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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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.

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6.1.2: Deriving the Equation of a Parabola

Key Concepts, continued

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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).

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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

Key Concepts, continued

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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

Key Concepts, continued

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6.1.2: Deriving the Equation of a Parabola

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

6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

Guided Practice: Example 2, continued

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6.1.2: Deriving the Equation of a Parabola

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.

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).

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6.1.2: Deriving the Equation of a Parabola

Guided Practice: Example 2, continued

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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.

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6.1.2: Deriving the Equation of a Parabola

Guided Practice: Example 2, continued

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6.1.2: Deriving the Equation of a Parabola

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.

Guided Practice: Example 2, continued

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6.1.2: Deriving the Equation of a Parabola

Square the binomials on the right side.

Distribute the negative sign.

Simplify.

Guided Practice: Example 2, continued

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

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6.1.2: Deriving the Equation of a Parabola

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

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, 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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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.

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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

✔

Guided Practice: Example 2, continued

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6.1.2: Deriving the Equation of a Parabola

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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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.

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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.

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6.1.2: Deriving the Equation of a Parabola

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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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.

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

6.1.2: Deriving the Equation of a Parabola

Standard form for a parabola that opens down

Substitute the values for h, p, and k.

Simplify.

Guided Practice: Example 4, continued

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

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6.1.2: Deriving the Equation of a Parabola

Standard equation

Multiply both sides by

Add –1.3 to both sides.

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, 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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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

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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6.1.2: Deriving the Equation of a Parabola

Guided Practice: Example 4, continued

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6.1.2: Deriving the Equation of a Parabola

✔

Guided Practice: Example 4, continued

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6.1.2: Deriving the Equation of a Parabola