Introduction
Earlier we studied the circle, which is the set of all points
in a plane that are equidistant from a given point in that
plane. We have investigated how to translate between
geometric descriptions and algebraic descriptions of
circles. Now we will investigate how to translate between
geometric descriptions and algebraic descriptions of
parabolas.
1
6.1.2: Deriving the Equation of a Parabola
Key Concepts
• A quadratic function is a function that can be written
in the form f(x) = ax2 + bx + c, where a ≠ 0.
• The graph of any quadratic function is a parabola that
opens up or down.
• A parabola is the set of all points that are equidistant
from a given fixed point and a given fixed line that are
both in the same plane as the parabola.
• That given fixed line is called the directrix of the
parabola.
• The fixed point is called the focus.
2
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
• The parabola, directrix, and focus are all in the same
plane.
• The vertex of the parabola is the point on the
parabola that is closest to the directrix.
• The illustration on the following slide shows the parts
of a parabola.
3
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
4
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
• Parabolas can open in any direction. In this lesson,
we will work with parabolas that open up, down, right,
and left.
• As with circles, there is a standard form for the
equation of a parabola; however, that equation
differs depending on which direction the parabola
opens (right/left or up/down).
5
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
Parabolas That Open Up or Down
• The standard form of an equation of a parabola that
opens up or down and has vertex (h, k) is
(x – h)2 = 4p(y – k), where p ≠ 0.
6
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
7
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
Parabolas That Open Right or Left
• The standard form of an equation of a parabola that
opens right or left and has vertex (h, k) is
(y – k)2 = 4p(x – h), where p ≠ 0.
8
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
9
6.1.2: Deriving the Equation of a Parabola
Key Concepts, continued
All Parabolas
• For any parabola, the focus and directrix are each |p|
units from the vertex.
• Also, the focus and directrix are 2|p| units from each
other.
• If the vertex is (0, 0), then the standard equation of
the parabola has a simple form.
• (x – 0)2 = 4p(y – 0) is equivalent to the simpler
form x 2 = 4py.
• (y – 0)2 = 4p(x – 0) is equivalent to the simpler
form y 2 = 4px.
6.1.2: Deriving the Equation of a Parabola
10
Key Concepts, continued
• Either of the following two methods can be used to
write an equation of a parabola:
• Apply the geometric definition to derive the
equation.
• Or, substitute the vertex coordinates and the value
of p directly into the standard form.
11
6.1.2: Deriving the Equation of a Parabola
Common Errors/Misconceptions
• confusing vertical and horizontal when graphing onevariable linear equations
• subtracting incorrectly when negative numbers are
involved
• using the wrong sign for the last term when writing the
result of squaring a binomial
• omitting the middle term when writing the result of
squaring a binomial
• using the incorrect standard form of the equation based
on the opening direction of the parabola
12
6.1.2: Deriving the Equation of a Parabola
Guided Practice
Example 2
Derive the standard equation of the parabola with focus
(–1, 2) and directrix x = 7 from the definition of a
parabola. Then write the equation by substituting the
vertex coordinates and the value of p directly into the
standard form.
13
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
1. To derive the
equation, begin by
plotting the focus.
Label it F (–1, 2).
Graph the directrix
and label it x = 7.
Sketch the parabola.
Label the vertex V.
14
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
2. Let A (x, y) be any point on the parabola.
Point A is equidistant from the focus and the directrix.
The distance from A to the directrix is the horizontal
distance AB, where B is on the directrix directly to the
right of A. The x-value of B is 7 because the directrix
is at x = 7. Because B is directly to the right of A, it
has the same y-coordinate as A. So, B has
coordinates (7, y).
15
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
16
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
3. Apply the definition of a parabola to derive
the standard equation using the distance
formula.
Since the definition of a parabola tells us that
AF = AB, use the graphed points for AF and AB to
apply the distance formula to this equation.
éë x - ( -1) ùû + ( y - 2) =
2
2
( x - 7) + ( y - y )
2
2
17
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
( x + 1) + ( y - 2)
2
2
=
( x - 7)
( x + 1) + ( y - 2) = ( x - 7)
2
2
( y - 2) = ( x - 7) - ( x + 1)
2
2
2
2
2
Simplify.
Square both sides.
Subtract (x + 1)2 from
both sides to get all x
terms on one side and all
y terms on the other side.
18
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
( y - 2) = ( x
2
2
) (
2
2
y
2
=
x
14x
+
49
x
- 2x - 1
( )
2
( y - 2)
2
)
- 14x + 49 - x 2 + 2x + 1
= -16x + 48
Square the
binomials on the
right side.
Distribute the
negative sign.
Simplify.
19
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
( y - 2)
2
= -16 ( x - 3 )
Factor on the right side to
obtain the standard form
(y – k)2 = 4p(x – h).
The standard equation is (y – 2)2 = –16(x – 3).
20
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
4. Write the equation using standard form.
To write the equation using
the standard form, first
determine the coordinates
of the vertex and the value
of p.
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6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
For any parabola, the distance between the focus and
the directrix is 2|p|. In this case, 2|p| = 7−(−1) = 8,
thus |p| = 4 . The parabola opens left, so p is
negative, and therefore p = –4. For any parabola, the
distance between the focus and the vertex is |p|.
Since |p| = 4 , the vertex is 4 units right of the focus.
Therefore, the vertex coordinates are (3, 2).
22
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
5. Use the results found in step 4 to write the
equation.
(y - k)
( y - 2)
( y - 2)
2
2
2
= 4p ( x - h )
= 4 ( -4 ) ( x - 3 )
= -16 ( x - 3 )
Standard form for a
parabola that opens
right or left
Substitute the values
for h, p, and k.
Simplify.
23
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
The standard equation is (y – 2)2 = –16(x – 3).
The results shown in steps 3 and 5 match; so, either
method of finding the equation (deriving it using the
definition or writing the equation using the standard
form) will yield the same equation.
✔
24
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 2, continued
25
6.1.2: Deriving the Equation of a Parabola
Guided Practice
Example 4
Write the standard equation of the parabola with focus
(–5, –6) and directrix y = 3.4. Then use a graphing
calculator to graph your equation.
26
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
1. Plot the focus and graph the directrix.
Sketch the parabola. Label the vertex V.
27
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
2. To write the equation, first determine the
coordinates of the vertex and the value of p.
28
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
The distance between the focus and the directrix is
2|p|. So 2|p| = 3.4 − (−6) = 9.4, and |p| = 4.7. The
parabola opens down, so p is negative, and therefore
p = –4.7. The distance between the focus and the
vertex is |p|, so the vertex is 4.7 units above the
focus. Add to find the y-coordinate of the vertex:
–6 + 4.7 = –1.3. The vertex coordinates are
(–5, –1.3).
29
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
3. Use the results found in step 2 to write the
equation.
( x - h)
2
= 4p ( y - k )
éë x - ( -5 ) ùû = 4 ( -4.7 ) éë y - ( -1.3 ) ùû
2
( x + 5)
2
= -18.8 ( y + 1.3 )
Standard form for a
parabola that opens
down
Substitute the values
for h, p, and k.
Simplify.
The standard equation is (x + 5)2 = –18.8(y + 1.3).
30
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
4. Solve the standard equation for y to
obtain a function that can be graphed.
( x + 5)
-
-
2
= -18.8 ( y + 1.3 )
1
2
1
2
x + 5)
(
18.8
x + 5)
(
18.8
= y + 1.3
- 1.3 = y
Standard equation
Multiply both sides by
1
.
18.8
Add –1.3 to both
sides.
31
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
5. Graph the function using a graphing
calculator.
On a TI-83/84:
Step 1: Press [Y=].
Step 2: At Y1, type in [(][(–)][1][ ÷ ][18.8][)]
[(][X,T,θ,n][+][5][)][x2][–][1.3].
Step 3: Press [WINDOW] to change the viewing
window.
Step 4: At Xmin, enter [(–)][12].
Step 5: At Xmax, enter [12].
32
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
Step 6: At Xscl, enter [1].
Step 7: At Ymin, enter [(–)] [8].
Step 8: At Ymax, enter [8].
Step 9: At Yscl, enter [1].
Step 10: Press [GRAPH].
33
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
On a TI-Nspire:
Step 1: Press the [home] key.
Step 2: Arrow to the graphing icon and press [enter].
Step 3: At the blinking cursor at the bottom of the
screen, enter [(][(–)][1][ ÷
][18.8][)][(][x][+][5][)][x2][–][1.3].
Step 4: Change the viewing window by pressing
[menu], arrowing down to number 4:
Window/Zoom, and clicking the center button
of the navigation pad.
34
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
Step 5: Choose 1: Window settings by pressing the
center button.
Step 6: Enter in an appropriate XMin value, –12, by
pressing [(–)] and [12], then press [tab].
Step 7: Enter in an appropriate XMax value, [12],
then press [tab].
Step 8: Leave the XScale set to “Auto.” Press [tab]
twice to navigate to YMin and enter an
appropriate YMin value, –8, by pressing [(–
)] and [8].
35
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
Step 9: Press [tab] to navigate to YMax. Enter [8].
Press [tab] twice to leave YScale set to
“auto” and to navigate to “OK.”
Step 10: Press [enter].
Step 11: Press [menu] and select 2: View and
5: Show Grid.
36
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
✔
37
6.1.2: Deriving the Equation of a Parabola
Guided Practice: Example 4, continued
38
6.1.2: Deriving the Equation of a Parabola