Conics parabola

CONICS:
PARABOLA
SIDI GABERSON D. EMPAS, LPT

OBJECTIVE
• Define parabola
• Graph the parabola given an
equation in vertex form

What difficulties
did you encounter
in doing the
previous activities?

PRE-ASSESSMENT
Direction: Choose the letter of the correct answer.

QUESTION 1
A conic that consists all points in the
plane equidistant from a fixed point called
focus F and a fixed line l called directrix,
not containing F.
A. CIRCLE B. PARABOLA
C. HYPERBOLA D. ELLIPSE

QUESTION 1
QUESTION 2 The opening of the parabola given by
the equation 𝑦 = − 𝑥 − 3 2
− 5.
A. upward B. downward
C. To the right D. To the left

QUESTION 1
QUESTION 2
QUESTION 3
The graph of a parabola that has a line of symmetry
𝑦 = −4.
A. B.
C. D.

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
The equivalent vertex form of the equation 𝑥 = 𝑦2 + 4𝑦 − 10 is
____________.
A. 𝑥 = 𝑦 + 2 2
− 14 B. 𝑥 = 𝑦 + 2 2 + 14
C.𝑥 = 𝑦 − 2 2
− 14 D.𝑥 = 𝑦 − 2 2 + 14

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
The coordinates of the vertex of the
parabola that is represented by the
Equation 𝑥 = −𝑦2
− 10.
A. 0,10 B. 0, −10
C. −10,0 D. 10,0

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
Find the value of k if the point (0, −3) is on the graph
of 𝑥 = 3𝑦2 + 𝑘𝑦.
A. 3 B. −3
C. 9 D. −9

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
The coordinates of the vertex of the
parabola that is represented by the
equation (𝑥 + 1)2
= 𝑦 + 4.
A. (1, −4) B. (−1, −4)
C. (−1,4) D. (−1,4)

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
The length of the latus rectum of the parabola
represented by the equation 𝑦 = 4𝑥2
.
A. 2 𝑢𝑛𝑖𝑡𝑠 B. 4 𝑢𝑛𝑖𝑡𝑠
C.
1
2
𝑢𝑛𝑖𝑡𝑠 D.1 𝑢𝑛𝑖𝑡

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
The graph of the parabola represented by the
equation 𝑥 = − 𝑦 − 5 2
+ 3.
A. B.
C. D.

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
QUESTION 10
The equation of the directrix of the parabola
given by the equation 𝑥 = −16𝑦2
.
A. 𝑦 = 4 B. 𝑦 = −4
C. 𝑥 = 4 D.𝑥 = −4

QUADRATIC
FUNCTION
A function of the form𝑦 =
𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a quadratic
function where 𝑎, 𝑏, 𝑐 are
real numbers and 𝑎 ≠ 0.

VERTEX FORM
OF A
QUADRATIC
FUNCTION
A quadratic function of the form 𝑦 =
𝑎𝑥2 + 𝑏𝑥 + 𝑐 can be transformed in the
vertex form using completing the
square.

VERTEX
FORM OF A
QUADRATIC
FUNCTION
𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐
𝑦 = 𝑎 𝑥2 +
𝑏
𝑎
𝑥 +
𝑏
2𝑎
2
+ 𝑐 − 𝑎
𝑏
2𝑎
2
𝑦 = 𝑎 𝑥 +
𝑏
2𝑎
2
+ 𝑐 −
𝑏2
4𝑎
𝑦 = 𝑎 𝑥 +
𝑏
2𝑎
2
+
4𝑎𝑐 − 𝑏2
4𝑎
𝑦 = 𝑎 𝑥 − −
𝑏
2𝑎
2
+
4𝑎𝑐 − 𝑏2
4𝑎
So, we have
ℎ = −
𝑏
2𝑎
; 𝑘 =
4𝑎𝑐−𝑏2
4𝑎

VERTEX OF A
QUADRATIC FUNCTION
The point (ℎ ,𝑘) the parabola
which is the graph of 𝑦 =
− 𝑎 𝑥 − ℎ 2
+ 𝑘 is the vertex,
where ℎ = −
𝑏
2𝑎
, and 𝑘 =
4𝑎𝑐−𝑏2
4𝑎

ACTIVITY 1:
This activity will enable you to review the vertex of a parabola.
Determine the highest and lowest value of each function by
identifying the y-coordinate of the vertex. Match Column A with
Column B
1. 𝑦 = 𝑥2 + 3𝑥 − 6
2. 𝑦 = −𝑥2
− 5𝑥 + 7
3. 𝑦 = −3(𝑥 − 5)2 + 2
4. 𝑦 = (𝑥 + 1)2−3
5. 𝑦 = −(𝑥 − 7)2 − 11
𝑦 = −8.25, 𝑙𝑜𝑤𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡
𝑦 = 13.25, ℎ𝑖𝑔ℎ𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡
𝑦 = 2, ℎ𝑖𝑔ℎ𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡
𝑦 = −3, 𝑙𝑜𝑤𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡
𝑦 = −11, 𝑙𝑜𝑤𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡

DEFINITION OF A
PARABOLA (Desmos)
A parabola is the set
of all points in the
plane equidistant
from a fixed point, F,
and a fixed line l not
containing F.

Parts of the Parabola (Desmos)

REMEMBER:
Standard Form of the Equation of a Parabola
with Vertex at the Origin
FOCUS EQUATION
PARABOLAS
OPENS
DIRECTRIX
AXIS OF
SYMMETRY
(𝑎, 0) 𝑦 = 4𝑎𝑥2 Upward 𝑦 = 𝑎 𝑥 = 0
(−𝑎, 0) 𝑦 = −4𝑎𝑥2 Downward 𝑦 = −𝑎 𝑥 = 0
(0, 𝑎) 𝑥 = 4𝑎𝑦2 To the right 𝑥 = 𝑎 𝑦 = 0
(0, −𝑎) 𝑥 = −4𝑎𝑦2 To the left 𝑥 = −𝑎 𝑦 = 0

Example 1
Determine the coordinates of
the vertex, axis of symmetry,
focal distance (𝑎), length of
latus rectum and endpoints
of the latus rectum of the
parabola 𝑥2 = − 4𝑦 . Sketch
the graph.

Solution
Vertex 𝑉(0,0)
Focus 𝐹(0, −2)
Focal distance (a) 2
Latus rectum (𝑙𝑙) 4
endpoints of the latus
rectum
−2, −1 ,
(2, −1)
Axis of symmetry 𝑦 = 0
𝐹
𝑙𝑙
𝑉
𝑦 = 0
𝐸1 𝐸2

Example 2
Determine the vertex, axis of
symmetry, focus, focal
distance (𝑎), length and
rectum of the parabola 𝑦2 =
12𝑥. Graph the parabola.

Solution
Vertex 𝑉(0,0)
Focus 𝐹(3,0)
rectum
−2, −1 ,
(2, −1)
Axis of symmetry 𝑦 = 0
𝐹
𝑉
𝐸1
𝐸2
𝑦 = 0
𝑙𝑙

ACTIVITY 2:
Give My Parts
Direction: Choose from inside
the box the corresponding
parts of the given graph of a
parabola. The equation of the
parabola is 3𝑦2 + 4𝑥 − 24𝑦 +
44 = 0.

Remember:
Parabolas whose vertex is at (𝒉,𝒌)
PARABOLAS
OPENS
EQUATION FOCUS DIRECTRIX
To the right (𝑦 − 𝑘)2= 4𝑎(𝑥 − ℎ) (ℎ + 𝑎, 𝑘) 𝑥 = ℎ − 𝑎
To the left (𝑦 − 𝑘)2
= −4𝑎(𝑥 − ℎ) (ℎ − 𝑎, 𝑘) 𝑥 = ℎ + 𝑎
Upward (𝑥 − ℎ)2
= 4𝑎(𝑦 − 𝑘) (ℎ, 𝑘 + 𝑎) 𝑦 = 𝑘 − 𝑎
Downward (𝑦 − ℎ)2= 4𝑎(𝑦 − 𝑘) (ℎ, 𝑘 − 𝑎) 𝑦 = 𝑘 + 𝑎

ACTIVITY 3: Find my Pair
Match the equation of the parabola to the correct graph.
Equations
Equations
1. 𝑦2
= 𝑥 − 4
2. 𝑦 = −𝑥2
+ 6𝑥
3. 𝑦 = 𝑥2
− 10𝑥 + 29
4. 𝑥 = −𝑦2
+ 2𝑦 + 1
5. 𝑦 = 𝑥2
A
B
C
D
E

EXAMPLE 4 (Desmos)
Determine the vertex, focus,
focal distance (𝑎), length of
latus rectum, endpoints of the
latus rectum and axis of
symmetry of the parabola
(𝑦 − 1)2 = 8(𝑥 − 4). Graph
the parabola.
Vertex 𝑉(4,1)
Focus 𝐹(6,1)
endpoints of the latus rectum 6,5 , (6, −3)
Axis of symmetry 𝑥 = 2

QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
The equivalent vertex form of the equation 𝑥 = 𝑦2 + 4𝑦 − 10 is
____________.
A. 𝑥 = 𝑦 + 2 2
− 14 B. 𝑥 = 𝑦 + 2 2 + 14
C.𝑥 = 𝑦 − 2 2 − 14 D.𝑥 = 𝑦 − 2 2 + 14

Problem:
Write an equation of the
parabola with vertical axis of
symmetry, vertex at the
point (5,1), and passing
through the point (1,3).

DIRECTION
IN ANSWERING THE PROBLEM, USE A SEPARATE
CLEAN SHEET OF PAPER AND SHOW YOUR
SOLUTION. ORGANIZED, NEAT AND CLEAN
PRESENTATION OF THE SOLUTION IS A PLUS IN
YOUR POINTS.

Conics parabola

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