INTRO TO CONIC SECTIONS. IT ALL DEPENDS ON HOW YOU SLICE IT! Start with a cone:

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INTRO TO CONIC SECTIONS

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

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

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

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

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

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

10.2 - PARABOLAS

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?

SOLUTIONS

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.

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

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.

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:

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

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

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

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

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

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)

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)

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

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.

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