Unit 2: Day 6 Continue

Unit 2: Day 6 Continue

Unit 2: Day 8 Equation of a Parabola with Focus and DirectX

Introduction

Parabolas Things you should already know about a parabola. Forms of equations y = a(x – h)2 + k opens up if a is positive opens down if a is negative vertex is (h, k) y = ax2 + bx + c opens up if a is positive V opens down if a is negative -b 2a -b 2a vertex is , f( ) Thus far in this course we have studied parabolas that are vertical - that is, they open up or down and the axis of symmetry is vertical

Parabola A parabola is a set of points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus. For any point Q that is on the parabola, d2 = d1 Q d2 Focus Focus: a point on the axis of symmetry that is “p” units from the vertex d1 Directrix Directx: a line that is perpendicular to the axis of symmetry “p” units from the vertex . It is outside of the parabola

Parabola A parabola is a set of points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus. For any point Q that is on the parabola, d2 = d1 Q d2 Focus The latus rectum of a parabola is a line segment that passes through the focus, is parallel to the directrix and has its endpoints on the parabola. d1 Directrix The length of the latus rectum is |4p| where p is the distance from the vertex to the focus.

Equations of a Parabola y = ax2 + bx + c Vertical Hyperbola -b 2a Vertex: x = If a > 0, opens up If a < 0, opens down The directrix is horizontal 1 a = 4p Remember: |p|is the distance from the vertex to the focus the directrix is the same distance from the vertex as the focus is

Standard Form Equation of a Parabola (x – h)2 = 4p(y – k) Vertical Parabola Vertex: (h, k) If 4p > 0, opens up If 4p < 0, opens down The directrix is horizontal and the vertex is midway between the focus and directrix Remember: |p| is the distance from the vertex to the focus

EX 1: Find the standard form of the equation of the parabola given: the vertex is (2, -3) and focus is (2, -5) Because of the location of the vertex and focus this must be a vertical parabola that opens down Equation: (x – h)2 = 4p(y – k) |p| = 2 V Equation: (x – 2)2 = -8(y + 3) F The vertex is midway between the focus and directrix, so the directrix for this parabola is y = -1

Ex 2: find the standard form equation of the parabola if… the focus is (2,5) and the directrix is y =3 Because of the location of the vertex and focus this must be a vertical parabola that opens up F Equation: (x – h)2 = 4p(y – k) Vertex is the mid point between focus and directrix (2, 4) |p| = 1 Equation: (x – 2)2 = 4(y - 4) Foil and solve for y.

Ex 3: find the standard form equation of the parabola if… the focus is (-1,-4) and the directrix is y =-6 Because of the location of the vertex and focus this must be a vertical parabola that opens up Equation: (x – h)2 = 4p(y – k) Vertex is the mid point between focus and directrix ( -1, -5) |p| = 1 Equation: (x +1)2 = 4(y +6) Foil and solve for y. F

Ex 4: find the equation of the parabola if… the focus is (2,3) and the directrix is y =-1 Because of the location of the vertex and focus this must be a vertical parabola that opens up Equation: (x – h)2 = 4p(y – k) Vertex is the mid point between focus and directrix (2 , 1) F |p| = 4 Equation: (x -2)2 = 4(4)(y -1) Foil and solve for y.

So… finding an equation from the focus and directrix. What is important to remember? • You can sketch the graph from Focus and directrix. • The distance from Focus to vertex to directrix is the same “p”. • The equation is (x – h)2 = 4p(y – k) . • Solve for y

Can you tell whether the parabola opens up or down by the location of the directrix? How?

Consider the following functions.

Try This….

Applications A satellite dish is in the shape of a parabolic surface. The dish is 12 ft in diameter and 2 ft deep. How far from the base should the receiver be placed? 12 2 (-6, 2) (6, 2) Consider a parabola cross-section of the dish and create a coordinate system where the origin is at the base of the dish. Since the parabola is vertical and has its vertex at (0, 0) its equation must be of the form: x2 = 4py The receiver should be placed 4.5 feet above the base of the dish. At (6, 2), 36 = 4p(2) so p = 4.5 thus the focus is at the point (0, 4.5)

Application The towers of a suspension bridge are 800 ft apart and rise 160 ft above the road. The cable between them has the shape of a parabola, and the cable just touches the road midway between the towers. (400, 160) (300, h) 300 100 What is the height of the cable 100 ft from a tower? Since the parabola is vertical and has its vertex at (0, 0) its equation must be of the form: At (300, h), 90,000 = 1000h x2 = 4py h = 90 At (400, 160), 160,000 = 4p(160) The cable would be 90 ft long at a point 100 ft from a tower. 1000 = 4p p = 250 thus the equation is x2 = 1000y

Parabolas In this unit we will also study parabolas that are horizontal – that is, they open right or left and the axis of symmetry is horizontal V In these equations it is the y-variable that is squared. x = a(y – k)2 + h or x = ay2 + by + c

Find the standard form of the equation of the parabola given: the focus is (2, 4) and the directrix is x = - 4 The vertex is midway between the focus and directrix, so the vertex is (-1, 4) The directrix is vertical so the parabola must be horizontal and since the focus is always inside the parabola, it must open to the right V F Equation: (y – k)2 = 4p(x – h) |p| = 3 Equation: (y – 4)2 = 12(x + 1)

Graphing a Parabola (y + 3)2 = 4(x + 1) Find the vertex, focus and directrix. Then graph the parabola Vertex: (-1, -3) The parabola is horizontal and opens to the right 4p = 4 p = 1 Focus: (0, -3) V Directrix: x = -2 F x = ¼(y + 3)2 – 1 x y 0 0 3 3 -1 -5 1 -7 -

Converting an Equation Convert the equation to standard form Find the vertex, focus, and directrix y2 – 2y + 12x – 35 = 0 y2 – 2y + ___ = -12x + 35 + ___ 1 1 (y – 1)2 = -12x + 36 F (y – 1)2 = -12(x – 3) V The parabola is horizontal and opens left Vertex: (3, 1) 4p = -12 Focus: (0, 1) p = -3 Directrix: x = 6

Unit 2: Day 6 Continue