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

IntroductionEarlier 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.

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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.

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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.

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Key Concepts, continued

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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).

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Key Concepts, continuedParabolas 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.

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Key Concepts, continuedParabolas 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.

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Key Concepts, continuedAll 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. 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.

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Common Errors/Misconceptions• confusing vertical and horizontal when graphing one-

variable 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

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Guided Practice

Example 2Derive 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.

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Guided Practice: Example 2, continued

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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.


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).

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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.

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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.


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Square the binomials on the right side.

Distribute the negative sign.

Simplify.


The standard equation is (y – 2)2 = –16(x – 3).

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Factor on the right side to obtain the standard form (y – k)2 = 4p(x – h).


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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Guided Practice: Example 2, continuedFor 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).

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5. Use the results found in step 4 to write the equation.

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Standard form for a parabola that opens right or left

Substitute the values for h, p, and k.

Simplify.

Guided Practice: Example 2, continuedThe 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.

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http://www.walch.com/ei/00283

Guided Practice

Example 4Write the standard equation of the parabola with focus (–5, –6) and directrix y = 3.4. Then use a graphing calculator to graph your equation.

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1. Plot the focus and graph the directrix. Sketch the parabola. Label the vertex V.

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2. To write the equation, first determine the coordinates of the vertex and the value of p.

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Guided Practice: Example 4, continuedThe 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).

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3. Use the results found in step 2 to write the equation.

The standard equation is (x + 5)2 = –18.8(y + 1.3). 30


Standard form for a parabola that opens down

Substitute the values for h, p, and k.

Simplify.


4. Solve the standard equation for y to obtain a function that can be graphed.

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Standard equation

Multiply both sides by

Add –1.3 to both sides.


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


Guided Practice: Example 4, continuedStep 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].

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Guided Practice: Example 4, continuedOn 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.

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Guided Practice: Example 4, continuedStep 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].

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Guided Practice: Example 4, continuedStep 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.

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Documents

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