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

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.

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

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

5. 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.

6. 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

7. 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

8. 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

9. 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

10. 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)

11. 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

12. 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 -

13. Converting an Equation
Convert the equation to standard form
Find the vertex, focus, and directrix
y2 – 2y + 12x – 35 = 0
y2 – 2y + ___ = -12x ___ 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

14. 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)

15. 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), ,000 = 1000h
x2 = 4py
h = 90
At (400, 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