SAT Multiple Choice Question(s)

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Chapter 7 Analyzing Conic Sections

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Chapter 7 Analyzing Conic Sections

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SAT Multiple Choice Question(s) For the two intersecting lines above, which of the following must be true? I. II. III. I only II only I and II only II and III only I, II, and III

SAT Multiple Choice Question(s) For the two intersecting lines above, which of the following must be true? I. II. III. I only II only I and II only II and III only I, II, and III

1. Center: (1,2) vertices: (4,2),(-2,2),(1,4),(1,0) Answers to HW: 1. Center: (1,2) vertices: (4,2),(-2,2),(1,4),(1,0) foci: (3.2, 2),(-1.2, 2) 2. Center: (-3,2) vertices: (-7,2),(1,2),(-3,7),(-3,-3) foci: (-3,-1),(-3,5) 3. Standard Form: (x-5)2/48 + (y-5)2/64 = 1 Center: (5,5) foci: (5,9),(5,1) vertices: (5,13),(5,-3),(-1.9, 5),(11.9, 5) 4. Standard Form: (x+2)2/64 + (y-1)2/48 = 1 Center: (-2,1) foci: (-6,1),(2,1) vertices: (-10,1),(6,1),(-2, 7.9),(-2, -5.9)

Center: (5, -2) foci: (3.6, -2),(6.4, -2) Answers to HW (Cont): 5.Standard Form: Center: (5, -2) foci: (3.6, -2),(6.4, -2) vertices: (3, -2),(7, -2),(5, -.6),(5, -3.4) 4. Standard Form: Center: (2, -1) foci: (-.2, -1),(4.2, -1) vertices: (5,-1),(-1,-1),(2, 1),(2, -3)

Unit Question: How do I recognize, write and graph equations of conic sections? How do I write equations of parabolas and graph all of its pieces and parts?

PARABOLAS Standard form (the distance from the sides of the parabola through the focus)

Latus focus vertex directrix

Parabolas 2 h=0 k=0 p = 1) x2 – 8y = 0 x is squared x2 = 8y 4p = 8 Vertex: (h,k) (0, 0) opens up Parabolas Axis of symmetry: Focus: (h, k+p) (0, 2) Directrix: y = k – p y = -2 Latus:

GRAPH x2 = 8y focus: (0, 2 di vertex: (0, 0) rectrix: y = - axis up ) 8 units across

Graph. k = 2 (x – 1)2 = -4(y – 2) h = 1 p =? -4 = 4p (1, 2) focus: vertical, down k = 2 (x – 1)2 = -4(y – 2) h = 1 p =? vertex: -4 = 4p (1, 2) focus: = (1, 1) p = -1 directrix: y = 3 latus:

Graph. (x – 1)2 = -4(y – 2)

Graph. horizontal, right (y + 2)2 = 8(x – 3) h = 3 k = -2 p =? 8 = 4p vertex: (3, -2) 2 = p focus: = (5, -2) directrix: x = 1 latus:

Graph. (y + 2)2 = 8(x – 3) vertex: (3, -2) directrix: x = 1 latus: 8 focus: (5, -2) semi latus: 4

Graph. (x + 1)2 = 12(y + 1) h = -1 k = -1 p =? 12 = 4p (-1, -1) 3 = p vertical, up h = -1 k = -1 p =? 12 = 4p vertex: (-1, -1) 3 = p focus: = (-1, 2) directrix: y = -4 axis: x = -1 latus: semi latus: 6

Graph. (x + 1)2 = 12(y + 1) vertex: (-1, -1) directrix: y = -4 latus: 12 focus: (-1, 2) axis: x = -1 semi latus: 6

Put in standard form 36 36

Find the equation of the parabola with focus (2,3) and directrix x=7. (y – 3)2 = -2(x – 4.5)