The document summarizes key aspects of parabolas as conic sections: 1) A parabola is defined as the set of points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). 2) The standard form of the equation of a parabola is y=ax^2, where the vertex is at the origin, the focus is on the y-axis, and the directrix is the x-axis. 3) Examples are worked through to find the equation, focus, directrix, and other properties of parabolas given information like the vertex or standard form equation.

1. Chapter 10 Conic Sections
(Be sure you have printed out your Conics Help Sheet)
Revelation 1:7 "Behold, he is coming with the clouds, and
every eye will see him, even those who pierced him, and all
tribes of the earth will wail on account of him. Even so. Amen."

2. Conic Sections are the curves we obtain when
intersecting a plane with a double-cone at
various angles.
use K on keyboard to Pause/Play

3. Conic Sections are the curves we obtain when
intersecting a plane with a double-cone at
various angles.
use K on keyboard to Pause/Play

4. 10.1 The Parabola

5. 10.1 The Parabola
Another way to define the
parabola is this:
A Parabola is the set of
points in the plane
equidistant from a fixed
point, F (the Focus), and
a fixed line, l (the
Directrix).

6. 10.1 The Parabola
Another way to define the
parabola is this:
A Parabola is the set of
points in the plane
equidistant from a fixed
point, F (the Focus), and
a fixed line, l (the
Directrix).

7. axis of symmetry
vertex
The parts of a parabola

8. Consider ... the vertex at the Origin

9. Consider ... the vertex at the Origin

10. Consider ... the vertex at the Origin
FP = PT

11. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p

12. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p

13. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p
2
x = 4 py

14. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p
2
x = 4 py
vertex : ( 0, 0 )

15. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p
2
x = 4 py
vertex : ( 0, 0 )
focus : ( 0, p )

16. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p
2
x = 4 py
vertex : ( 0, 0 )
focus : ( 0, p )
directrix : y = − p

17. Consider ... the vertex at the Origin
FP = PT
2 2
( x − 0) + ( y − p) = y+ p
2 2 2 2 2
x + y − 2 py + p = y + 2 py + p
2
x = 4 py
vertex : ( 0, 0 )
focus : ( 0, p )
directrix : y = − p
p > 0, opens up
p < 0, opens down

18. If we take the inverse (interchange x and y) the
parabola is reflected over the y=x line ...

19. If we take the inverse (interchange x and y) the
parabola is reflected over the y=x line ...

20. If we take the inverse (interchange x and y) the
parabola is reflected over the y=x line ...
2
y = 4 px
vertex : ( 0, 0 )
focus : ( p, 0 )
directrix : x = − p
p > 0, opens right
p < 0, opens left

21. If we take the inverse (interchange x and y) the
parabola is reflected over the y=x line ...
2
y = 4 px
vertex : ( 0, 0 )
focus : ( p, 0 )
latus rectum directrix : x = − p
p > 0, opens right
p < 0, opens left
The Latus Rectum (also called the focal diameter) is the
segment with endpoints on the parabola, is perpendicular
to the axis of symmetry and contains the focus point.
Its length is 4 p .

22. 1. Find the equation of the parabola with
vertex V ( 0, 0 ) and focus F ( 0, − 8 ) .

23. 1. Find the equation of the parabola with
vertex V ( 0, 0 ) and focus F ( 0, − 8 ) .
2
x = 4 py

24. 1. Find the equation of the parabola with
vertex V ( 0, 0 ) and focus F ( 0, − 8 ) .
2
x = 4 py
p = −8

25. 1. Find the equation of the parabola with
vertex V ( 0, 0 ) and focus F ( 0, − 8 ) .
2
x = 4 py
p = −8
2
x = −32y

26. 2. Find the focus and directrix of the parabola
2
y = −5x

27. 2. Find the focus and directrix of the parabola
2
y = −5x
2
x = 4 py

28. 2. Find the focus and directrix of the parabola
2
y = −5x
2
x = 4 py
2 1
x =− y
5

29. 2. Find the focus and directrix of the parabola
2
y = −5x
2
x = 4 py
2 1
x =− y
5
1
∴ 4p = −
5

30. 2. Find the focus and directrix of the parabola
2
y = −5x
2
x = 4 py
2 1
x =− y
5
1
∴ 4p = −
5
1
p=−
20

31. 2. Find the focus and directrix of the parabola
2
y = −5x
2
x = 4 py
12
x =− y
5
1
∴ 4p = −
5
1
p=−
20
⎛ 1 ⎞ 1
F ⎜ 0, − ⎟ dir : y =
⎝ 20 ⎠ 20

32. 3. Find the focus and directrix of the parabola
2
2x + y = 0

33. 3. Find the focus and directrix of the parabola
2
2x + y = 0
2
y = −2x

34. 3. Find the focus and directrix of the parabola
2
2x + y = 0
2
y = −2x
4 p = −2

35. 3. Find the focus and directrix of the parabola
2
2x + y = 0
2
y = −2x
4 p = −2
1
p=−
2

36. 3. Find the focus and directrix of the parabola
2
2x + y = 0
2
y = −2x
4 p = −2
1
p=−
2
⎛ 1 ⎞ 1
F ⎜ − , 0 ⎟ dir : x =
⎝ 2 ⎠ 2

37. 4. Find the focus, directrix and focal diameter of
4 2
y= x
9

38. 4. Find the focus, directrix and focal diameter of
4 2
y= x
9
9
4p =
4

39. 4. Find the focus, directrix and focal diameter of
4 2
y= x
9
9
4p = this is the focal diameter
4

40. 4. Find the focus, directrix and focal diameter of
4 2
y= x
9
9
4p = this is the focal diameter
4
9
p=
16

41. 4. Find the focus, directrix and focal diameter of
4 2
y= x
9
9
4p = this is the focal diameter
4
9
p=
16
⎛ 9 ⎞ 9 9
F ⎜ 0, ⎟ dir : y = − foc. dia :
⎝ 16 ⎠ 16 4

42. When you are asked to graph these ...
use your calculator and then put them on graph paper.
HW #1
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the quality which guarantees all others.”
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