard
equation of each parabola from the given information.
a.
b.
c.
3.
focus  1, 0  , directrix y = 10
vertex (5, 7), directrix x  1
vertex 4 ,  4  , focus 0,  4 
vertex (0, 0), focus  4 , 0 
Write the standard form of the parabola equation. Determine p and the coordinates
of the vertex, and focus. Write the equation of the directrix.
a.
b.
c.
d.
5.
focus (0, 5), directrix y  5
vertex (0, 0), focus (3, 0)
focus (2, 1), directrix x  2
Write the standard equation of each parabola with the given properties.
a.
b.
c.
d.
4.
focus (6, 1), directrix, x  2
focus (6, 1), directrix y = 3
focus (0, 0), vertex 0,  3 
y2  x  2 y
y 2  8 x  4 y  20  0
x2  6x  8 y 7  0
x2  4 x  2 y  6  0
How far away is the focus from the vertex of a parabola with vertex (0, 0), if (5, 4)
and  5, 4  are also on the curve?
6. A parabolic reflector is 40 cm in diameter and 5 cm deep. How far from the vertex
should the light source be placed?
Mathematics C30
346
Lesson 5
Answers to Exercises
Exercise 5.1 1.
a.
b.
c.
d.
e.
f.
2.
a.
b.
c.
d.
e.
3.
a.
b.
c.
d.
4.
a.
b.
c.
d.
Mathematics C30
225°, 315°
135°, 315°
210°, 330°
150°, 210°
270°
0°, 180°, 360°

11 
 2 n  and
 2 n , n  I
6
6
3
  2 n , n  I
2

5
 2 n  and
 2 n , n  I
4
4

5
 2 n  and
 2 n , n  I
6
6
3
7
  2 n  and   2 n , n  I
4
4
 2 4 
,
,
,
3 3
3
0,  , 2 
 2 4 
,
,
,
3 3
3
 2 4 
,
,
,
3 3
3
5
3
5
3
5
3
129.8° + n(360°) and 230.2° + n(360°), n  I
89.3° + n(360°) and 269.3° + n(360°), n  I
50° + n(360°), 130° + n(360°), 230° + n(360°) and
310° + n(360°) n  I
66.4° + n(360°) and 293.6° + n(360°), n  I
347
Lesson 5
5.
In decimal form, the restriction on x is
0 .7854  x  2 .3562 .
By calculator, x  0 .7463 and   0.7463  2.3953 .
Therefore, no solution fits the restriction.
Exercise 5.2 1.
2.
a.
2
  n 2   , n  I
3
4
x    n 2  , n  I
3
b.
x
a.
7
2
  n , x 
9
3
For n  1, 0 , 1,
7
13
1
x  ,
, ,
9
9
9
x
10 
 4 n , n  I
3
14 
x
 4 n , n  I
3
x
b.
8
14
2
,
, 
9
9
9
16
  8n 
3
3.
x
4.
x
Mathematics C30
8
2
  n , n  I
9
3

 2 n , n  I
2
348
Lesson 5
Exercise 5.3 1.
a.
b.
c.
d.
2
4
1
5
, , , 
3
3
4
4

3
0, , , 
2
2
 3
 5
, , , 
2 2
3 3
 5 3
,
, 
6 6 2
e.
3 sin   1sin   2   0
f.
1
,
3
  0.3398 ,   0.3398
sin   2 (no solution)
2
4
0 , , 
3
3
 5
,
6 6
 5
, 
4 4
sin  
g.
h.
2.
a.
22 .5  n 180 , 157 .5  n180 , n  I
b.
sin  x  90  
c.
33.75° + n90°, 78.75° + n90°, n  I
d.
  n180  , n  I
e.
60° + n360°, 120° + n360°,
3
, x  30   n 360 , n  I
2
240° + n360°, 300° + n360°, n  I
Mathematics C30
349
Lesson 5
Exercise 5.4 1.
a.
b.
c.
d.
e.
9
1
9
1
16
25
4
1
2
2.
a.
3  103

y  x   
4
16

b.
y  2x  1  9
c.
1
11

y  3 x    1
6
12

a.
1 
1
1

x   y    4
2 
2
2

b.
 x  1 2   y  4 2
c.
2  x  1   3  y  2   15
d.
 x  1 2   y  3 2
e.
2
4

2
5 x    2  y  1   3
5
5

f.
x 2  2  y  1   2
2
2
2
3.
2
 17
2
2
 17
2
Mathematics C30
2
350
Lesson 5
4.
a.
b.
c.
 x  2 2  y 2  4
 x  3 2   y  4 2
 25
x  y  41
2
2
d.
e.
 x  3  2   y  3 2
f.
 x  2  2   y  1 2
 25
x  y 9
2
2
9
y
x
(2, –1)
5.
a.
b.
c.
This is a linear equation and its graph is a straight line.
There is a negative between x 2 and y 2 .
The coefficient of x 2 and y 2 are not equal.
6.
Completing the squares gives  x  1    y  1   98 .
2
2
The radius squared cannot be a negative, so this is not the equation of
a circle.
Mathematics C30
351
Lesson 5
Exercise 5.5 1.
a.
b.
c.
V(2, 1), arms to the right
V(6, 2), arms down
directrix, y  6 , arms up
2.
a.
b.
x 2  20 y
y 2  12 x
c.
 y  1 2
a.
 x  1 2  20  y  5 
 y  7 2  24  x  5 
 y  4 2  16  x  4 
3.
b.
c.
d.
4.
y 2  16 x
1
 3
, F  ,
4
 4
 8  x  2 , V 2, 2 , p  2, F 4 , 2 ,
a.
 y  1
b.
c.
 y  2 2
 x  3 2  8  y  2 ,
d.
 x  2
2
2
 x  1, V  1, 1  , p 
1

1  , x  1
4

x 0
V 3, 2 , p  2, F 3, 0 , y  4
1
1
1

 2  y  1  , V  2, 1  , p  , F  2, 1  , y 
2
2
2

25
9
1
16
16
5.
p
6.
20 cm
Mathematics C30
 8x
352
Lesson 5
Mathematics C30
Module 1
Assignment 5
Mathematics C30
353
Lesson 5
Mathematics C30
354
Lesson 5
Optional insert: Assignment #5 frontal sheet here.
Mathematics C30
355
Lesson 5
Mathematics C30
356
Lesson 5
Assignment 5
Values
(40)
A.
Multiple Choice: Select the best answer for each of the following and place a
check () beside it.
1.
2.
The particular solutions to sin
____
a.
____
b.
____
c.
____
d.
Mathematics C30

4

3

6
a.
b.
c.
d.
x
3
have a reference angle ***.
4
3
2
The solutions to cos x  
____
____
____
____
3.

2
1
in [ 0 , 2 ) are ***.
2
120°,  120 
120°, 240°
60°,  60 
30°, 60°
The general solution to sin x  1 is ***.
____
a.
____
b.
____
c.
____
d.

 2 n , n  I
2
  2 n , n  I

3
 2 n ,   2 n , n  I
4
4

 2 n , n  I
2

357
Lesson 5
4.
The number of solutions that sin
____
____
____
____
5.
6.
7.
____
a.
____
b.
____
c.
____
d.
x  2 sin x  0 has in [ 0 , 2 ) is ***.
0
1
2
4
The general solutions of cos
2
 2 n ,
3
4
 2 n ,
3
4
 4 n ,
3
4 8
,
3
3
1
1
x   are ***.
2
2
4
 2 n , n  I
3
8
 2 n , n  I
3
8
 4 n , n  I
3
All the values of x for which csc 2 x  is not defined are ***.
____
a.
____
b.
____
____
c.
d.

n, n  I
2

n, n  I
4
0, 
n , n  I
1
1

The number of solutions that the equation cos 2  x      has in
2
2

[ 0 , 2 ) is ***.
____
____
____
____
Mathematics C30
a.
b.
c.
d.
2
a.
b.
c.
d.
0
1
2
4
358
Lesson 5
8.
9.


The particular solutions to tan  2 x    1 are ***.
4

____
a.
____
b.
____
c.
____
d.
If sin
____
____
____
____
10.
11.
Mathematics C30
2

 n , n  I
2
 5
,
4 4
0, 

0,
2
n ,
x  sin x  6  0 , then x ***.
a.
b.
c.
d.
equals 2,  3
equals  2 , 3
equals 1,  6
has no solution
If 2 cos 2 x  sin x  1  0 , then sin x equals ***.
____
____
a.
b.
____
c.
____
d.
1
0, 1
1
 ,1
2
1
, 1
2
The solution to sin x  cos x in 0, 2  is ***.
____
a.
____
b.
____
c.
____
d.

,
4

,
2

,
4

4
3 5 7
,
,
4
4
4
3
2
5
4
359
Lesson 5
12.
The equation cos 5 x cos 3 x  sin 5 x sin 3 x  0 is equivalent to ***.
____
____
____
____
13.
cos 8 x  0
cos 2 x  0
sin 8 x  0
sin 2 x  0


To complete the square in 5 x 2  6 x  A  B , the values of A and B are
***.
____
____
____
____
14.
a.
b.
c.
d.
a.
b.
c.
d.
A =5, B = 5
A = 9, B = 9
A = 9, B = 45
A = 5, B = 25
The equation x 2  3 x  3 y 2  12 y  0 is equivalent to ***.
2
a.
3
1

2
 x     y  2  6
2
4

____
b.
3
3

2
 x    3  y  2   9
2
4

____
c.
____
d.
____
2
2
15.
The circle whose equation is x 2  y 2  8 y  0 has center and radius ***.
____
____
____
____
Mathematics C30
3
3

2
 x    3  y  2   9
2
4

3
1

 x    3 y  2   7
2
2

a.
b.
c.
d.
(0, 4), 4
0,  4 , 16
0,  4 , 4
 4, 0 , 16
360
Lesson 5
16.
The one equation of a circle is ***.
____
____
____
____
17.
Mathematics C30
x2  2x  y  4
x2  2 y2  3x  3 y  0
x2  2x  3  y2  0
2
a.
b.
c.
d.
3
4
6
12
The equation of the directrix of a parabola with vertex at 5,  7  and
focus at  3,  7  is ***.
____
____
____
____
19.
x 2  y 2  100
In the parabola whose equation is  x  4   12  y  2 , the distance
from the focus to the vertex is ***.
____
____
____
____
18.
a.
b.
c.
d.
a.
b.
c.
d.
x  13
x  13
y  13
y  7
The vertex of the parabola whose equation is x  5 y 2  2 y  1 is at ***.
____
a.
____
b.
____
c.
____
d.
1
 4
 ,  
5
 5
4 1
 , 
5 5
3 1
 , 
4 5
 17 1 
 , 
 20 5 
361
Lesson 5
20.
Mathematics C30
The general equation of the parabola with vertex at 1, 1 and focus
one unit to the left is ***.
____
a.
____
b.
____
c.
____
d.
 x  1 2   y  1 2
1
1
1
3
x   y2  y 
4
2
4
2
 x  1   4  y  1 
 y  1 2
 4  x  1 
362
Lesson 5
Answer Part B and Part C in the space provided. Evaluation of your solution
to each problem will be based on the following:
(8)
B.
•
A correct mathematical method for solving the problem is
shown.
•
The final answer is accurate and a check of the answer is shown
where asked for by the question.
•
The solution is written in a style that is clear, logical, wellorganized, uses proper terms, and states a conclusion.
1.
a.
Determine the general solution for  in degrees.
3 cot   3  0
b.
Determine the general solution for all first quadrant values
for  in radians.
3 cos 2   2 .586  0
Mathematics C30
363
Lesson 5
(8)
(8)
2.
3.
Find a solution in 0, 2 for each trigonometric equation.
a.
1

sin  x     0
2

b.

1
csc  x    1
6
4
Find the general solutions to each trigonometric equation.
a.
Mathematics C30
sin 2   3 cos   3  0
example 1, 5.3)
364
Determine  in degrees. (Refer to
Lesson 5
b.
(5)
4.
Mathematics C30
1
Determine  in radians.
2
(Hint: use addition and subtraction identity)
sin 3 x cos 2 .5 x  sin 2 .5 x cos 3 x  
Write the standard form of the equation for each circle and sketch.
a.
center (3, 1) diameter 6
b.
center  1, 1 passing through point 4 ,  3 
365
Lesson 5
(8)
(5)
5.
6.
Find the center and radius of each of the circles whose general
equation is given.
a.
x2  y2  4 x  6 y  7  0
b.
3 x 2  36 x  3 y 2  0
Given the standard form of the equation for a parabola, determine the
coordinates of the focus, vertex, and equation of the directrix and axis
of symmetry.
 x  5 2
Mathematics C30
 12  y  7 
366
Lesson 5
(8)
7.
Write the standard equation of each parabola which has the given
properties. Sketch the graph, and label the vertex, focus, and directrix,
with the coordinates and equation.
a.
focus  2, 3  ; directrix y  3
y
x
b.
 1
focus  0 ,  ; directrix x = 7.
 4
y
x
Mathematics C30
367
Lesson 5
(10)
C.
1.
Write a point form summary of the lesson, including essential
definitions.
_____
(100)
Mathematics C30
368
Lesson 5