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INTRO TO CONIC SECTIONS

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IT ALL DEPENDS ON HOW YOU SLICE IT! Start with a cone:

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IF WE SLICE THE CONE, PARALLEL TO THE BASE, WHAT DO WE GET? A Circle!

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IF WE SLICE THE CONE AT AN ANGLE, WHAT DO WE GET NOW? An Ellipse!

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IF WE JUST TAKE A SLICE FROM THE LATERAL FACE OF THE CONE, WHAT DO WE GET? A Parabola!

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FINALLY, LET’S TAKE A SLICE FROM THE LATERAL FACE, PERPENDICULAR TO THE BASE A Hyperbola!

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THESE SHAPES ARE CALLED CONIC SECTIONS. WE CAN USE ALGEBRA TO DESCRIBE THE EQUATIONS AND GRAPHS OF THESE SHAPES.

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10.2 - PARABOLAS

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REVIEW: NO CALCS! Find the distance between the given points. 1.(2, 3) and (4, 1)2. (4, 6) and (3, –2)3. (–1, 5) and (2, –3) HMMM... Remember something called the Distance Formula?

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SOLUTIONS

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LET’S REVIEW PARTS OF A PARABOLA This is the y-intercept, It is where the parabola crosses the y-axis This is the vertex, V (h, k) This is the called the axis of symmetry, a.o.s. Here a.o.s. is the line x = 2 These are the roots Roots are also called: -zeros -solutions - x-intercepts The Vertex Form of a Parabola looks like this: y = a(x-h)² + k ‘a’ describes how wide or narrow the parabola will be.

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NEW VOCABULARY A parabola is a set of all points that are the same distance form a fixed line and a fixed point not on the line The fixed point is called the focus of the parabola. The fixed line is called the directrix. The focus A point in the arc of the parabola such that all points on the parabola are equal distance away from the focus and the directrix The line segment through a focus of a parabola, perpendicular to the major axis, which has both endpoints on the curve is called the Latus Rectum. C is the distance between the focus and the vertex

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Write an equation for a parabola with a vertex at the origin and a focus at (0, –7). EXAMPLE Step 1:Determine the orientation of the parabola. Does it point up or down? Make a sketch. Since the focus is located below the vertex, the parabola must open downward. Use y = ax 2. Step 2:Find a.

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WRITE AN EQUATION FOR A PARABOLA WITH A VERTEX AT THE ORIGIN AND A FOCUS AT (0, –7). An equation for the parabola is y = – x 2. 1 28 Since the parabola opens downward, a is negative. So a = –. 1 28 | a | =Note: c is the distance from the vertex to the focus. =Since the focus is a distance of 7 units from the vertex, c = 7. = 1 4(7) 1 28 1 4c To find a, we use the formula:

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WRITING EQUATIONS GIVEN … WRITE THE EQ GIVEN VERTEX(-2,4) AND FOCUS(-2,2) Draw a graph with given info Use given info to get measurements C = distance from Vertex to Focus, so c = 2 (4-2 of the vertical coordinates) Also, parabola will open down because the focus is below the vertex Use standard form Y = a(x – h)² + k Need values for h,k, and a (h, k) = (-2, 4) To find a, use formula a = 1/4c Therefore a = 1/8 Plug into formula Y = -1/8 (x + 2)² + 4 V(-2,4) F(-2,2) C = 2

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WRITING EQUATIONS GIVEN … WRITE THE EQ GIVEN F(3,2) AND DIRECTRIX IS X = -5 Draw a graph with given info Use given info to get measurements Vertex is in middle of directrix and focus. The distance from the directrix to the focus is 8 units. That means V = (-1, 2) C = distance from V to F, so c = 4 Also, parabola will open right Use standard form x = a(y – k)² + h Need values for h,k, and a (h, k) = (-1, 2) To find a, use formula a = 1/4c Therefore a = 1/16 Plug into formula x = 1/16 (y – 2)² – 1 F(3,2) C = 4 X = -5 Vertex is in middle of directrix and F So V = (-1, 2) Distance = 8

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A parabolic mirror has a focus that is located 4 in. from the vertex of the mirror. Write an equation of the parabola that models the cross section of the mirror. The distance from the vertex to the focus is 4 in., so c = 4. Find the value of a. LET’S TRY ONE a = 1 4c = 1 4(4) = Since the parabola opens upward, a is positive. 1 16 The equation of the parabola is y = x 2. 1 16

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Write an equation for a graph that is the set of all points in the plane that are equidistant from point F(0, 1) and the line y = –1. LET’S TRY ONE

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Write an equation for a graph that is the set of all points in the plane that are equidistant from point F(0, 1) and the line y = –1. LET’S TRY ONE An equation for a graph that is the set of all points in the plane that are equidistant from the point F (0, 1) and the line y = –1 is y = x 2. 1414

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EXAMPLE: GIVEN THE EQUATION OF A PARABOLA, GRAPH AND LABEL ALL PARTS Start with the vertex V = (h,k) = (3,-1) Find the focus. C = 1/4a Since a = 1/8, then C = 2 Focus is at (3, -1 + 2) (3,1) Label the directrix and A.O.S. Directrix is at y = -3 (the y coordinate of the vertex, minus the value of c) A.O.S. x = 3 Find the latus rectum The length of the l.r. is | 1/a | Since a = 1/8, the l.r. =8 Plot and label everything Y = -3 X = 3 F(3,1) V(3,-1)

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GRAPH AND LABEL ALL PARTS Start with the vertex V = (h,k) = (2,-1) Find the focus. C = 1/4a Since a = ¼, then C = 1 Focus is at (2+1, -1) (3,-1) Label the directrix and A.O.S. Directrix is at x = 1 A.O.S. y = -1 Find the latus rectum The length of the l.r. is | 1/a | Since a = ¼, the l.r. =4 Plot and label everything Y = -1 X = 1 F(3,-1) V(2,-1)

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Identify the focus and directrix of the graph of the equation x = – y 2. EXAMPLE The parabola is of the form x = ay 2, so the vertex is at the origin and the parabola has a horizontal axis of symmetry. Since a < 0, the parabola opens to the left. 4c = 8 The focus is at (–2, 0). The equation of the directrix is x = 2. c = 2 1818 | a | = 1 4c | – | = 1 4c 1818

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Identify the vertex, the focus, and the directrix of the graph of the equation x 2 + 4x + 8y – 4 = 0. Then graph the parabola. EXAMPLE x 2 + 4x + 8y – 4 = 0 8y = –x 2 – 4x + 4Solve for y, since y is the only term. 8y = –(x 2 + 4x + 4) + 4 + 4Complete the square in x. y = – (x + 2) 2 + 1vertex form 1818 The parabola is of the form y = a(x – h) 2 + k, so the vertex is at (–2, 1) and the parabola has a vertical axis of symmetry. Since a < 0 (negative), the parabola opens downward.

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EXAMPLE (CONT) The vertex is at (–2, 1) and the focus is at (–2, –1). The equation of the directrix is y = 3. Locate one or more points on the parabola. Select a value for x such as –6. The point on the parabola with an x-value of –6 is (–6, –1). Use the symmetric nature of a parabola to find the corresponding point (2, –1). | – | =Substitute – for a. 1818 | a | = 4c = 8 Solve for c. c = 2 1818 1 4c 1 4c

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