1. Parabola
Conic section

2. Warm-up 3. y = x2 - 5 1. y = x2 - 6x + 8 2. y = –x2 + 4x – 4
Graph the following parabola using:
I Finding the solution of the equations (Factoring)
1. y = x2 - 6x + 8
II Finding the VERTEX (Using formula)
2. y = –x2 + 4x – 4
III Graphing on y-axis (using vertex)
3. y = x2 - 5

3. Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix.

4. Note the line through the focus, perpendicular to the directrix
Axis of symmetry
Note the point midway between the directrix and the focus
Vertex 4

5. If p > 0, the parabola opens up.
The equation of a parabola with vertex (0, 0) and focus on the y-axis is x2 = 4py.
The coordinates of the focus are (0, p).
The equation of the directrix is y = -p.
If p > 0, the parabola opens up.
If p < 0, the parabola opens down.

6. Standard equation of a PARABOLA
The equation of a parabola with vertex (0, 0) and focus on the x-axis is y2 = 4px.
The coordinates of the focus are (p, 0).
The equation of the directrix is x = -p.
If p > 0, the parabola opens right.
If p < 0, the parabola opens left.

7. Finding the Equation of a Parabola with Vertex (0, 0)
A parabola has vertex (0, 0) and the focus on an axis.
Write the equation of each parabola.
a) The focus is (-6, 0).
Since the focus is (-6, 0), the equation of the parabola is y2 = 4px.
p is equal to the distance from the vertex to the focus, therefore p = -6.
The equation of the parabola is y2 = -24x.
b) The directrix is defined by x = 5.
The equation of the directrix is x = -p, therefore -p = 5 or p = -5.
Since the focus is on the x-axis, the equation of the parabola is y2 = 4px.
The equation of the parabola is y2 = -20x.
c) The focus is (0, 3).
Since the focus is (0, 3), the equation of the parabola is x2 = 4py.
p is equal to the distance from the vertex to the focus, therefore p = 3.
The equation of the parabola is x2 = 12y.

8. Practice Finding the Equation of a Parabola with Vertex (0, 0)
A parabola has vertex (0, 0) and the focus on an axis.
Write the equation of each parabola.
a) The focus is (8, 0).
The equation of the parabola is y2 = 32x.
b) The directrix is defined by x = 3.
The equation of the parabola is y2 = -12x.
c) The focus is (0, -5).
The equation of the parabola is x2 = -20y.

9. Finding the FOCUS DIRECTRIX
y = 4x2
x = -3y2
y = 4(4py)
y = 16py
1 = 16p
1/16 = p
x = -3(4px)
x = -12px
1 = -12p
-1/12 = p
FOCUS: (0, 1/16)
FOCUS: (-1/12, 0)
Directrix Y = - 1/16
Directrix x = 1/12

10. Practice y = 8x2 x = -4y2 FOCUS: (0, 1/32) FOCUS: (-1/16, 0)
Directrix Y = - 1/32
Directrix x = 1/16

11. Parabola
Conic section

12. Find the focus and directrix of the following:
WARM -UP
Find the focus and directrix of the following:
2. (try this one on your own)
x = 8y2
1. (try this one on your own)
y = -6x2
FOCUS
(0, -1/24)
Directrix
y = 1/24
FOCUS
(1/32, 0)
Directrix
x = -32

13. Horizontal Axis, directrix: x = h-p
(y-k)2 =4p(x-h),p≠0
Horizontal Axis, directrix: x = h-p
The equation of the
axis of symmetry is y = k.
The coordinates of the focus are (h + p, k).
The equation of the directrix is x = h - p.

14. Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h)
h = 6 and k = 5
(y - 5)2 = 4(-3)(x - 6)
(y - 5)2 = -12(x - 6)
Standard form
y2 - 10y + 25 = -12x + 72
y2 + 12x - 10y - 47 = 0
General form

15. (y-6)2 = -8(x-6) Standard Form
Practice
Write the equation of the parabola with a focus at (4, 6) and the directrix at x = 8, in standard form and general form.
Vertex: (6,6)
(y-6)2 = -8(x-6) Standard Form
y2 + 8x -12y -12 General Form

16. Standard Equation of a Parabola with vertex at (h,k)
(x-h)2 =4p(y-k),p≠0
Vertical Axis, directrix: y = k-p
The equation of the
axis of symmetry is x = h.
The coordinates of the focus are (h, k + p).
The equation of the directrix is y = k - p.
The general form of the parabola
is Ax2 + Cy2 + Dx + Ey + F = 0
where A = 0 or C = 0.

17. Find the equation of the parabola that has a
min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.
The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equation
and solve for p:
(x - h)2 = 4p(y - k)
x = 2 and y = 8
(2 - (-2))2 = 4p(8 - 6)
16 = 8p
2 = p
(x - h)2 = 4p(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
Standard form
x2 + 4x + 4 = 8y - 48
x2 + 4x - 8y + 52 = 0
General form

18. (x-3)2 = -36(y-6) Standard Form
homework
Find the equation of the parabola that has a
maximum at (3, 6) and passes through the point (9, 5).
Vertex: (3,6)
(x-3)2 = -36(y-6) Standard Form
x2 - 6x +36y -207 General Form

19. (x-h)2 = 4p (y-k) h=2, k=1, p= 4-1 = 3 (x-2)2 = 4(3) (y-1)
Find the equation of the parabola that has a
vertex at (2,1) and focus (2,4).
(x-h)2 = 4p (y-k)
h=2, k=1, p= 4-1 = 3
(x-2)2 = 4(3) (y-1)
(x-2)2 = 12 (y-1)
Standard Form
General Form
X2 - 4x -12y + 16 = 0

20. The equation of the
axis of symmetry is y = k.
The equation of the
axis of symmetry is x = h.
The coordinates of the focus are (h + p, k).
The coordinates of the focus are (h, k + p).
The equation of the directrix is x = h - p.
The equation of the directrix is y = k - p.

21. Analyzing a Parabola Find the coordinates of the vertex and focus,
the equation of the directrix, the axis of symmetry,
and the direction of opening of y2 - 8x - 2y - 15 = 0.
y2 - 8x - 2y - 15 = 0
y2 - 2y + _____ = 8x _____ 1 1
(y - 1)2 = 8x + 16
(y - 1)2 = 8(x + 2)
Standard
form
4p = 8
p = 2
The vertex is (-2, 1).
The focus is (0, 1).
The equation of the directrix is x + 4 = 0.
The axis of symmetry is y - 1 = 0.
The parabola opens to the right.

22. Finding the FOCUS DIRECTRIX
y = 4x2
x = -3y2
y = 4(4py)
y = 16py
1 = 16p
1/16 = p
x = -3(4px)
x = -12px
1 = -12p
-1/12 = p
FOCUS: (0, 1/16)
FOCUS: (-1/12, 0)
Directrix Y = - 1/16
Directrix x = 1/12

23. Find the equation of the parabola that has a
min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.
The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equation
and solve for p:
(x - h)2 = 4p(y - k)
x = 2 and y = 8
(2 - (-2))2 = 4p(8 - 6)
16 = 8p
2 = p
(x - h)2 = 4p(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
Standard form
x2 + 4x + 4 = 8y - 48
x2 + 4x - 8y + 52 = 0
General form

24. Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h)
h = 6 and k = 5
(y - 5)2 = 4(-3)(x - 6)
(y - 5)2 = -12(x - 6)
Standard form
y2 - 10y + 25 = -12x + 72
y2 + 12x - 10y - 47 = 0
General form