1. Lesson 11 – 4 Day 1 The Parabola
Pre-calculus

2. Learning Objective
To write an equation of a parabola given vertex, directrix, and/or foci

3. Conic Section Parabola: Basic look of a parabola
The set of all points P in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line.
Basic look of a parabola
Focus c
parabola c
directrix
*Think: food goes in a bowl on a desk
*order works even if parabola is sideways or upside down

4. Conic Section Parabola centered at (ℎ, 𝑘) 𝑎 (𝑥−ℎ) 2 =𝑦−𝑘
A of S 2c F 2c c c
V(h, k) d
𝑎 (𝑥−ℎ) 2 =𝑦−𝑘
Focus: (ℎ, 𝑘+𝑐)
𝑎= 1 4𝑐
Directrix: 𝑦=𝑘−𝑐
Vertex: (ℎ, 𝑘)
Opens: Up if 𝑎>0
Down if 𝑎<0
Axis of Symmetry: 𝑥=ℎ

5. Conic Section Parabola centered at (ℎ, 𝑘) 𝑎 (𝑦−𝑘) 2 =𝑥−ℎ 𝑎= 1 4𝑐
A of S V
(h, k) F 2c
𝑎 (𝑦−𝑘) 2 =𝑥−ℎ
𝑎= 1 4𝑐
Axis of Symmetry: 𝑦=𝑘
Vertex: (ℎ, 𝑘)
Focus: (ℎ+𝑐, 𝑘)
Opens: Right if 𝑎>0
Left if 𝑎<0
Directrix: 𝑥=ℎ−𝑐

6. 1. Determine the vertex, the axis of symmetry, the focus, & the directrix of (𝑦−3) 2 =8(𝑥−2). Graph it.
Conic Section
1 8 (𝑦−3) 2 =𝑥−2
Vertex: (2, 3)
𝑦 2 & 𝑎 is (+)
 Opens right
1 8 = 1 4𝑐
 𝑐=2
(plot points 2 right & 2 left)
Focus: (4, 3)
Directrix: 𝑥=0
A of S: 𝑦=3

7. 2. Find the vertex, the axis of symmetry, the focus, & the directrix of
2 𝑥 2 −4𝑥+𝑦+4=0
Conic Section
𝑦+4=−2 𝑥 2 +4𝑥
Make a sketch!
𝑦 = −2( 𝑥 2 −2𝑥 ) 4 4 −
1 8
= −2 2 =4
(1, –2)
1 8
𝑦+2=−2 (𝑥−1) 2
Vertex: (1, −2)
Opens down
Directrix: 𝑦=− 
A of S: 𝑥=1
−2= 1 4𝑐
Directrix: 𝑦=− 15 8
−8𝑐=1
Focus: 1,−2− 
Focus: 1, − 17 8
𝑐=− 1 8

8. Conic Section Halfway: 2−4 2 = −2 2 =−1 𝑦+1=− 1 12 (𝑥−3) 2
3. Determine the equation of the parabola with focus (3, −4) and directrix 𝑦=2
Conic Section
*Think about food–bowl–desk
*Make a sketch!
y = 2
Vertex in between!
V(3, –1)
F(3, –4)
Halfway: 2−4 2 = −2 2 =−1
Vertex: (3, −1)
𝑦+1=− (𝑥−3) 2
𝑐=3  𝑎= 1 4(3)
 𝑎= 1 12
Opens down so 𝑎 is (–)

9. Conic Section Opens down so 𝑎 is (–) 𝑐= 1 6  𝑎= 1 4 1 6  𝑎= 3 2
4. Determine the equation of the parabola with Vertex 7, and Focus 7, 2 3
Conic Section
*Make a sketch!
V 7, 5 6
Opens down so 𝑎 is (–)
F 7, 2 3
𝑐= 1 6  𝑎=
 𝑎= 3 2
𝑦− 5 6 =− 3 2 (𝑥−7) 2

10. 5. Determine the equation of the parabola that passes through (0, 9), (1, 1), and (2, 1)
Conic Section
𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐
9=𝑎 (0) 2 +𝑏(0)+𝑐
 9=𝑐
1=𝑎 (1) 2 +𝑏(1)+𝑐
 1=𝑎+𝑏+9
1=𝑎 (2) 2 +𝑏(2)+𝑐
 1=4𝑎+2𝑏+9
𝑎+𝑏=−8 4𝑎+2𝑏=−8
( )(−2)
 −2𝑎−2𝑏=16 4𝑎+2𝑏=−8
2𝑎=8
𝑎=4
𝑦=4 𝑥 2 −12𝑥+9
1=4+𝑏+9
𝑏=−12

11. 1. Find the focus 5 𝑦 2 +10𝑦−7𝑥−2=0
Check – up
Focus: − , −1

12. Assignment
Pg. 561 #1, 5, 9, 13, 17, 22, 31, 35, 39, 41, 43