1. Intro to Conic Sections

2. It all depends on how you slice it!
Start with a cone:

3. IF we slice the cone, parallel to the base, what do we get?
A Circle!

4. IF we slice the cone at an angle, what do we get now?
An Ellipse!

5. IF we just take a slice from the lateral face of the cone, what do we get?
A Parabola!

6. Finally, let’s take a slice from the lateral face, perpendicular to the base
A Hyperbola!

7. These shapes are called Conic Sections
These shapes are called Conic Sections. We can use Algebra to describe the equations and graphs of these shapes.

8. Parabolas

9. 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?

10. Solutions

11. Let’s review Parts of a parabola
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.
These are the roots
Roots are also called:
-zeros
-solutions
- x-intercepts
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

12. New Vocabulary
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
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 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

13. Example
Write an equation for a parabola with a vertex at the origin and a focus at (0, –7).
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 = ax2.
Step 2: Find a.

14. To find a, we use the formula:
Write an equation for a parabola with a vertex at the origin and a focus at (0, –7).
To find a, we use the formula:
| 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) 28 4c
Since the parabola opens downward, a is negative.
So a = – 1 28
An equation for the parabola is y = – x2. 1 28

15. 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)
C = 2
F(-2,2)

16. 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
X = -5
C = 4
F(3,2)
Distance = 8
Vertex is in middle
of directrix and F
So V = (-1 , 2)

17. Let’s try one
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.
a = 1 4c = 1
4(4)
= Since the parabola opens upward, a is positive. 1 16
The equation of the parabola is y = x2. 1 16

18. 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.

19. 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.
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 = x2. 1 4

20. 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, )  (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
X = 3
F(3,1)
Y = -3
V(3,-1)

21. 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
X = 1
V(2,-1)
Y = -1
F(3,-1)

22. Example Identify the focus and directrix of the graph of the
equation x = – y2. 1 8
The parabola is of the form x = ay2, 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.
| a | = 1 4c
| – | = 1 4c 8
4c = 8
c = 2
The focus is at (–2, 0). The equation of the directrix is x = 2.

23. Example
Identify the vertex, the focus, and the directrix of the graph of the equation x x + 8y – 4 = 0. Then graph the parabola.
x x + 8y – 4 = 0
8y = –x2 – 4x Solve for y, since y is the only term.
8y = –(x x + 4) Complete the square in x.
y = – (x + 2) vertex form 1 8
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.

24. Example (Cont) | a | = | – | = Substitute – for a. 4c = 8 Solve for c.1 8
| a | =
4c = 8 Solve for c.
c = 2 4c
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).