1. 8.2 Graph and Write Equations of Parabolas
Where is the focus and directrix compared to the vertex?
How do you know what direction a parabola opens?
How do you write the equation of a parabola given the focus/directrix?
What is the general equation for a parabola?

2. Parabolas
A parabola is defined in terms of a fixed point, called the focus,
focus
and a fixed line,
called the directrix.
A parabola is the set of all points P(x,y) in the plane whose distance to the focus
equals its distance to the directrix.
directrix
axis of symmetry

3. Horizontal Directrix
Standard Equation of a parabola with its vertex at the origin is x y
D(x, –p)
P(x, y)
F(0, p)
y = –p O
x2 = 4py
p > 0: opens upward
p < 0: opens downward
focus: (0, p)
directrix: y = –p
axis of symmetry: y-axis

4. Vertical Directrix
Standard Equation of a parabola with its vertex at the origin is x y
D(x, –p)
P(x, y)
F(p, 0)
x = –p O
y2= 4px
p > 0: opens right
p < 0: opens left
focus: (p, 0)
directrix: x = –p
axis of symmetry: x-axis

5. Example 1 Graph . Label the vertex, focus, and directrix. y2 = 4px
Identify p. -4 -2 2 4
y2 = 4(1)x
So, p = 1
Since p > 0, the parabola opens to the right.
Vertex: (0,0)
Focus: (1,0)
Directrix: x = -1

6. Example 1 Graph . Label the vertex, focus, and directrix. Y2 = 4x y x
Use a table to sketch a graph -4 -2 2 4 y x 2 4 -2 -4 1 4 1 4

7. Graph x = –⅛y 2. Identify the focus, directrix, and axis
of symmetry.
SOLUTION
STEP 1
Rewrite the equation in standard form. 18
x = –
Write original equation.
– 8x = y 2
Multiply each side by – 8.
STEP 2
Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = – 2. The focus is (p, 0), or (– 2, 0). The directrix is x = – p, or x = 2. Because y is squared, the axis of symmetry is the x - axis.

8. STEP 3
Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values.

9. Rewrite the equation in standard form. Y 2 = 4 (– )x 3 2 STEP 2
Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.
1. Y 2 = –6x
SOLUTION
STEP 1
Rewrite the equation in standard form.
Y 2 = 4 (– )x 3 2
STEP 2
Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = – The focus is (p, 0), or (– , 0). The directrix is x = – p, or x = . Because y is squared, the axis of symmetry is the x - axis. 3 2

10. STEP 3
Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values.
2.45
4.24
4.90
5.48

11. Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.14
3. y = – x 2
SOLUTION
STEP 1
Rewrite the equation in standard form.
y = – x 2 14
Write original equation.
Multiply each side by – 4.
– 4y = x 2
equation
STEP 2
focus
directrix
axis of symmetry
x 2 = – 4
0, –1
y = 1
Vertical x = 0

12. STEP 3
Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative y - values. y x 2
4.47
2.83
3.46 4

13. Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.
4. x = – y 2 13
SOLUTION
STEP 1
Rewrite the equation in standard form. 13
x =
Y 2
Write original equation.
Multiply each side by 3.
3x = y 2
equation
STEP 2
focus
directrix
axis of symmetry
Y 2 = x 3 4 0, 3 4
x = – 3 4
Horizontal y = 0

14. STEP 3
Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values. y x
1.73
3.87
2.45 3
3.46

15. Example 2
Write the standard equation of the parabola with its vertex at the origin and the directrix y = -6.
Since the directrix is below the vertex, the parabola opens up
Since y = -p and y = -6,
p = 6
x2=4(6)y
x2 = 24y

16. Write an equation of the parabola shown.
SOLUTION
The graph shows that the vertex is (0, 0) and the
directrix is y = – p = for p in the standard form of the equation of a parabola.
3 2 –
x2 = 4py
Standard form, vertical axis of symmetry
x2 = 4•
( )y
3 2
Substitute for p
3 2
x 2 = 6y
Simplify.

17. Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus.
5. Directrix: y = 2
SOLUTION
x 2 = 4py
Standard form, vertical axis of symmetry
x 2 = 4 (–2)y
Substitute –2 for p
x 2 = – 8y
Simplify.

18. Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus.
8. Focus: (0, 3)
SOLUTION
x 2 = 4py
Standard form, vertical axis of symmetry
x 2 = 4 (3)y
Substitute 3 for p
x 2 = 12y
Simplify.

19. Solar Energy
The EuroDish, developed to provide electricity in remote areas, uses a parabolic reflector to concentrate sunlight onto a high-efficiency engine located at the reflector’s focus. The sunlight heats helium to 650°C to power the engine.
• Write an equation for the EuroDish’s cross section with its vertex at (0, 0).
• How deep is the dish?
Read more on page 498

20. Where is the focus and directrix compared to vertex?
The focus is a point on the line of symmetry and the directrix is a line below the vertex. The focus and directrix are equidistance from the vertex.
How do you know what direction a parabola opens?
x2, graph opens up or down, y2, graph opens right or left
How do you write the equation of a parabola given the focus/directrix?
Find the distance from the focus/directrix to the vertex (p value) and substitute into the equation.
What is the general equation for a parabola?
x2= 4py (opens up [p>0] or down [p<0]), y2 = 4px (opens right [p>0] or left [p<0])

21. 8.2 Assignment
p. 499, odd,
27-33 odd,
39-45 odd

22. 8.2 Graph and Write Equations of Parabolas day 2
What does it mean if a parabola has a translated vertex?
What general equations can you use for a parabola when the vertex has been translated?

23. Standard Equation of a Translated Parabola
Vertical axis:
(x − h)2 = 4p(y − k)
vertex: (h, k)
focus: (h, k + p)
directrix: y = k – p
axis of symmetry: x = h

24. Standard Equation of a Translated Parabola
Horizontal axis:
(y − k)2 = 4p(x − h)
vertex: (h, k)
focus: (h + p, k)
directrix: x = h - p
axis of symmetry: y = k

25. Example 3
Write the standard equation of the parabola with a focus at F(-3,2) and directrix y = 4. Sketch the info.
The parabola opens downward, so the equation is of the form
(x − h)2 = 4p(y − k)
vertex: (-3,3)
h = -3, k = 3
p = -1
(x + 3)2 = 4(−1)(y − 3)

26. Draw a curve through the points.
Graph (x – 2)2 = 8 (y + 3).
SOLUTION
STEP 1
Compare the given equation to the standard form of an equation of a parabola . You can see that the graph is a parabola with vertex at (2, – 3) ,focus (2, – 1) and directrix y = – 5
Draw the parabola by making a table of value and plot y point. Because p > 0, he parabola open to the right. So use only points x- value
STEP 2 x 1 2 3 4 5 y
–2.875 –3
– 2.875
–2.5
– 1.875
STEP 3
Draw a curve through the points.

27. Example 4
Write an equation of a parabola whose vertex is at (−2,1) and whose focus is at (−3, 1).
Begin by sketching the parabola. Because the parabola opens to the left, it has the form
(y −k)2 = 4p(x − h)
Find h and k: The vertex is at (−2,1) so h = −2 and k = 1
Find p: The distance between the vertex (−2,1) and the focus (−3,1) by using the distance formula.
p = −1 (y − 1)2 = −4(x + 2)

28. Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3.
Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3).
SOLUTION
STEP 1
Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (y – k)2 = 4p(x – h) where p < 0.
STEP 2
Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3.
STEP 3
Find p. The vertex (– 2, 3) and focus (4, 3) both lie on the line y = 3, so the distance between them is p | = | – 4 – (– 2) | = 2, and thus p = +2. Because p < 0, it follows that p = – 2, so 4p = – 8.
The standard form of the equation is (y – 3)2 = – 8(x + 2).

29. Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1.
Write the standard form of a parabola with vertex at (3, – 1) and focus at (3, 2).
SOLUTION
STEP 1
Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (x – h)2 = 4p(y – k) where p > 0.
STEP 2
Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1.
STEP 3
Find p. The vertex (3, – 1) and focus (3, 2) both lie on the line x = 3, so the distance between them is p | = | – 2 – (– 1) | = 3, and thus p = + 3. Because p > 0, it follows that p = 3, so 4p = 12.
The standard form of the equation is (x – 3)2 = 12(y + 1)

30. What does it mean if a parabola has a translated vertex?
It means that the vertex of the parabola has been moved from (0,0) to (h,k).
What general equations can you use for a parabola when the vertex has been translated?
(y-k)2 =4p(x-h) (x-h)2 =4p(y-k)

31. 8.2 Assignment day 2 p. 499, 26-32 even, 40-46 even