Parabolas GEO HN CCSS: G.GPE.2   Parabolas GEO HN CCSS: G.GPE.2

Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others.   4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning.

Essential Question: What is the relationship between focus and directrix to the equation of a parabola?

CCSS: G.GPE.2 DERIVE the equation of a parabola given a focus and directrix. 

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 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. Focus d1 Directrix The length of the latus rectum is |4p| where p is the distance from the vertex to the focus.

Things you should already know about a parabola. 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

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

Remember: |p| is the distance from the vertex to the focus Equations of a Parabola x = ay2 + by + c y = ax2 + bx + c Horizontal Hyperbola Vertical Hyperbola -b 2a -b 2a vertex: y = Vertex: x = If a > 0, opens right If a > 0, opens up If a < 0, opens left If a < 0, opens down The directrix is vertical 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

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

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)

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

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) Directrix: x = -2 V F x = ¼(y + 3)2 – 1 x y 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

The receiver should be placed 4.5 feet above the base of the dish. 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)

The cable would be 90 ft long at a point 100 ft from a tower. 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