8.2 Graph and Write 8.2 Graph and Write Equations of ParabolasEquations of Parabolas8.2 Graph and Write 8.2 Graph and Write

Equations of ParabolasEquations of Parabolas

•Where is the focus and directrix compared to the Where is the focus and directrix compared to the vertex?vertex?•How do you know what direction a parabola opens?How do you know what direction a parabola opens?•How do you write the equation of a parabola given How do you write the equation of a parabola given the focus/directrix?the focus/directrix?•What is the general equation for a parabola?What is the general equation for a parabola?

Parabolas

focus

axis of symmetry

directrix

A parabola is defined in terms of a fixed point, called the 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.

Horizontal Directrix

p > 0: opens upward

focus: (0, p)

directrix: y = –p

axis of symmetry: y-axisx

y

D(x, –p)

P(x, y)

F(0, p)

y = –pO

p < 0: opens downward

Standard Equation of a parabola with its vertex at the origin is

x2 = 4py

Vertical Directrix

p > 0: opens right

focus: (p, 0)

directrix: x = –p

axis of symmetry: x-axis

p < 0: opens left

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

Example 1Graph . Label the vertex, focus, and directrix. y2 = 4px

21x y

4

-4 -2

2

42

4

-4

-2

Identify p.

So, p = 1

Since p > 0, the parabola opens to the right.

Vertex: (0,0)Focus: (1,0)Directrix: x = -1

y2 = 4(1)x

Example 1Graph . Label the vertex, focus, and directrix. Y2 = 4x

21x y

4

-4 -2

2

42

4

-4

-2

Use a table to sketch a graph

y

x0

0

2

1

4

4

-2

1

-4

4

SOLUTION

STEP 1

Rewrite the equation in standard form.18

x = – Write original equation.

Graph x = –⅛y 2. Identify the focus, directrix, and axis

of symmetry.

– 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.

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.

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 (– )x32

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.

32

32 3

2

STEP 3Draw 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 2.45 4.24 4.90 5.48

SOLUTION

STEP 1 Rewrite the equation in standard form.Write original equation.

– 4y = x 2Multiply each side by – 4.

Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

3. y = – x 214

y = – x 214

focus directrix axis of symmetry

x 2 = – 4 0, –1 y = 1 Vertical x = 0

equation

STEP 2

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 4.472.83 3.46 4

SOLUTION

STEP 1 Rewrite the equation in standard form.

Write original equation.

3x = y 2 Multiply each side by 3.

Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

4. x = – y 2

13

13

x = Y 2

focus directrix axis of symmetry

Horizontal y = 0

equationSTEP 2

Y 2 = 4 x34

0, 34

x = – 34

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 3.872.45 3 3.46

Example 2Write 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 upSince y = -p and y = -6,p = 6

x2=4(6)y x2 = 24y

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• ( )y3 2 Substitute for p3

2 x 2 = 6y Simplify.

Write an equation of the parabola shown.

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

x 2 = 4py Standard form, vertical axis of symmetry

x 2 = 4 (–2)y Substitute –2 for p

x 2 = – 8y Simplify.

SOLUTION

Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus.

x 2 = 4py Standard form, vertical axis of symmetry

x 2 = 4 (3)y Substitute 3 for p

x 2 = 12y Simplify.

8. Focus: (0, 3)

SOLUTION

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

• 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])

8.2 Assignment

p. 499, 3-17 odd, 27-33 odd, 39-45 odd

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?

Standard Equation of a Translated Parabola

Vertical axis:

vertex: (h, k)

focus: (h, k + p)

directrix: y = k – p

axis of symmetry: x = h

(x − h)2 = 4p(y − k)

Standard Equation of a Translated Parabola

Horizontal axis:

vertex: (h, k)

focus: (h + p, k)

directrix: x = h - p

axis of symmetry: y = k

(y − k)2 = 4p(x − h)

Example 3Write 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

vertex: (-3,3)

h = -3, k = 3

p = -1

(x − h)2 = 4p(y − k)

(x + 3)2 = 4(−1)(y − 3)

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 = – 5Draw 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

x 1 2 3 4 5

y –2.875 –3 – 2.875 –2.5 – 1.875

STEP 3 Draw a curve through the points.

STEP 2

Example 4Write 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)

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).

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)

• 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)

8.2 Assignment day 2

p. 499, 26-32 even, 40-46 even

p. 531, 3, 9, 15-16