The Cavite Mutiny of 1872 was an uprising of military personnel of Fort San Felipe, the Spanish arsenal in Cavite, Philippines on January 20, 1872, about 200 Filipino military personnel of Fort San Felipe Arsenal in Cavite, Philippines, staged a mutiny which in a way led to the Philippine Revolution in 1896. The 1872 Cavite Mutiny was precipitated by the removal of long-standing personal benefits to the workers such as tax (tribute) and forced labor exemptions on order from the Governor General Rafael de Izquierdo. Many scholars believe that the Cavite Mutiny of 1872 was the beginning of Filipino nationalism that would eventually lead to the Philippine Revolution of 1896.

1. Math 111-Precalculus
Remegia L. Ganot-Math Faculty
Department of Mathematics &
Statistics, USeP-CAS
Free template by

2. 2
Module 5: Functions and Relations
Definition of Terms
Domain and Range
Graphical Representation of Functions
Operations on Functions
Graphs and Equations

3. 3
Theorem 5.56. In a right triangle, if 𝑎 and 𝑏 are the lengths of the perpendicular
sides and 𝑐 is the length of the hypotenuse, then 𝑐2
= 𝑎2
+ 𝑏2
.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
𝑦
𝑎
𝑏
𝑐

4. 4
Theorem 5.57. The distance between two points 𝑃1(𝑥1, 𝑦1) and 𝑃2(𝑥2, 𝑦2) is given
by 𝑃1𝑃2 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
𝑦
𝑃2(𝑥2,𝑦2)
𝑃1(𝑥1, 𝑦1)
𝑦2 − 𝑦1
𝑥2 − 𝑥1
Illustration.
The distance between the two points
𝑃1(−2, 1) and 𝑃2(4, 7) is
𝑃1𝑃2 = 4 − (−2) 2 + 7 − 1 2
= 62 + 62
= 6 2 units.

5. 5
The distance between two points 𝑃1(𝑥1, 𝑦1) and 𝑃2(𝑥2, 𝑦2) in a horizontal segment
is given by 𝑃1𝑃2 = 𝑥2 − 𝑥1
2 + 𝑦1 − 𝑦1
2 = 𝑥2 − 𝑥1
2 = 𝑥2 − 𝑥1 .
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
𝑦
𝑃2(𝑥2, 𝑦1)
𝑃1(𝑥1, 𝑦1)
𝑃1𝑃2
Illustration.
The distance between the two points
𝑃1(−7, 7) and 𝑃2(5, 7) is
𝑃1𝑃2 = 5 − (−7)
= 12 units.

6. 6
The distance between two points 𝑃1(𝑥1, 𝑦1) and 𝑃2(𝑥1, 𝑦2) in a vertical segment is
given by 𝑃1𝑃2 = 𝑥1 − 𝑥1
2 + 𝑦1 − 𝑦1
2 = 𝑦2 − 𝑦1
2 = 𝑦2 − 𝑦1 .
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
𝑦
𝑃2(𝑥1, 𝑦2)
𝑃1(𝑥1, 𝑦1)
𝑃1𝑃2
Illustration.
The distance between the two points
𝑃1(7, 5) and 𝑃2(7, −7) is
𝑃1𝑃2 = −7 − 5
= 12 units.

7. 7
Theorem 5.58. If 𝑎, 𝑏 and 𝑐 are the lengths of the sides of a triangle and 𝑐2 =
𝑎2 + 𝑏2, then the triangle is a right triangle, and 𝑐 is the length of the hypotenuse
side.
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
𝑥
𝑦
Example 5.59. Show that the points 𝑃(1, 2), 𝑄(4, 7), and 𝑅(−9, 8) are vertices of
a right triangle.
𝑃(1, 2)
𝑄(4, 7)
𝑅(−9, 8)
Solution: Applying Theorem 5.58, we have 𝑃𝑄 =
4 − 1 2 + 7 − 2 2 = 34,
𝑃𝑅 = −9 − 1 2 + 8 − 2 2 = 136, and
𝑄𝑅 = −9 − 4 2 + 8 − 7 2 = 170. Observe
that
𝑃𝑄
2
+ 𝑃𝑅
2
= 34
2
+ 136
2
= 170
= 𝑄𝑅
2
.
Thus the given 3 points are vertices of a right triangle.

8. 8
Definition 5.60 (Midpoint). If 𝑀(𝑥, 𝑦) is the midpoint of the line segment from
𝑃1(𝑥1, 𝑦1) and 𝑃2(𝑥2, 𝑦2), then 𝑥 =
𝑥1+𝑥2
2
and 𝑦 =
𝑦1+𝑦2
2
.
Definition 5.60 (Median). A median of a triangle is a line segment from a vertex
to the midpoint of the opposite side.
Example 5.61. Find the length of the medians of the triangle having vertices
𝐴(2, 3), 𝐵(3, −3) and 𝐶(−1, −1).
Solution.
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
𝑥
𝐴(2, 3)
𝐵(3, −3)
𝐶(−1,−1)
Let the midpoint of 𝐴𝐵, 𝐴𝐶 and 𝐵𝐶 be respectively, 𝑀,
𝑁 and 𝑂. Then
𝑀 𝑥, 𝑦 = 𝑀
2+3
2
,
3+(−3)
2
= 𝑀
5
2
, 0 ,
𝑁 𝑥, 𝑦 = 𝑁
2+(−1)
2
,
3+(−1)
2
= 𝑁
1
2
, 1 ,
𝑀
5
2
, 0
𝑁
1
2
, 1
𝑂(1,−2)

9. 9
and 𝑂 𝑥, 𝑦 = 𝑂
3+(−1)
2
,
−3+(−1)
2
= 𝑂(1, −2).
Thus the lengths of the median are
𝐴𝑂 = 1 − 2 2 + −2 − 3 2 = 26 units,
𝐵𝑁 =
1
2
− 3
2
+ 1 − (−3) 2 =
89
4
=
1
2
89 units, and
𝐶𝑀 =
5
2
− (−1)
2
+ 0 − (−1) 2 =
53
4
=
1
2
53 units.
Definition 5.62 (Slope). If 𝑃1(𝑥1, 𝑦1) and 𝑃2(𝑥2, 𝑦2) are any two distinct points on
line 𝑙, which is not parallel to the 𝑦 axis, then the slope of 𝑙, denoted by 𝑚, is given
by
𝑚 =
𝑦2−𝑦1
𝑥2−𝑥1
.

10. 10
Note:
1. From Definition 5.62 we see that if 𝑃(𝑥, 𝑦) and 𝑃1(𝑥1, 𝑦1) are any points on a
line then the point-slope form of an equation of the line is
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1 .
2. If in the point-slope form we choose the particular point (0, 𝑏), that is, the point
where the line intersects the 𝑦 axis, for the point 𝑥1, 𝑦1 , we have
𝑦 − 𝑏 = 𝑚 𝑥 − 0 ⟺ 𝑦 = 𝑚𝑥 + 𝑏
and this is called the slope-intercept form of an equation of the line.
Theorem 5.63.
(i) An equation of the vertical line having 𝑥 intercept 𝑎 is 𝑥 = 𝑎.
(ii) An equation of the horizontal line having 𝑦 intercept 𝑏 is 𝑦 = 𝑏.

11. 11
Theorem 5.63.
(i) An equation of the vertical line having 𝑥 intercept 𝑎 is 𝑥 = 𝑎.
(ii) An equation of the horizontal line having 𝑦 intercept 𝑏 is 𝑦 = 𝑏.
Theorem 5.64. The graph of the equation 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 where 𝐴, 𝐵, and 𝐶 are
constants and where not both 𝐴 and 𝐵 are zero is line.
Theorem 5.65. If 𝑙1 and 𝑙2 are two distinct nonvertical lines having slopes 𝑚1 and
𝑚2, respectively, then 𝑙1 and 𝑙2 are parallel if and only if 𝑚1 = 𝑚2.
Theorem 5.66. Two nonvertical lines 𝑙1 and 𝑙2, having slopes 𝑚1 and 𝑚2,
respectively, are perpendicular if and only if 𝑚1𝑚2 = −1.

12. 12
Example 5.67. Determine by means of slopes whether the points 𝐴 −3, −4 ,
𝐵(2, −1) and 𝐶(7, 2) are collinear.
Solution: Points are said to be collinear if they lie on the same line. Thus the
given points are collinear if we can show that the slope 𝑚1 of 𝐴𝐵 and the slope
𝑚2 of 𝐵𝐶 are equal with 𝐵 as common point. Now,
𝑚1 =
−1−(−4)
2−(−3)
=
3
5
and 𝑚2 =
2−(−1)
7−2
=
3
5
.
Since 𝑚1 = 𝑚2, it follows that the line through 𝐴 and 𝐵 and the line through 𝐵
and 𝐶 have the same slope and contain the common point 𝐵. Therefore the given
points 𝐴, 𝐵 and 𝐶 are collinear.
Example 5.68. Given the line 𝑙 having the equation 5𝑥 + 4𝑦 − 20 = 0, find an
equation of the line through the point (2, −3) that is (a) parallel to 𝑙, and (b)
perpendicular to 𝑙.

13. 13
Solution: To visualize, we sketch the graph first.
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
𝑥
𝑦
𝑃(2, −3)
Solving for 𝑥 and 𝑦
intercepts of the linear
equation 5𝑥 + 4𝑦 − 20 = 0
we have,
𝑥-intercept:
𝑦 = 0 ⟹ 𝑥 = 4
𝑦-intercept:
𝑥 = 0 ⟹ 𝑦 = 5
5𝑥 + 4𝑦 − 20 = 0
Perpendicular line passing
through the point 𝑃(2, −3)
Parallel line passing
through the point 𝑃(2, −3)

14. 14
Solution:
First we need to get the slope of 5𝑥 + 4𝑦 − 20 = 0, that is
4𝑦 = −5𝑥 + 20
𝑦 = −
5
4
𝑥 + 5
which implies that the slope is 𝑚 = −
5
4
. Let 𝑚∥ and 𝑚⊥ be the slope of parallel
and perpendicular lines respectively. Then
(a) 𝑚∥ = −
5
4
and since the required line contains the point (2, −3), we use the
point-slope form which gives
𝑦 + 3 = −
5
4
𝑥 − 2 ⟺ 5𝑥 + 4𝑦 + 2 = 0.
(b) 𝑚⊥ =
4
5
and so we have
𝑦 + 3 =
4
5
𝑥 − 2 ⟺ 4𝑥 − 5𝑦 − 23 = 0.

15. 15
Example 5.69. Prove that points 𝐴 6, −13 , 𝐵 −2, 2 , 𝐶(13, 10), and 𝐷(21, −5)
are vertices of a square.
Solution:
-30 -25 -20 -15 -10 -5 5 10 15 20 25 30
-30
-20
-10
10
20
30
𝑥
𝑦
𝐴 6,−13
𝐵 −2,2
𝐶(13, 10)
𝐷(21,−5)
To prove that points 𝐴, 𝐵, 𝐶 and 𝐷 are vertices of
a square, we need to show the following:
(i) 𝐴𝐵 = 𝐴𝐷 = 𝐵𝐶 = 𝐶𝐷
(ii) Let 𝑚1, 𝑚2, 𝑚3, and 𝑚4 be the respective
slope of 𝐴𝐵, 𝐴𝐷, 𝐵𝐶, and 𝐶𝐷, then show
that 𝑚1𝑚2 = −1, 𝑚1𝑚3 = −1,
𝑚2𝑚4 = −1, and 𝑚3𝑚4 = −1.
𝑚1
𝑚2
𝑚3
𝑚4

16. 16
Now, 𝐴𝐵 = −2 − 6 2 + 2 − (−13) 2 = 17
𝐴𝐷 = 21 − 6 2 + −5 − (−13) 2 = 17
𝐵𝐶 = 13 − (−2) 2 + 10 − 2 2 = 17 and
𝐶𝐷 = 21 − 13 2 + −5 − 10 2 = 17.
Therefore, 𝐴𝐵 = 𝐴𝐷 = 𝐵𝐶 = 𝐶𝐷 .
Computing for 𝑚1, 𝑚2, 𝑚3, and 𝑚4, we have
𝑚1 =
2−(−13)
−2−6
=
15
−8
, 𝑚2 =
−5−(−13)
21−6
=
8
15
,
𝑚3 =
10−2
13−(−2)
=
8
15
, and 𝑚4 =
−5−10
21−13
=
−15
8
.
Observe that
𝑚1𝑚2 = −
15
8
8
15
= −1, 𝑚1𝑚3 = −
15
8
8
15
= −1,
𝑚2𝑚4 =
8
15
−
15
8
= −1, and 𝑚3𝑚4 =
8
15
−
15
8
= −1.

17. 17
Therefore the given four points are vertices of a square.
Solving for a diagonal, we obtain
𝐴𝐶 = 13 − 6 2 + 10 − (−13) 2 = 17 2.
Example 5.70. Given the line 𝑙 having the equation 2𝑦 − 3𝑥 − 4 = 0 and the
point 𝑃(1, −3), find (a) an equation of the line through 𝑃 and perpendicular to 𝑙
and (b) the shortest distance from 𝑃 to 𝑙.
Solution:
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
Let us first solve the 𝑥 and 𝑦 intercepts for the
given linear equation 2𝑦 − 3𝑥 = 4
𝑥-intercept:
𝑦 = 0 ⟹ 𝑥 = −
4
3
𝑦-intercept:
𝑥 = 0 ⟹ 𝑦 = 2
2𝑦 − 3𝑥 = 4
𝑃(1,−3)
𝑄(𝑥, 𝑦)

18. 18
(a) 2𝑦 − 3𝑥 − 4 = 0
2𝑦 = 3𝑥 + 4
𝑦 =
3
2
𝑥 +
4
2
∴ 𝑚𝑙 =
3
2
⟹ 𝑚⊥ = −
2
3
.
Thus, the equation of the perpendicular line to 𝑙 through point 𝑃(1, −3) is
𝑦 − −3 = −
2
3
(𝑥 − 1)
𝑦 + 3 = −
2
3
(𝑥 − 1)
3𝑦 + 9 = −2𝑥 + 2
2𝑥 + 3𝑦 + 7 = 0.

19. 19
(b) To find the shortest distance we need to solve the point of intersection between
given line 𝑙 which is 2𝑦 − 3𝑥 − 4 = 0 and the line perpendicular to 𝑙 which is
2𝑥 + 3𝑦 + 7 = 0. Now solving for the values of 𝑥 and 𝑦 of the system below,
−3𝑥 + 2𝑦 = 4
2𝑥 + 3𝑦 = −7
we have 𝑥 = −2 and 𝑦 = −1, that is, 𝑄(−2, −1). Therefore the shortest distance
containing points 𝑃(1, −3) and 𝑄(−2, −1) is
𝑃𝑄 = −2 − 1 2 + −1 − (−3) 2 = 13 units.

20. 20
Example 5.71. If two vertices of an equilateral triangle are (−4, 3) and (0, 0), find
the third vertex.
Solution:
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
𝑥
𝐵(−4,3)
𝐴(0, 0)
Let 𝐴(0, 0), 𝐵(−4, 3) and 𝐶(𝑥, 𝑦). Since we are
dealing with equilateral triangle, it follows that
𝐴𝐵 = 𝐵𝐶 = 𝐴𝐶

21. 21
(1) 𝐴𝐵 = −4 − 0 2 + 3 − 0 2 = 5
(2) 𝐵𝐶 = 𝑥 − (−4) 2 + 𝑦 − 3 2 = 𝑥 − 4 2 + 𝑦 − 3 2
(3) 𝐴𝐶 = 𝑥 − 0 2 + 𝑦 − 0 2 = 𝑥2 + 𝑦2
Recall: 𝐴𝐵 = 𝐵𝐶 = 𝐴𝐶
Equations (2) & (3): 𝐵𝐶 = 𝐴𝐶
𝑥 − 4 2 + 𝑦 − 3 2 = 𝑥2 + 𝑦2
𝑥 − 4 2 + 𝑦 − 3 2
2
= 𝑥2 + 𝑦2
2
𝑥 − 4 2 + 𝑦 − 3 2 = 𝑥2 + 𝑦2

22. 22
𝑥2 + 8𝑥 + 16 + 𝑦2 − 6𝑦 + 9 = 𝑥2 + 𝑦2
8𝑥 − 6𝑦 + 25 = 0
6𝑦 = 8𝑥 + 25
4 ∴ 𝑦 =
8𝑥+25
6
and 𝑥 =
6𝑦−25
8
Equations (1) & (3): 𝐵𝐶 = 𝐴𝐶
𝑥2 + 𝑦2 = 5
𝑥2 + 𝑦2
2
= 5 2
𝑥2
+ 𝑦2
= 25

23. 23
Equations (4) & (5):
𝑥2 +
8𝑥 + 25
6
2
= 25
𝑥2 +
64𝑥2
+ 400𝑥 + 625
36
= 25
𝑥2
+
64𝑥2 + 400𝑥 + 625
36
= 25 (36)
36𝑥2 + 64𝑥2 + 400𝑥 + 625 = 900
36𝑥2 + 64𝑥2 + 400𝑥 − 275 = 0
36𝑥2
+ 64𝑥2
+ 400𝑥 − 275 = 0
1
25

24. 24
4𝑥2 + 16𝑥 − 11 = 0
Quadratic Formula: 𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
𝑥 =
−16 ± 162 − 4(4)(−11)
2(4)
𝑥 =
−16 ± 432)
8
𝑥 = −2 ±
3
2
3
6𝑦 − 25
8
2
+ 𝑦2
= 25
Solving for 𝑦:

25. 25
36𝑦2 − 300𝑦 + 625
64
+ 𝑦2
= 25
36𝑦2
− 300𝑦 + 625
64
+ 𝑦2 = 25 (64)
36𝑦2
− 300𝑦 + 625 + 64𝑦2
= 1600
100𝑦2
− 300𝑦 − 975 = 0
100𝑦2
− 300𝑦 − 975 = 0
1
25
4𝑦2
− 12𝑦 − 39 = 0

26. 26
Quadratic Formula: 𝑦 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
𝑦 =
−(−12) ± (−12)2−4(4)(−39)
2(4)
𝑦 =
−(−12) ± (−12)2−4(4)(−39)
2(4)
𝑦 =
12 ± 768
8
𝑦 =
12 ± 768
8
=
3
2
± 2 3

27. 27
Therefore the third vertex of an equilateral triangle is either in first quadrant at the
point
𝐶 −2 +
3
2
3,
3
2
+ 2 3
or in the third quadrant at the point
𝐶 −2 −
3
2
3,
3
2
− 2 3

28. 28
Definition 5.72. The parabola is defined to be the set of all points in a plane equidistant from
the fixed point which is called the focus and the fixed line which is called the directrix. The
line through the focus perpendicular to the directrix is called the axis of the parabola.
The general/standard equation of the parabola which is either concave upward or
downward is of the form
(𝑥 − ℎ)2= 4𝑝(𝑦 − 𝑘)
which has the following properties:
(a) Vertex: 𝑉(ℎ, 𝑘);
(b) when 𝑝 > 0, the parabola concaves upward; otherwise concave downwards when 𝑝 < 0;
(c) The location of the focus is at point 𝐹(ℎ, 𝑘 + 𝑝);
(d) The Latus Rectum is the chord passes through the focus whose length is
𝐿𝑅 = 4𝑝 ;
(e) The equation of the directrix is 𝑦 = 𝑘 − 𝑝.
(f) The equation of the axis is 𝑥 = ℎ.

29. 29
Example 5.73. Let us draw a sketch of the parabola having an equation 𝑥2
− 4𝑥 − 8𝑦 −
28 = 0 and determine the properties (a)-(e) in the previous slide. The complete solution is
shown below.
𝑥2
− 4𝑥 − 8𝑦 − 28 = 0
𝑥2
− 4𝑥 = 8𝑦 + 28
𝑥2 − 4𝑥 +
−4
2
2
= 8𝑦 + 28 + 4 (by completing the square method)
(𝑥 − 2)2= 8𝑦 + 32
(𝑥 − 2)2= 8(𝑦 + 4)
Thus,
(a) Vertex: 𝑉 2, −4 ;
(b) Concavity: 4𝑝 = 8 ⟹ 𝑝 = 2 > 0, parabola concaves upward;
(c) Focus: 𝐹 2, −4 + 2 = 𝐹(2, −2);
(d) Length of latus rectum: 𝐿𝑅 = 8 = 8 with extreme points at 𝐸′(−2, −2) and 𝐸(6, −2);
(e) Directix: 𝑦 = 𝑘 − 𝑝 ⟹ 𝑦 = −4 − 2 = −6; and
(f) Axis: 𝑥 = 2.

30. 30
Graph
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
x
y
𝑉(2, −4)
𝐹(2, −2)
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥: 𝑦 = −6
𝐸(6,−2)
𝐸′(−2,−2)
𝐿𝑎𝑡𝑢𝑠 𝑅𝑒𝑐𝑡𝑢𝑚
𝐴𝑥𝑖𝑠: 𝑥 = 2
Note:
* Function
* Domain= 𝑥 𝑥 ∈ ℝ
= (−∞, +∞)
* Range= 𝑦 𝑦 ≥ −4
= [−4, −∞)

31. 31
Definition 5.73. Suppose we have an equation of a parabola is 𝑦 +
1
4
𝑥2
− 𝑥 − 6 = 0. Then
we have
𝑦 +
1
4
𝑥2
− 𝑥 − 6 = 0
1
4
𝑥2
− 𝑥 = −𝑦 + 6 (4)
𝑥2
− 4𝑥 = −4𝑦 + 24
𝑥2 − 4𝑥 +
−4
2
2
= −4𝑦 + 24 + 4 (by completing the square)
𝑥 − 2 2
= −4𝑦 + 28
𝑥 − 2 2
= −4(𝑦 − 7)
Hence,
(a) Vertex: 𝑉 2,7 ;
(b) Concavity: 4𝑝 = −4 ⟹ 𝑝 = −1 < 0, parabola concaves downward;
(c) Focus: 𝐹 2, 7 + (−1) = 𝐹 2,6 ;
(d) Length of latus rectum: 𝐿𝑅 = −4 = 4 with extreme points at 𝐸′(0,6) and 𝐸(4, 6);
(e) Directix: 𝑦 = 𝑘 − 𝑝 ⟹ 𝑦 = 7 − (−1) = 8; and
(f) Axis; 𝑥 = 2.

32. 32
Graph
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
x
y
𝑉(2,7)
𝐹(2,6)
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥: 𝑦 = 8
𝐸(4, 6)
𝐸′(0, 6)
𝐿𝑎𝑡𝑢𝑠 𝑅𝑒𝑐𝑡𝑢𝑚
𝐴𝑥𝑖𝑠: 𝑥 = 2
Note:
* Function
* Domain= 𝑥 𝑥 ∈ ℝ
= (−∞, +∞)
* Range= 𝑦 𝑦 ≤ 7
= (−∞, 7]

33. 33
The general/standard equation of the parabola which is either concave to the
right or to the left is
(𝑦 − 𝑘)2
= 4𝑝(𝑥 − ℎ)
which has the following properties:
(a) Vertex: 𝑉(ℎ, 𝑘);
(b) when 𝑝 > 0, the parabola concaves to the right; otherwise concave to the
left when 𝑝 < 0;
(c) The location of the focus is at point 𝐹(ℎ + 𝑝, 𝑘);
(d) The Latus Rectum is the chord passes through the focus whose length is
𝐿𝑅 = 4𝑝 ;
(e) The equation of the directrix is 𝑥 = ℎ − 𝑝.
(f) Axis: 𝑦 = 𝑘

34. 34
Definition 5.74. Given the equation of a parabola 𝑥 − 2𝑦2
− 8𝑦 − 11 = 0. Then we have
−2𝑦2
− 8𝑦 = −𝑥 + 11 −
1
2
𝑦2
+ 4𝑦 =
1
2
𝑥 −
11
2
𝑦2
+ 4𝑦 +
4
2
2
=
1
2
𝑥 −
11
2
+ 4 (by completing the square)
𝑦 + 2 2
=
1
2
𝑥 −
3
2
𝑦 + 2 2
=
1
2
𝑥 − 3
Thus,
(a) Vertex: 𝑉 3, −2 ;
(b) Concavity: 4𝑝 =
1
2
⟹ 𝑝 =
1
8
> 0, parabola concaves to the right;
(c) Focus: 𝐹 3 +
1
8
, −2 = 𝐹
25
8
, −2 ;
(d) Length of latus rectum: 𝐿𝑅 =
1
2
=
1
2
with extreme points at 𝐸′ 25
8
, −
7
4
and 𝐸
25
8
, −
9
4
;
(e) Directix: 𝑥 = ℎ − 𝑝 ⟹ 𝑦 = 3 −
1
8
=
23
8
; and
(f) Axis: 𝑦 = −2.

35. 35
Graph
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
x
y
𝑉(3,−2)
𝐹
25
8
, −2
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥: 𝑥 =
23
8
𝐸′
25
8
, −
9
4
𝐿𝑎𝑡𝑢𝑠 𝑅𝑒𝑐𝑡𝑢𝑚
𝐸
25
8
, −
7
4
Note:
* Not a function
* Domain= 𝑥 𝑥 ≥ 3
= [3, +∞)
* Range= 𝑦 𝑦 ∈ ℝ
= (−∞, +∞)

36. 36
Example 5.75. Suppose we have an equation of a parabola is 𝑦2 + 6𝑥 + 6𝑦 + 39 = 0. Then
we have 𝑦2
+ 6𝑦 = −6𝑥 − 39
𝑦2
+ 6𝑦 +
6
2
2
= −6𝑥 − 39 + 9 (by completing the square)
𝑦 + 3 2
= −6(𝑥 + 5)
Hence,
(a) Vertex: 𝑉 −5, −3
(c) Concavity: 4𝑝 = −6 ⟹ 𝑝 = −
3
2
< 0, parabola concaves to the left
(d) Focus: 𝐹 −5 + −
3
2
, −3 = 𝐹 −
13
2
, −3
(b) Length of latus rectum: 𝐿𝑅 = −6 = 6 with extreme points at 𝐸′ −13
2
, −6 and
𝐸
−13
2
, 0
(e) Directix: 𝑥 = ℎ − 𝑝 ⟹ 𝑦 = −5 − −
3
2
= −
7
2
(f) Axis: 𝑦 = −3

37. 37
Graph
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
x
y
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥: 𝑥 = −
7
2
𝐿𝑎𝑡𝑢𝑠 𝑅𝑒𝑐𝑡𝑢𝑚
𝐴𝑥𝑖𝑠: 𝑦 = −3
𝑉 −5,−3
𝐹 −
13
2
, −3
𝐸
−13
2
,0
𝐸′
−13
2
, −6
Note:
* Not a function
* Domain= 𝑥 𝑥 ≤ −5
= (−∞, −5]
* Range= 𝑦 𝑦 ∈ ℝ
= (−∞, +∞)

38. 38
Example 5.76 (Application). The cable of a suspension suspension bridge hangs in the form
of a parabola when the load is uniformly distributed horizontally. The distance between two
towers are 150 meters, the points of support of the cable on the towers are 22 meters above
the roadway, and the lowest point is 7 meters above the roadway. Find the vertical distance to
the cable from a point in the roadway 15 meters from the foot of a tower.
Solution:
𝑉 0,7
0, 0
150 m
22 m
75 m
15 m
𝑃2 60,𝑦
𝑃1 75,22
Apply: 𝑥 − ℎ 2 = 4𝑝(𝑦 − 𝑘)
*𝑉 0,7 : 𝑥 − 0 2
= 4𝑝(𝑦 − 7)
𝑥2
= 4𝑝(𝑦 − 7)
*𝑃1(75, 22): (75)2
= 4𝑝(22 − 7)
4𝑝 = 375
*𝑃2(60, 𝑦): (60)2= 375(𝑦 − 7)
𝑦 =
83
5
= 16.6
Therefore the vertical distance to the cable
from the point in the roadway 15 meters from
the foot of a tower is 16.6 meters.

39. 39
Example 5.77 (Application). A reflecting telescope has a parabolic mirror for which the
distance from the vertex to the focus is 30 feet. If the distance across the top of the mirror is
64 inches, how deep is the mirror at the center?
Solution:
𝑃(32,𝑦)
𝐹(0, 360)
(0,0)
30 ft
12 𝑖𝑛
1𝑓𝑡
= 360 𝑖𝑛
Apply: 𝑥 − ℎ 2 = 4𝑝(𝑦 − 𝑘)
* 𝑉(0, 0): 𝑥 − 0 2
= 4𝑝(𝑦 − 0)
𝑥2
= 4𝑝𝑦
* 𝐹 0, 360 ⟹ 𝑝 = 360
* 𝑃(32, 𝑦): (32)2= 4(360)𝑦
𝑦 =
32
45
≈ 0.71 in

40. 40
Definition 5.78 (Circle). A circle is the set of all points in a plane equidistant from a fixed
point. The fixed point is called the center of the circle and the constant equal distance is
called the radius of the circle.
The general equation of the circle is
𝑥2
+ 𝑦2
+ 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0
where 𝐷, 𝐸, and 𝐹 are constants which can be transformed in to
(𝑥 − ℎ)2
+(𝑦 − 𝑘)2
= 𝑟2
of which the center is at the point 𝐶(ℎ, 𝑘) and the radius is 𝑟.
Example 5.79. Determine whether the graph is a circle, point, or the empty set.
a. 𝑥2
+ 𝑦2
+ 6𝑥 − 2𝑦 − 15 = 0
b. 𝑥2 + 𝑦2 − 4𝑥 + 10𝑦 + 29 = 0
c. 𝑥2
+ 𝑦2
+ 8𝑥 − 6𝑦 + 30 = 0
b. 4𝑥2
+ 4𝑦2
+ 24𝑥 − 4𝑦 + 1 = 0
We take note that after transforming is 𝑥2 + 𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 in to
(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟2 and if 𝑟 = 0, then the graph is a point; otherwise an empty set if 𝑟 is
negative.

41. 41
(a) 𝑥2
+ 𝑦2
+ 6𝑥 − 2𝑦 − 15 = 0
𝑥2 + 6𝑥 + 𝑦2 − 2𝑦 = 15
𝑥2 + 6𝑥 +
6
2
2
+ 𝑦2 − 2𝑦 +
−2
2
2
= 15 + 9 + 1
𝑥 + 3 2 + 𝑦 − 1 2 = 25
Therefore the center is at 𝐶(−3,1) with 𝑟 = 5.
SOLUTION:

42. 42
Graph
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4,
8
12
x
y
𝐶 −3,1 2,1
−8,1
−3,6
−3,−4
Therefore the center is at 𝐶 −3,
1
2
with radius 𝑟 = 3 units where the graph is shown below.
Note:
* Not a function
* Domain= [−8,2]
* Range= −4,6

43. 43
(b) 𝑥2
+ 𝑦2
− 4𝑥 + 10𝑦 + 29 = 0
𝑥2
− 4𝑥 + 𝑦2
+ 10𝑦 = −29
𝑥2 − 4𝑥 + −
4
2
2
+ 𝑦2 + 10𝑦 +
10
2
2
= −29 + 4 + 25
(by completing the square method)
𝑥 − 2 2 + 𝑦 + 5 2 = 0
Hence 𝐶(2, −5) and observe that 𝑟 = 0 implies that the graph is a point as shown
in the next slide.

44. 44
Graph
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
x
y
𝐶 2, −5

45. 45
(c) 𝑥2
+ 𝑦2
+ 8𝑥 − 6𝑦 + 30 = 0
(𝑥2 + 8𝑥) + 𝑦2 − 6𝑦 = −30
𝑥2 + 8𝑥 +
8
2
2
+ 𝑦2 − 6𝑦 +
−6
2
2
= −30 + 16 + 9
𝑥 + 4 2
+ 𝑥 − 3 2
= −5
Notice that after transforming 𝑥2
+ 𝑦2
+ 8𝑥 − 6𝑦 + 30 = 0 into
𝑥 + 4 2
+ 𝑥 − 3 2
= −5, we see that 𝑟 = −5 and this indicates the we are
dealing with an empty set.

46. 46
(d) Transforming the equation 4𝑥2 + 4𝑦2 + 24𝑥 − 4𝑦 + 1 = 0 into
(𝑥 − ℎ)2+(𝑦 − 𝑘)2= 𝑟2, we have
4𝑥2
+ 4𝑦2
+ 24𝑥 − 4𝑦 + 1 = 0
4𝑥2 + 4𝑦2 + 24𝑥 − 4𝑦 = −1
1
4
𝑥2+ 𝑦2 + 6𝑥 − 𝑦 = −
1
4
𝑥2+6𝑥 +
6
2
2
+ 𝑦2−𝑦 + −
1
2
2
= −
1
4
+ 9 +
1
4
(by
completing the square method)
𝑥 + 3 2 + 𝑥 −
1
2
2
= 9
𝑥 + 3 2 + 𝑥 −
1
2
2
= 32.

47. 47
Graph
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
x
y
𝐶 −3,
1
2 0,
1
2
−6,
1
2
−3,
7
2
−3, −
5
2
Therefore the center is at 𝐶 −3,
1
2
with radius 𝑟 = 3 units where the graph is shown below.
Note:
* Not a function
* Domain= [−6,0]
* Range= −
5
2
,
7
2

48. 48
Definition 5.79 (Ellipse). An ellipse is the set of points in a plane, the sum of
whose distances from the two fixed points is a constant. Each fixed point is called a
focus.
The general equation of an ellipse is
𝑥−ℎ 2
𝑎2 +
𝑦−𝑘 2
𝑏2 = 1.
(a) 𝑎 > 𝑏 ⟹Principal axis (P.A.) is
horizontal and minor axis (M.A.) is
vertical;
𝑏 > 𝑎 ⟹Principal axis (P.A.) is vertical
and minor axis (M.A.) is horizontal;
(b) The center is at 𝐶(ℎ, 𝑘) and the
equation of the principal axis is 𝑦 = 𝑘.
Center: 𝐶(ℎ, 𝑘) and the equation of the
principal axis is 𝑥 = ℎ.
Let us tabulate some informations on ellipse or to compare where the
principal axis is either horizontal or vertical.

49. 49
(c) Let us label the vertices of P.A. and
extremities of M.A. by the following:
Vertices of P.A.:
𝑉1 ℎ − 𝑎, 𝑘 and 𝑉2 ℎ + 𝑎, 𝑘 ;
Extremities of M.A.:
𝐸1 ℎ, 𝑘 + 𝑏 and 𝐸2(ℎ, 𝑘 − 𝑏).
Vertices of P.A.:
𝑉1 ℎ, 𝑘 + 𝑏 and 𝑉2(ℎ, 𝑘 − 𝑏);
Extremities of M.A.:
𝐸1 ℎ − 𝑎, 𝑘 and 𝐸2 ℎ + 𝑎, 𝑘 .
(d) Location of the Foci:
𝐹1(ℎ − 𝑐, 𝑘), 𝐹2(ℎ + 𝑐, 𝑘)
where solving for the value of 𝑐, we use
𝑐2 = 𝑎2 − 𝑏2.
Location of the Foci:
𝐹1(ℎ, 𝑘 + 𝑐), 𝐹2(ℎ, 𝑘 − 𝑐)
where solving for the value of 𝑐 we use
the formula
𝑐2
= 𝑏2
− 𝑎2
.
(e) Distance between the foci:
𝐹1𝐹2 = 2𝑐
𝐹1𝐹2 = 2𝑐

50. 50
(f) Distance between the vertices in the
P.A.:
𝑉1𝑉2 = 2𝑎
𝑉1𝑉2 = 2𝑏
(g) From the definition of an ellipse we
see that it is the set of points in a plane,
the sum of whose distances from the
two fixed points is a constant, so we
must have
𝐹1𝑃 + 𝐹2𝑃 = 2𝑎
where point 𝑃(𝑥, 𝑦) is any point on the
graph of an ellipse.
𝐹1𝑃 + 𝐹2𝑃 = 2𝑏

51. 51
Example 5.80. Let us label all the parts of the graph of an ellipse (by using the
information tabulated above) having and equation
6𝑥2
+ 9𝑦2
− 24𝑥 − 54𝑦 + 51 = 0.
Converting 6𝑥2
+ 9𝑦2
− 24𝑥 − 54𝑦 + 51 = 0 into
𝑥−ℎ 2
𝑎2 +
𝑦−𝑘 2
𝑏2 = 1, we have
6𝑥2 + 9𝑦2 − 24𝑥 − 54𝑦 + 51 = 0
6𝑥2 − 24𝑥 + 9𝑦2 − 54𝑦 = −51
6 𝑥2 − 4𝑥 + 9 𝑦2 − 6𝑦 = −51
6 𝑥2 − 4𝑥 + −
4
2
+ 9 𝑦2 − 6𝑦 + −
6
2
= −51
6 𝑥 − 2 2
+ 9 𝑦 − 3 2
= −51 + 24 + 81

52. 52
6 𝑥 − 2 2 + 9 𝑦 − 3 2 = 54
1
54
𝑥 − 2 2
9
+
𝑦 − 3 2
6
= 1
𝑥 − 2 2
32 +
𝑦 − 3 2
6
2 = 1
Thus, 𝑎 = 3 and 𝑏 = 6 and so to verify the facts about an ellipse which are
listed in the table above, we have
(a) 𝑎 = 3 > 𝑏 = 6 ⟹Principal axis (P.A.) is horizontal and minor axis (M.A.)
is vertical;

53. 53
(b) Center: 𝐶 ℎ, 𝑘 = 𝐶(2, 3) and the equation of the principal axis is 𝑦 = 3;
(c) Vertices of P.A.: 𝑉1 2 − 3, 3 = 𝑉1 −1, 3 and 𝑉2 2 + 3, 3 = 𝑉2 5, 3 ;
Extremities of M.A.: 𝐸1 2, 3 + 6 and 𝐸2 2, 3 − 6 ;
(d) The foci are located at 𝐹1 2 − 3, 3 and 𝐹2(2 + 3, 3) because
𝑐2
= 9 − 6 ⟹ 𝑐 = 3;
(e) Distance between the foci: 𝐹1𝐹2 = 2 − 3 − 2 + 3
2
+ 3 − 3 2
= −2 3
2
+ 0 2
= −2 3
= 2 3
= 2𝑐;

54. 54
(f) Distance between the vertices in the P.A.:
𝑉1𝑉2 = −1 − 5 2 + 3 − 3 2
= −6 2 + 0 2
= −6
= 6
= 2(3)
= 2𝑎;

55. 55
(g) From the definition of an ellipse we see that it is the set of points in a plane, the
sum of whose distances from the two fixed points is a constant, so we must
have 𝐹1𝑃 + 𝐹2𝑃 = 2𝑎. Now, take the point 𝑃 𝑥, 𝑦 = 𝐸1 2, 3 + 6 and so
𝐹1𝑃 + 𝐹2𝑃
= 2 + 3 − 2
2
+ 3 − 3 + 6
2
+ 2 − 2 − 3
2
+ 3 + 6 − 3
2
= 3
2
+ 6
2
+ 3
2
+ 6
2
= 9 + 9
= 3 + 3
= 6
= 2(3)
= 2𝑎.

56. 56
Graph
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-6
-4
-2
2
4
6
x
y
𝐶 2,3 𝑉2 5,3
𝐸1 2,3 + 6
𝑉1 −1,3
𝐸2 2,3 − 6
𝐹1 2 − 3, 3 𝐹2 2 + 3,3
Note:
* Not a function
* Domain= [−1, 5]
* Range= 3 − 6, 3 + 6

57. 57
Example 5.81. Similarly let us label all the parts of the graph of an ellipse (by
using the information tabulated above) having and equation
4𝑥2
+ 𝑦2
+ 8𝑥 − 4𝑦 − 92 = 0.
Determining the important parts and verify the formulas in the table above, we
have
4𝑥2
+ 𝑦2
+ 8𝑥 − 4𝑦 − 92 = 0
4𝑥2 + 8𝑥 + 𝑦2 − 4𝑦 = 92
4 𝑥2
+ 2𝑥 +
2
2
2
+ 𝑦2
− 4𝑦 + −
4
2
2
= 92 + 4 + 4
4 𝑥 + 1 2 + 𝑥 − 2 2 = 100
1
100

58. 58
𝑥 + 1 2
25
+
𝑥 − 2 2
100
= 1
𝑥 + 1 2
52
+
𝑥 − 2 2
102
= 1
Therefore,
(a) 𝑎 = 5 < 𝑏 = 10 ⟹Principal axis (P.A.) is vertical and minor axis (M.A.) is
horizontal;
(b) Center: 𝐶(−1, 2) and the equation of the principal axis is 𝑥 = −1;
(c) Vertices of P.A.:
𝑉1 −1, 2 + 10 = 𝑉1 −1, 12 and 𝑉2 −1, 2 − 10 = 𝑉2 −1, −8 ;
Extremities of M.A.: 𝐸1 −1 − 5, 2 = 𝐸1 −6, 2 and 𝐸2 4, 2 ;

59. 59
(d) The Foci are located at 𝐹1(−1, 2 + 5 3), 𝐹2(−1, 2 − 5 3) because
𝑐2
= 100 − 25 = 75 ⟹ 𝑐 = 5 3;
(e) Distance between the foci:
𝐹1𝐹2 = −1 − (−1) 2 + 2 − 5 3 − 2 + 5 3
2
= 0 2 + −10 3
2
= −10 3
2
= −10 3
= 10 3
= 2 5 3
= 2𝑐;

60. 60
(f) Distance between the vertices in the P.A.:
𝑉1𝑉2 = −1 − (−1) 2 + −8 − 12 2
= 0 2 + −20 2
= −20 2
= 20
= 2(10)
= 2𝑏;

61. 61
(g) Let 𝑃 𝑥, 𝑦 = 𝐸1 −6, 2 . Then
𝐹1𝑃 + 𝐹2𝑃 = −6 − (−1) 2 + 2 − 2 + 5 3
2
+
−6 − (−1) 2 + 2 − 2 + 5 3
2
= −5 2 + −5 3
2
+ −5 2 + 5 3
2
= 25 + 75 + 25 + 75
= 100 + 100
= 10 + 10
= 20
= 2(10)
= 2𝑏;

62. 62
Graph
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
x
y
𝐶 −1,2
𝑉2 −1,−8
𝐸1 −6,2
𝑉1 −1,12
𝐸2 4,2
𝐹1 −1,2 + 5 3
𝐹2 −1,2 − 5 3
Note:
* Not a function
* Domain= [−6, 4]
* Range= −8, 12

63. 63
Definition 5.80 (Hyperbola). A hyperbola is the set of points in a plane, the absolute value
of the difference of whose distances from the two fixed points is a constant. The two fixed
points are called the foci.
The general equation of the hyperbola is
𝑥−ℎ 2
𝑎2 −
𝑦−𝑘 2
𝑏2 = 1.
Note:
1. The line through the foci is called the principal axis of the hyperbola.
2. The points where the graphs intersects the principal axis are called the vertices.
3. If the principal axis is horizontal, then the equation of the hyperbola is
𝑥 − ℎ 2
𝑎2 −
𝑦 − 𝑘 2
𝑏2 = 1
where the:
(a) center 𝐶(ℎ, 𝑘) is the midpoint between the vertices;

64. 64
(b) vertices are located at 𝑉1(ℎ − 𝑎, 𝑘) and 𝑉2(ℎ + 𝑎, 𝑘) an;
(c) line segment joining the vertices is called the transverse axis and its
length is 𝑻𝑨 = 𝑉1𝑉2 = 𝟐𝒂 units.
(d) line segment having extremities at 𝐸1(ℎ, 𝑘 + 𝑏) and 𝐸2(ℎ, 𝑘 − 𝑏) is called
the conjugate axis;
(e) length of conjugate axis is 𝑪𝑨 = 𝐸1𝐸2 = 𝟐𝒃 units;
(f) foci are located at 𝐹1(ℎ − 𝑐, 𝑘) and 𝐹2(ℎ + 𝑐, 𝑘) where 𝑐2
= 𝑎2
+ 𝑏2
;
(g) 𝐹1𝑃 − 𝐹2𝑃 = 2𝑎 where point 𝑃(𝑥, 𝑦) is any point on the hyperbola;
(h) equation of the asymptotes
𝑥−ℎ 2
𝑎2 −
𝑦−𝑘 2
𝑏2 0.

65. 65
4. If the principal axis is vertical, then the equation of the hyperbola is
−
𝑥 − ℎ 2
𝑎2
+
𝑦 − 𝑘 2
𝑏2
= 1
where the:
(a) center 𝐶(ℎ, 𝑘) is the midpoint between the vertices;
(b) vertices are located at 𝑉1(ℎ, 𝑘 + 𝑏) and 𝑉2(ℎ, 𝑘 − 𝑏);
(c) line segment joining the vertices is called the transverse axis and its
length is 𝑻𝑨 = 𝑉1𝑉2 = 𝟐𝒃 units.
(d) line segment having extremities at 𝐸1(ℎ − 𝑎, 𝑘) and 𝐸2(ℎ + 𝑎, 𝑘) is called
the conjugate axis;
(e) and its length is 𝑪𝑨 = 𝐸1𝐸2 = 𝟐𝒂 units;
(f) foci are located at 𝐹1(ℎ, 𝑘 + 𝑐) and 𝐹2(ℎ, 𝑘 − 𝑐) where 𝑐2 = 𝑎2 + 𝑏2;
(g) 𝐹1𝑃 − 𝐹2𝑃 = 2𝑏 where point 𝑃(𝑥, 𝑦) is any point on the hyperbola; and
(h) equation of the asymptotes
−
𝑥−ℎ 2
𝑎2 +
𝑦−𝑘 2
𝑏2 0.

66. 66
Example 5.81. Sketch the graph of 4𝑥2 − 𝑦2 − 8𝑥 − 12 = 0 and label its parts as
listed in the previous slide.
Solution:
(4𝑥2
− 8𝑥) − 𝑦2
= 12
4 𝑥2 − 2𝑥 +
−2
2
2
− 𝑦2 = 12 + 4
4 𝑥 − 1 2
− 𝑦 − 0 2
= 16
1
16
𝑥 − 1 2
4
−
𝑦 − 0 2
16
= 1 ⟺
𝑥 − 1 2
22
−
𝑦 − 0 2
42
= 1
(a) Center: 𝐶(1,0) with 𝑎 = 2 and 𝑏 = 4
(b) Vertices: 𝑉1 ℎ − 𝑎, 𝑘 = 𝑉1 1 − 2, 0 = 𝑉1 −1, 0 ;

67. 67
𝑉2 ℎ + 𝑎, 𝑘 = 𝑉2 1 + 2, 0 = 𝑉2 3, 0 ;
(c) 𝑻𝑨 = 𝑉1𝑉2 = 3 − (−1) 2 + 0 − 0 2 = 4 = 2 2 = 2𝑎;
(e) 𝑪𝑨 = 𝐸1𝐸2 = 1 − 1 2 + −4 − 4 2 = 8 = 2 4 = 2𝑏;
(d) 𝐸1(ℎ, 𝑘 + 𝑏) = 𝐸1 (1,4)and 𝐸2 ℎ, 𝑘 − 𝑏 = 𝐸2 1, −4
(f) 𝐹1 ℎ − 𝑐, 𝑘 = 𝐹1 1 − 2 5, 0 and 𝐹2 ℎ + 𝑐, 𝑘 = 𝐹2(1 + 2 5, 0)
because 𝑐2 = 𝑎2 + 𝑏2 = 4 + 16 = 20 ⟹ 𝑐 = 2 5.
(g) Take 𝑃 𝑥, 𝑦 = 𝑉2 3, 0 . Then
𝐹1𝑃 − 𝐹2𝑃 = 3 − 1 − 2 5
2
+ 0 − 0 2 − 3 − 1 − 2 + 5
2
+ 0 − 0 2
= 2 − 2 5 − 2 + 2 5 = 2 5 − 2 − (2 + 2 5) = 4 = 2 2 = 2a.

68. 68
(h) Equations of the asymptotes:
𝑥 − 1 2
4
−
𝑦 − 0 2
16
= 0 ⟺
𝑥 − 1
2
−
𝑦 − 0
4
𝑥 − 1
2
+
𝑦 − 0
4
= 0
𝑥 − 1
2
−
𝑦 − 0
4
= 0 ⟺ 2𝑥 − 𝑦 − 2 = 0
𝑥 − 1
2
+
𝑦 − 0
4
= 0 ⟺ 2𝑥 + 𝑦 − 2 = 0

69. 69
-12 -10 -8 -6 -4 -2 2 4 6 8 10 12
-12
-8
-4
4
8
12
x
y
𝐶 1,0
Note:
* Not a function
* Domain= −∞, − ∪ [3, +∞)
* Range= (−∞, +∞)
𝑉1 −1,0 𝑉2 3,0
𝐸1 1,4
𝐸2 1,−4
𝐹2 1 + 2 5, 0
𝐹1 1 − 2 5, 0

70. 70
Example 5.82. Sketch the graph of 9𝑥2
− 18𝑦2
+ 54𝑥 − 36𝑦 + 79 = 0 and label
its parts as listed in the previous slide.
The solution will be discuss during google class meeting!

71. 71

72. 72

73. 73

74. 74

75. 75
REFERENCES
[1] Barnett, Raymond A., Ziegler, Michael R., Byleen, Karl E., Sobecki, D.
Precalculus 7th Edition. McGraw-Hill, c 2011
[2] Hart, William L. Plane and Spherical Trigonometry. Boston: D.C. Heath and
Company, c1964
[3] Johnson, Richard E., et. al. Algebra and Trigonometry 2nd edition. California:
Addison – Wesley Publishing Company, c1971
[4] Leithold, Louis College Algebra and Trigonometry. Massachusetts: Addison –
Wesley Publishing Company, c1989
[5] Miller, Charles D. Fundamentals of College Algebra. New York: Harper
Collins College Publishers, c1994
[6] Robinson N. Elements of Plane and Spherical Trigonometry. American Book
Company, c1970
[7]Spiegel, Murray, Moyer Robert E. College Algebra. New York. McGraw –
Hill, c1998

76. 76
[8] Sullivan, Michael. Trigonometry: A Unit Circle Approach. Prentice Hall, c
2012
[9] Vance, Elbridge P. Modern Algebra and Trigonometry. Massachusetts:
Addison – Wesley Publishing Company, c1975