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Circles Ellipse Parabolas Hyperbolas

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Presentation on theme: "Circles Ellipse Parabolas Hyperbolas"— Presentation transcript:

1 Circles Ellipse Parabolas Hyperbolas
Conic Sections Circles Ellipse Parabolas Hyperbolas

2 Trick to looking at an equation and knowing what the shape will look like:
Ellipse – you have 2 squared terms that are added together Circle – you have 2 squared terms with the same coefficient and are added together Hyperbola – 2 squared terms that are subtracted Parabola – 1 square term

3 parabolas

4 PARABOLA FORMULA Opens Opens Vertex (h , k) Vertex (h , k)
Up if p > 0 Down if p < 0 Vertex (h , k) Directrix: y = k - p Axis of Symmetry (AOS): x = h Focus (h , k + p) Opens Right if p > 0 Left if p < 0 Vertex (h , k) Directrix: x = h - p Axis of Symmetry (AOS): y = k Focus (h + p , k)

5 To find p you take 4p=4 and solve.
Example 1 Find the required parts for the parabola 𝑦 2 =4𝑥 Rewrite it with all parts in order 𝑦−0 2 =4(𝑥−0) To find p you take 4p=4 and solve. p=1 Opens: right Vertex: 0,0 Directrix: x= -1 Axis of Symmetry: y=0 Focus: 1,0

6 To find p you take 4p=8 and solve.
Example 2 Find the required parts for the parabola 8 𝑦−1 = 𝑥+2 2 Rewrite it with all parts in order 𝑥+2 2 =8(𝑦−1) To find p you take 4p=8 and solve. p=2 Opens: up Vertex: −2,1 Directrix: y=-1 Axis of Symmetry: x=-2 Focus: −2,3

7 Example 3 Find the required parts for the parabola 𝑥=−3 𝑦 2 +12𝑦−9 𝑥+9=−3 𝑦 2 +12𝑦 𝑥+9=−3( 𝑦 2 −4𝑦+4) 𝑥+9−12=−3 𝑦−2 2 (𝑥−3)=−3 𝑦−2 2 − 1 3 (𝑥−3)= 𝑦−2 2 Your 4p must be on the side without the squared term. To find p you take 4𝑝=− 1 3 and solve. 𝑝=− 1 12 Opens: Left Vertex: 3,2 Directrix: 𝑥= 37 12 Axis of Symmetry: y=2 Focus: 35 12 ,2

8 Practice Problems x = y² - 6y + 8 y = 4x² y = -x² + 4x – 2 y² + x = 0


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