MARITIME ACADEMY OF ASIA AND THE PACIFIC-KAMAYA POINT
DGE303/EGE303 DEPARTMENT OF ACADEMICS
DIFFERENTIAL CALCULUS WITH
Associated Marine Officers and Seamen’s Union of the Philippines – PTGWO-ITF
ANALYTIC GEOMETRY
Kamaya Point, Alas-asin, Mariveles, Bataan
TITLE CALCULUS1 MODULE
ISSUE NO. 0 REV. 0 DATE EFFECTIVE: PAGE 1 OF 5
Course: Differential Calculus with Analytic Geometry
Module Number: 3
Topic: Parabola
Learning Outcomes: The students should be able to:
LO1.1. identify an equation of a parabola.
2. write the equation of a parabola in standard and in general form.
3. draw the conic section by using the special features like vertex, focus, ends of the latus rectum, directrix.
4. solve real world problems involving properties of a parabola.
Keywords and Concepts:
A. Conic sections B. Parabola
Cutting plane Vertex
Standard equation Axis of symmetry
General equation Focus
Ends of the latus rectum
Directrix
Eccentricity
Lecture
A: Conic Sections
Conics or conic sections are made by cutting a two-napped right circular cone by planes.
If the cutting plane is perpendicular to the axis of the cone, the conic section formed is a circle.
If the cutting plane is inclined with the axis but does not intersect with the base, the conic section formed
is an ellipse.
If the cutting plane is inclined with the axis and parallel to the side of the cone, the conic section formed is
a parabola.
If the cutting plane is inclined with the axis and intersects the two bases, or perpendicular to the base but does
not intersect with the axis, the conic section formed is a hyperbola.
Parabola
Your turn: How are the following conic sections formed?
a. Point
b. Line/ Lines
MARITIME ACADEMY OF ASIA AND THE PACIFIC-KAMAYA POINT
DGE303/EGE303 DEPARTMENT OF ACADEMICS
DIFFERENTIAL CALCULUS WITH
Associated Marine Officers and Seamen’s Union of the Philippines – PTGWO-ITF
ANALYTIC GEOMETRY
Kamaya Point, Alas-asin, Mariveles, Bataan
TITLE CALCULUS1 MODULE
ISSUE NO. 0 REV. 0 DATE EFFECTIVE: PAGE 2 OF 5
Conic sections are represented algebraically by the second degree equation:
AX2 + BY2 + CXY + DX + EY + F=0. For our topic coverage, we let C=0
B. Parabola
Definition: A parabola is a set of points equidistant from a fixed point and
a fixed line in the plane.
The fixed point is called focus and the fixed line is called directrix.
The eccentricity is the ratio between the distances from a point P to the focus and from the
same P to the directrix : e = PF/PD
The eccentricity of a parabola equals 1
The general equation of the parabola has only one second-degree term:
a) Ax2 + Dx + Ey + F=0 such that the Parabola has a vertical axis of symmetry.
b) By2 + Dx + Ey + F =0 such that the Parabola has a horizontal axis of symmetry.
The corresponding standard equations are:
a) (x- h)2 = 4p (y- k) Parabola opens upward when p>0 and downward when p<0
b) (y- k)2 = 4p (x- h) Parabola opens to the right when p>0 and to the left when p<0
Vertical Parabola:
(x – h)2 = 4p (y – k) p=+ ; parabola opens upward
p= – ; parabola opens downward
Vertex: V(h,k)
Focus: F(h, k + p)
Directrix : y = k – p
●P │p│ = │VF │= │VD │
Latus Rectum: EE’= │4p│
Horizontal Parabola:
(y – k)2 = 4p (x – h) p=+ ; parabola opens to the right
p= – ; parabola opens to the left
Vertex: V(h,k)
Focus: F(h + p, k)
Directrix : x = h – p
p = │VF│ = │ VD│
Latus Rectum: EE’= │4p│
MARITIME ACADEMY OF ASIA AND THE PACIFIC-KAMAYA POINT
DGE303/EGE303 DEPARTMENT OF ACADEMICS
DIFFERENTIAL CALCULUS WITH
Associated Marine Officers and Seamen’s Union of the Philippines – PTGWO-ITF
ANALYTIC GEOMETRY
Kamaya Point, Alas-asin, Mariveles, Bataan
TITLE CALCULUS1 MODULE
ISSUE NO. 0 REV. 0 DATE EFFECTIVE: PAGE 3 OF 5
The table below shows the summary of the features of a vertical and horizontal parabola
Vertical Parabola Horizontal Parabola
General form Ax2+ Dx +Ey +F =0 By2+ Dx +Ey + F=0
Standard form (x – h)2 = 4p (y – k) (y – k)2 = 4p (x – h)
Axis x=h y=k
Vertex (h,k) (h,k)
Focus (h, k+p) (h+p, k)
Directrix y= k-p x=h-p
Ends of Latus Rectum (h ±2p, k+p) (h+p, k ±2p)
Applications:
Parabola has numerous applications in radio communications, astronomy, submarine tracking, photography,
wind tunnel, and bridge construction
a. The chief applications of parabola involve their use as reflectors of light and radio waves. Rays originating
at the focus are reflected out of the parabola parallel to the axis, and rays coming into the parabola are
reflected to the focus.
b. Parabolas are often found in architecture, especially in the cables of suspension bridges. This is because
the stresses on the cables as the bridge is suspended from the top of the towers are most efficiently
distributed along a parabola. The bridge can remain stable against the forces that act against it.
Examples
1.Given: y2 + 8x – 6y + 25 =0
Required :a. Express the equation in standard form
b. Graph. Label the following: V ; F ; Ends of LR ; Directrix
Solution:
1. a. express the equation in standard form: y2 + 8x – 6y + 17 =0
y2-6y+9 = -8x - 17+9
Equation in standard form: (y-3)2 = - 8(x+1)
b. Features of the parabola
General form y2+ 8x - 6y + 17=0 Horizontal Parabola
Standard form (y – k)2 = 4p (x – h)
(y – 3)2 = - 8 (x + 1)
4p=- 8 p= - 2 parabola opens to the left
Axis of symmetry: y=k Horizontal : y=3
Vertex: V(h,k) (-1, 3)
Focus: F(h+p, k) (- 3, 3)
Directrix: D: x=h-p x=1
Ends of Latus Rectum:E E1( -3, 7)
(h+p, k ±2p) E2 (-3, -1)
c. Graph
Y
● E1(-3,7)
)
directrix
X=1
axis of symmetry
Y=3 ● ●
F(-3,3) V(-1,3)
) )
X
(0,0)
● E (-3,-1)
2
MARITIME ACADEMY OF ASIA AND THE PACIFIC-KAMAYA POINT
DGE303/EGE303 DEPARTMENT OF ACADEMICS
DIFFERENTIAL CALCULUS WITH
Associated Marine Officers and Seamen’s Union of the Philippines – PTGWO-ITF
ANALYTIC GEOMETRY
Kamaya Point, Alas-asin, Mariveles, Bataan
TITLE CALCULUS1 MODULE
ISSUE NO. 0 REV. 0 DATE EFFECTIVE: PAGE 4 OF 5
2. The cross section of a flashlight reflector is parabolic. The bulb is located 1 inch from the vertex.
If the flashlight is 2 inches deep, how wide is it at the brim?
Solution: The cross section of a flashlight reflector is in the shape of a parabola. Let the parabola open
upward and the vertex be at the origin. The position of the bulb is the position of the focus.
Diagram:
Y
width at the brim = 2x
● P(X,2)
Bulb
● F(0,1)
p=1
● X axis
V(0,0)
Standard equation of the parabola: (x-h)2=4p(y-k) where (h,k) ≈ (0,0) and p=1
x2 = 4y
The width of the brim of the flashlight reflector will be 2x inches, so we will solve for x when y=2.
x2 = 4y
x2 = 4(2)
x = 2.83
2x = 5.66 Therefore, the width of the reflector at the brim is 5.66 inches.
Exercises
1.Given : Parabola with F: ( 4, ─3) and Directrix : y = 5
Required: a. Equation in standard form
b. Graph. Label the following: V ; F ; Ends of LR
2. Given: Parabola with focus at (-2, -3) and the length of its latus rectum is 4 units
Required: a. Solve for the equation of the parabola.
b. Graph of the parabola. Label the following: V ; F ; Ends of LR
3. Given: Parabola with vertex at (-1,-2). Axis is vertical. Parabola passes thru (3,6 ).
Required:a. Equation of the curve.
b. Graph. Label the following: V, F, Directrix
4. A parabola has a horizontal axis. It intersects the y-axis at y=-4 and y=6.
It intersects the x-axis at x=-2.
Required: a. Solve for the equation of the parabola.
b. Graph of the parabola. Label the following: V ; F ; Ends of LR
3. Given: Parabola with horizontal axis. Passes through the points:
A ( 1, 1 ) B( 1, - 3 ) C( -2, 0 )
Required: a. Equation of the parabola
b. Graph. label the following: V, F, D
4. Put the equation in standard form and draw the corresponding graph of each Parabola.
Label the features: V, F, Ends of LR
a. x2 - 8y = 16
b. x2 + 4x+ 2y + 6 =0
MARITIME ACADEMY OF ASIA AND THE PACIFIC-KAMAYA POINT
DGE303/EGE303 DEPARTMENT OF ACADEMICS
DIFFERENTIAL CALCULUS WITH
Associated Marine Officers and Seamen’s Union of the Philippines – PTGWO-ITF
ANALYTIC GEOMETRY
Kamaya Point, Alas-asin, Mariveles, Bataan
TITLE CALCULUS1 MODULE
ISSUE NO. 0 REV. 0 DATE EFFECTIVE: PAGE 5 OF 5
c. y + 4x – 8 =0
2
d. y2 - 4y = 8x – 4
e. x2 + 2y = - 16x – 8
5. A cable suspended from two towers (of the same heights) that are 600 ft apart has a sag of 100 ft.
The lowest point is 10 f from the ground. The cable hangs in the form of a parabola. An object on the
cable is 50 ft from the ground. How far is it from the nearest tower?
6. A TV-satellite dish is parabolic and has its receiver 12 inches from its vertex. Solve for the equation of
the parabolic cross section of the dish which goes through the vertex.
7. The entrance to a park is in the form of a parabolic segment 8m on the base and 6m high at the
center. Solve for the equation of the parabolic curve of the entrance. How high is the entrance when it is
3m from the center.
Additional Resources:
a. The Calculus with Analytic Geometry by Louis Leithold
b. Calculus and Analytic Geometry by George B. Thomas Jr and Ross L.Finney
c. Differential Calculus with Analytic Geometry Modules
