Lesson 11 – 4 Day 1 The Parabola Pre-calculus

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

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

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

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

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

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+4+ = −2( 𝑥 2 −2𝑥+ ) 4 4 −4 2 2 1 8 = −2 2 =4 (1, –2) 1 8 𝑦+2=−2 (𝑥−1) 2 Vertex: (1, −2) Opens down Directrix: 𝑦=−2+ 1 8  A of S: 𝑥=1 −2= 1 4𝑐 Directrix: 𝑦=− 15 8 −8𝑐=1 Focus: 1,−2− 1 8  Focus: 1, − 17 8 𝑐=− 1 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=− 1 12 (𝑥−3) 2 𝑐=3  𝑎= 1 4(3)  𝑎= 1 12 Opens down so 𝑎 is (–)

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, 5 6 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= 1 4 1 6  𝑎= 3 2 𝑦− 5 6 =− 3 2 (𝑥−7) 2

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

1. Find the focus 5 𝑦 2 +10𝑦−7𝑥−2=0 Check – up Focus: − 13 20 , −1

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