7-1 parabolas - montville.net · the parabola. (x ±3) 2 = 12(y ±7) 62/87,21 the equation is in...

For each equation, identify the vertex, focus,

axis of symmetry, and directrix. Then graph

the parabola.

(x 3)2 = 12(y 7)

The equation is in standard form and the squared term is x, which means that the parabola opens

vertically. The equation is in the form (x h)2 = 4p

(y k), so h = 3 and k = 7. Because 4p = 12 and p= 3, the graph opens up.Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = (3, 7)focus: (h, k + p ) = (3, 10)directrix: y = k p or 4axis of symmetry: x = h or 3

Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape of the curve.

(x + 1)2 = 12(y 6)

(x + 1)2 = 12(y 6)


x h)2 =

4p (y k), so h = 1 and k = 6. Because 4p = 12andp = 3, the graph opens down.Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 1, 6)focus: (h, k + p ) = ( 1, 3)directrix: y = k p or 9

x = h or 1


(y 4)2 = 20(x + 2)

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7-1 Parabolas

(y 4)2 = 20(x + 2)

The equation is in standard form and the squared term is y , which means that the parabola opens

y k)2 =

4p (x h), so h = 2 and k = 4. Because 4p = 20 andp = 5, the graph opens to the right. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 2, 4)focus: (h + p , k) = (3, 4)directrix: x = h p or 7

y = k or 4


1(x + 7) = (y + 5)2

1(x + 7) = (y + 5)2


horizontally. The equation is in the form 4p (x h) =

(y k)2, so h = 7 and k = 5. Because 4p = 1

andp = 0.25, the graph opens to the left. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 7, 5)focus: (h + p , k) = ( 7.25, 5)directrix: x = h p or 6.75

y = k or 5


(x + 8)2 = 8(y 3)


7-1 Parabolas

(x + 8)2 = 8(y 3)


vertically. The equation is in the form (x h)2 = 4p

(y k), so h = 8 and k = 3. Because 4p = 8 and p= 2, the graph opens up. Use the values of h, k , andp to determine the characteristics of the parabola.


x = h or 8


40(x + 4) = (y 9)2

40(x + 4) = (y 9)2




andp = 10, the graph opens to the left. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 4, 9)focus: (h + p , k) = ( 14, 9)directrix: x = h p or 6

y = k or 9


(y + 5)2 = 24(x 1)


7-1 Parabolas

(y + 5)2 = 24(x 1)


horizontally. The equation is in the form (y k)2 =

4p (x h), so h = 1 and k = 5. Because 4p = 24 andp = 6, the graph opens to the right. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = (1, 5)focus: (h + p , k) = (7, 5)directrix: x = h p or 5

y = k or 5


2(y + 12) = (x 6)2

2(y + 12) = (x 6)2


vertically. The equation is in the form 4p (y k) =

(x h)2, so h = 6 and k = 12. Because 4p = 2 and

p = , the graph opens up. Use the values of h, k ,

andp to determine the characteristics of the parabola.

vertex: (h, k) = (6, 12)focus: (h + p , k) = (6, 11.5)directrix: y = k p or 12.5

x = h or 6


4(y +2) = (x + 8)2


7-1 Parabolas

4(y +2) = (x + 8)2


vertically. The equation is in the form 4p (y k) =

(x h)2, so h = 8 and k = 2. Because 4p = 4

andp = 1, the graph opens down. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 8, 2)focus: (h + p , k) = ( 8, 3)directrix: y = k p or 1

x = h or 8


10(x + 11) = (y + 3)2

10(x + 11) = (y + 3)2




andp = 2.5, the graph opens to the right. Use the values of h, k , and p to determine the characteristics of the parabola.

vertex: (h, k) = ( 11, 3)focus: (h + p , k) = ( 8.5, 3)directrix: x = h p or 13.5

y = k or 3


SKATEBOARDING A group of high school students designing a half pipe have decided that the ramps, or transitions, could be obtained by splitting a parabola in half. A parabolic cross

section of the ramps can be modeled by x2 = 8(y

2), where the values of x and y are measured in feet. Where is the focus of the parabola in relation to the ground if the ground represents the directrix?

Because the x term is squared and p is positive, the parabola opens up and the focus is located at (h, k + p ). The equation is provided in standard form, and h = 0 and k = 2. Because 4p = 8, p is 2. So, the location of the focus is (0, 2 + 2) or (0, 4). Therefore, the focus is 4 feet above the ground.

COMMUNICATION


7-1 Parabolas

COMMUNICATION

satellite television dish has a parabolic shape that focuses the satellite signals onto a receiver located at the focus of the parabola. The parabolic cross

section can be modeled by (x 6)2 = 12(y 10),

where the values of x and y are measured in inches. Where is the receiver located in relation to this particular cross section?

The receiver is located at the focus of the satellite dish. Because the x term is squared and p is positive, the parabola opens up and the focus is located at (h, k + p ). The equation is provided in standard form, and h = 6 and y = 10. Because 4p =12,p is 3. So, the location of the focus is (6, 10 + 3)or (6, 13). The location of the focus for the satellite dish is (6, 13). Therefore, the receiver is 3 inches above the vertex of the satellite dish.

BOATING

BOATING

water, it creates a wake in the shape of a

the stern of the boat. A swimmer on a wakeboard, attached by a piece of rope, is being pulled by the boat. When he is directly behind the boat, he is positioned at the focus of the parabola. The parabola formed by the wake can be modeled using

y2

180x + 10y + 565 = 0, where x and y are measured in feet.

a. Write the equation in standard form.

b. How long is the length of rope attaching the swimmer to the stern of the boat?

a.

b.The equation in standard form has y as the squared term, which means that the parabola openshorizontally. Because 4p = 180, p = 45 and the graph opens to the right. The equation is in the form

(y k)2 = 4p (x h), so h = 3 and k = 5. Since

the stern is located at the vertex of the parabola formed, it is at the point (h, k) or (3,

The swimmer is at the focus, located at (h + p , k),which is (3 + 45, 5) or (48, 5). The distance the swimmer is from the stern of the boat represents the length of rope needed to attach the swimmer to

Using the distance formula, the distance between these two points is

Therefore, the length of rope attaching the swimmer to the stern of the boat is 45 feet.

BASEBALL During Philadelphia Phillies baseball


7-1 Parabolas

BASEBALL During Philadelphia Phillies baseball games, the team s mascot, The Phanatic, launches hot dogs into the stands. The launching device propels the hot dogs into the air at an initial velocity of 64 feet per second. A hot dog s distance yabove ground after x seconds can be illustrated by

y = 16x2 + 64x + 6.

a. Write the equation in standard form.

b. What is the maximum height of a propelled hot dog?

a.

b.The equation in standard form has x as the squared term, which means that the parabola opens

vertically. Because and the

graph opens down. The equation is in the form 4p (y

k)2 = (x h)

2, so h = 2 and k = 70. The

maximum height a hot dog can reach is located at the vertex of the parabola formed. It is at the point (h, k) or (2, 70). Therefore, the maximum height a hot dog can reach is 70 feet.

Write each equation in standard form. Identify

the vertex, focus, axis of symmetry, and

Write each equation in standard form. Identify

the vertex, focus, axis of symmetry, and

directrix. Then graph the parabola.

x2

17 = 8y + 39

Because the x term is squared and p = 2, the graph opens up. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = (0, 7)focus: (h, k + p ) = (0, 5)directrix: y = k p or 9

x = h or 0

Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The curve should be symmetric about the axis of symmetry.

y2 + 33 = 8x 23


7-1 Parabolas

y2 + 33 = 8x 23

Because the y term is squared and p = 2, the graph opens to the left. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = ( 7, 0)focus: (h + p , k) = ( 9, 0)directrix: x = h p or 5

y = k or 0


3x2 + 72 = 72y

3x2 + 72 = 72y

Because the x term is squared and p = 6, the graph opens down. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = (0, 1)focus: (h, k + p ) = (0, 7)directrix: y = k p or 5axis of symmetry: x = h or 0


12y + 10 = x2

4x + 14


7-1 Parabolas

12y + 10 = x2

4x + 14


vertex: (h, k) = (2, 0)

focus: (h, k + p ) = (2, 3)directrix: y = k p or 3axis of symmetry: x = h or 2


60x 80 = 3y2 + 100

60x 80 = 3y2 + 100

Because the y term is squared and p = 5, the graph opens to the right. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = (3, 0)focus: (h + p , k) = (8, 0)

directrix: x = h p or 2axis of symmetry: y = k or 0


33 = x2

12y 6x


7-1 Parabolas

33 = x2

12y 6x

Because the x term is squared and p = 3, the graph opens up. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = (3, 2)focus: (h, k + p ) = (3, 5)

directrix: y = k p or 1axis of symmetry: x = h or 3


72 = 2y2

16y 20x

72 = 2y2

16y 20x

Because the y term is squared and p = 2.5, the graph opens to the right. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = (2, 4)focus: (h + p , k) = (4.5, 4)

directrix: x = h p or 0.5axis of symmetry: y = k or 4


y2

+ 21 = 20x 6y 68


7-1 Parabolas

y2

+ 21 = 20x 6y 68

Because the y term is squared and p = 5, the graph opens to the left. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = ( 4, 3)

focus: (h + p , k) = ( 9, 3)directrix: x = h p or 1

axis of symmetry: y = k or 3


x2

18y + 12x = 126

x2

18y + 12x = 126

Because the x term is squared and p = 4.5, the graph opens up. Use the standard form to determine the characteristics of the parabola.

vertex: (h, k) = ( 6, 9)

focus: (h, k + p ) = ( 6, 4.5)

directrix: y = k p or 13.5axis of symmetry: x = h or 6


34 = 2x2 + 20x + 8y


7-1 Parabolas

34 = 2x2 + 20x + 8y



axis of symmetry: x = h or 5


LIGHTING Stadium lights at an athletic field need

LIGHTING Stadium lights at an athletic field needto reflect light at maximum intensity. The bulb should be placed at the focal point of the parabolic globe surrounding it. If the shape of the globe is

given by x2 = 36y, where x and y are in inches,

how far from the vertex of the globe should the bulb be placed for maximum light?

The bulb is placed at the focus of the parabola. Because the x term is squared and p is positive, the parabola opens up and the focus is located at (h, k + p ). The equation is provided in standard form, and h = 0 and k = 0. Because 4p = 36, p is 9.So, the location of the focus is (0, 0 + 9) or (0, 9). The vertex of the globe is located at (0, 0). The

distance from the focus to the vertex is 9 0 or 9. Therefore, the bulb should be placed 9 inches from the vertex of the globe.

Write an equation for and graph a parabola

with the given focus F and vertex V.


7-1 Parabolas


with the given focus F and vertex V.

F( 9, 7), V( 9, 4)

Because the focus and vertex share the same xcoordinate, the graph is vertical. The focus is (h, k

+p ), so the value of p is 7 ( 4) or 3. Because p is negative, the graph opens down.

Write the equation for the parabola in standard form using the values of h,p , and k .

The standard form of the equation is (x + 9)2 = 12

(y + 4).

Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (y k) = (x h)2

4( 3)[y ( 4)] = [x ( 9)]2

12(y + 4) = (x + 9)2

F(2, 1), V( 4, 1)

F(2, 1), V( 4, 1)

Because the focus and vertex share the same ycoordinate, the graph is horizontal. The focus is (h+p , k), so the value of p is 2 ( 4) or 6. Because p is positive, the graph opens to the right.


The standard form of the equation is (y + 1)2 = 24

(x + 4).Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (x h) = (y k)2

4(6)[x ( 4)] = [y ( 1)]2

24(x + 4) = (y + 1)2

F( 3, 2), V(1, 2)


7-1 Parabolas

F( 3, 2), V(1, 2)

Because the focus and vertex share the same ycoordinate, the graph is horizontal. The focus is (h

+p , k), so the value of p is 3 1 or 4. Because p is negative, the graph opens to the left.



(x 1).Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (x h) = (y k)2

4( 4)(x 1) = [y ( 2)]2

16(x 1) = (y + 2)2

F( 3, 4), V( 3, 2)

F( 3, 4), V( 3, 2)


+p ), so the value of p is 4 2 or 2. Because p is positive, the graph opens up.


The standard form of the equation is (x + 3)2 = 8(y

2). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (y k) = (x h)2

4(2)(y 2) = [x ( 3)]2

8(y 2) = (x + 3)2

F( 2, 4), V( 2, 5)


7-1 Parabolas

F( 2, 4), V( 2, 5)


+p ), so the value of p is 4 ( 5) or 1. Because pis positive, the graph opens up.


The standard form of the equation is (x + 2)2 = 4(y

+ 5). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (y k) = (x h)2

4(1)[y ( 5)] = [x ( 2)]2

(y + 5) = (x + 2)2

F( 1, 4), V(7, 4)

F( 1, 4), V(7, 4)


+p , k), so the value of p is 1 7 or 8. Because p is negative, the graph opens to the left.


The standard form of the equation is (y 4)2 = 32

(x 7). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (x h) = (y k)2

4( 8)(x 7) = (y 4)2

32(x 7) = (y 4)2

F(14, 8), V(7, 8)


7-1 Parabolas

F(14, 8), V(7, 8)

Because the focus and vertex share the same ycoordinate, the graph is horizontal. The focus is (h+p , k), so the value of p is 14 7 or 7. Because pis positive, the graph opens to the right.



(x 7). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (x h) = (y k)2

4(7)(x 7) = [y ( 8)]2

28(x 7) = (y + 8)2

F(1, 3), V(1, 6)

F(1, 3), V(1, 6)


+p ), so the value of p is 3 6 or 3. Because p is negative, the graph opens down.


The standard form of the equation is (x 1)2 = 12

(y 6). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (y k) = (x h)2

4( 3)(y 6) = (x 1)2

12(y 6) = (x 1)2

F( 4, 9), V( 2, 9)


7-1 Parabolas

F( 4, 9), V( 2, 9)


+p , k), so the value of p is 4 ( 2) or 2.Because p is negative, the graph opens to the left.


The standard form of the equation is (y 9)2 = 8

(x + 2). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (x h) = (y k)2

4( 2)[x ( 2)] = (y 9)2

8(x + 2) = (y 9)2

F(8, 3), V(8, 7)

F(8, 3), V(8, 7)


+p ), so the value of p is 3 ( 7) or 4. Because pis positive, the graph opens up.


The standard form of the equation is (x 8)2 = 16

(y + 7). Graph the vertex and focus. Then make a table of values to graph the parabola.

4p (y k) = (x h)2

4(4)[y ( 7)] = (x 8)2

16(y + 7) = (x 8)2

Write an equation for and graph each parabola

with focus F and the given characteristics.


7-1 Parabolas

Write an equation for and graph each parabola

with focus F and the given characteristics.

F(3, 3); opens up; contains (23, 18)

Because the parabola opens up, the vertex is (3, 3

p ). Use the standard form of the equation of a vertical parabola and the point (23, 18) to find the equation.

Because the parabola opens up, the value of p must

be positive. Therefore, p = 5. The vertex is (3, 2),

and the standard form of the equation is (x 3)2 =

20(y + 2). Use a table of values to graph the parabola.

4p (y k) = (x h)2

4p [18 (3 p )] = (23 3)2

4p (15 + p ) = 400

4p2 + 60p = 400

p2

+ 15p 100 = 0

(p + 20)(p 5) = 0

p = 20 or 5

F(1, 2); opens down; contains (7, 2)

F(1, 2); opens down; contains (7, 2)

Because the parabola opens down, the vertex is (1,

2 p ). Use the standard form of the equation of a vertical parabola and the point (7, 2) to find the equation.

Because the parabola opens down, the value of p

must be negative. Therefore, p = 3. The vertex is (1, 5), and the standard form of the equation is (x

1)2 = 12(y 5). Use a table of values to graph the

parabola.

4p (y k) = (x h)2

4p [2 (2 p )] = (7 1)2

4p (p ) = 36

p2 = 9

p = 3

F(11, 4); opens right; contains (20, 16)


7-1 Parabolas

F(11, 4); opens right; contains (20, 16)

Because the parabola opens to the right, the vertex

is (11 p , 4). Use the standard form of the equation of a horizontal parabola and the point (20, 16) to find the equation.

Because the parabola opens to the right, the value ofp must be positive. Therefore, p = 3. The vertex is (8, 4), and the standard form of the equation is 12

(x 8) = (y 4)2. Use a table of values to graph

the parabola.

4p (x h) = (y k)2

4p [20 (11 p )] = (16 4)2

4p (9 + p ) = 144

4p2 + 36p = 144

p2

+ 9p 36 = 0

(p + 12)(p 3) = 0

p = 12 or 3

F( 4, 0); opens down; contains (4, 15)

F( 4, 0); opens down; contains (4, 15)

Because the parabola opens down, the vertex is

( 4, p ). Use the standard form of the equation of

a vertical parabola and the point (4, 15) to find theequation.

Because the parabola opens down, the value of p

must be negative. Therefore, p = 1. The vertex is ( 4, 1), and the standard form of the equation is (x

+ 4)2 = 4(y 1). Use a table of values to graph

the parabola.

4p (y k) = (x h)2

4p [ 15 ( p )] = [4 ( 4)]2

4p ( 15 + p ) = 64

4p2

60p = 64

p2

15p 16 = 0

(p 16)(p + 1) = 0

p = 16 or 1

F(1, 3); opens left; contains ( 14, 11)


7-1 Parabolas

F(1, 3); opens left; contains ( 14, 11)

Because the parabola opens to the left, the vertex

is (1 p , 3). Use the standard form of the equation of a horizontal parabola and the point ( 14, 11) to find the equation.

Because the parabola opens to the left, the value of

p must be negative. Therefore, p = 1. The vertex

is (2, 3), and the standard form of the equation is 4

(x 2) = (y 3)2. Use a table of values to graph

the parabola.

4p (x h) = (y k)2

4p [ 14 (1 p )] = (11 3)2

4p ( 15 + p ) = 64

4p2

60p = 64

p2

15p 16 = 0

(p 16)(p + 1) = 0

p = 1 or 16

F( 5, 9); opens right; contains (10, 1)

F( 5, 9); opens right; contains (10, 1)

Because the parabola opens to the right, the vertex

is ( 5 p , 9). Use the standard form of the

equation of a horizontal parabola and the point (10,1) to find the equation.

Because the parabola opens to the right, the value ofp must be positive. Therefore, p = 1. The vertex

is ( 6, 9), and the standard form of the equation is

4(x + 6) = (y + 9)2. Use a table of values to graph

the parabola.

4p (x h) = (y k)2

4p [10 ( 5 p )] = [ 1 ( 9)]2

4p (15 + p ) = 64

4p2 + 60p = 64

p2

+ 15p 16 = 0

(p + 16)(p 1) = 0

p = 1 or 16

F( 7, 6); opens left; contains ( 4, 10)


7-1 Parabolas

F( 7, 6); opens left; contains ( 4, 10)

Because the parabola opens to the left, the vertex

is ( 7 p , 6). Use the standard form of the equation of a horizontal parabola and the point ( 4,10) to find the equation.

Because the parabola opens to the left, the value of

p must be negative. Therefore, p = 4. The vertex

is ( 3, 6), and the standard form of the equation is

16(x + 3) = (y 6)2. Use a table of values to

graph the parabola.

4p (x h) = (y k)2

4p [ 4 ( 7 p )] = (10 6)2

4p (3 + p ) = 16

4p2 + 12p = 16

p2

+ 3p 4 = 0

(p + 4)(p 1) = 0

p = 4 or 1

F( 5, 2); opens up; contains ( 13, 2)

F( 5, 2); opens up; contains ( 13, 2)

Because the parabola opens up, the vertex is ( 5,

2 p ). Use the standard form of the equation of a

horizontal parabola and the point ( 13, 2) to find the equation.

Because the parabola opens up, the value of p must

be positive. Therefore, p = 4. The vertex is ( 5,6), and the standard form of the equation is (x +

5)2 = 16(y + 6). Use a table of values to graph the

parabola.

4p (y k) = (x h)2

4p [ 2 ( 2 p )] = [ 13 ( 5)]2

4p (p ) = 64

p2 = 16

p = 4

ARCHITECTURE air


7-1 Parabolas

ARCHITECTURE airplaza has a parabolic arch above two columns. Thelight in the center is located at the focus of the

a.

b. Graph the equation.

a. Assume the ground represents the x axis and the left column the y axis. Since the center of the entrance is at 29 feet, the vertex for the parabolic arch is located at the point (29, 28) and the focus is located at the point (29, 14.9). Because the graph isvertical, the focus is (h, k + p ), so the value of p is

14.9 (28) or 13.1.


The standard form of the equation is 52.4(y 28)

= (x 29)2.

Graph the vertex and focus. Then make a table of values to graph the parabola.

b. Use a table of values to graph the parabola.

4p (y k) = (x h)2

4( 13.1)(y 28) = (x 29)2

52.4(y 28) = (x 29)2

Write an equation for the line tangent to each

parabola at each given point.

Write an equation for the line tangent to each

parabola at each given point.

The graph opens vertically. Determine the vertex

and focus.

standard form. Because . The

vertex is ( 7, 3) and the focus is ( 7, 2.875). We need to determine d, the distance between the focus and the point of tangency, C.

This is one leg of the isosceles triangle.

Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is negative, the parabola opens down and A will be above the focus.

Points A and C both lie on the line tangent to the parabola. Find an equation of this line.

y2

= (x 4), (24, 2)


7-1 Parabolas

y2

= (x 4), (24, 2)

The graph opens horizontally. Determine the vertex

and focus. y2 = (x 4) is written in standard

form. Because 4p = ,p = . The vertex is (4,

0) and the focus is (4.05, 0). We need to determine d, the distance between the focus and the point of tangency, C.


Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is positive, the parabola opens to the right and A will be to the left of the focus.A = (4.05 20.05, 0) or ( 16, 0)


d =

=

= 20.05

m =

y y1 = m(x x1)

y 2 = x 24)

y 2 =

y = x +

(x + 6)2 = 3(y 2), (0, 14)

(x + 6)2 = 3(y 2), (0, 14)


and focus. (x + 6)2 = 3(y 2) is written in standard

form. Because 4p = 3, p = . The vertex is ( 6, 2)

and the focus is ( 6, 2.75). We need to determine d, the distance between the focus and the point of tangency, C.


Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is positive, the parabola opens up and A will be below the focus.

A = ( 6, 2.75 12.75) or ( 6, 10)


d =

=

= 12.75

m =

y y1 = m(x x1)

y 14 = 4(x 0)

y 14 = 4x

y = 4x + 14

(x 3)2 = y + 4, ( 1, 12)


7-1 Parabolas

(x 3)2 = y + 4, ( 1, 12)


and focus. (x 3)2 = y + 4 is written in standard

form. Because 4p = 1, p = . The vertex is (3, 4)

and the focus is (3, 3.75). We need to determine d, the distance between the focus and the point of tangency, C.


Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is positive, the parabola opens up and A will be below the focus.

A = (3, 3.75 16.25) or (3, 20)


d =

=

= 16.25

m = 8

y y1 = m(x x1)

y 12 = 8[x ( 1)]

y 12 = 8x 8

y = 8x + 4

0.25(x 6)2 = y 9, (10, 5)

0.25(x 6)2 = y 9, (10, 5)

The graph opens vertically. Determine the vertex and focus.

Because 4p = 4,p = 1. The vertex is (6, 9) and the focus is (6, 8). We need to determine d, the distance between the focus and the point of tangency, C.


Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is negative, the parabola opens down and A will be above the focus.A = (6, 8 + 5) or (6, 13)


0.25(x 6)2 = y 9

(x 6)2 = 4(y 9)

d =

=

= 5

m = 2

y y1 = m(x x1)

y 5 = 2(x 10)

y 5 = 2x + 20

y = 2x + 25

4x = (y + 5)2, (0, 5)


7-1 Parabolas

4x = (y + 5)2, (0, 5)

The graph opens horizontally. Determine the vertex

and focus. 4x = (y + 5)2 is written in standard

form. Because 4p = 4,p = 1. The vertex is (0,

5) and the focus is ( 1, 5). We need to determine d, the distance between the focus and the point of tangency, C.


Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is negative, the parabola opens to the left and A will be to the right of the focus.

A = ( 1 + 1, 5) or (0, 5)

Points A and C are the same point. Attempting to find the slope of a line using identical points results in an undefined slope.

Since the parabola opens horizontally and the slope of the tangent line is undefined, the line tangent to the graph at the vertex will be a vertical line. Since the x coordinate of the vertex is 0, the equation forthe line tangent to the parabola through this point is x = 0.

d =

=

= 1

m =

Determine the orientation of each parabola.

directrix y = 4, p = 2

Since the directrix is y = 4, the graph opens vertically. Since p is negative, the graph will open down.

y2

= 8(x 6)

y2

= 8(x 6)

Because the y term is squared, the parabola opens

horizontally. Since 4p = 8,p is 2. Since p is negative, the graph will open towards the left.

vertex ( 5, 1), focus ( 5, 3)


+p ), so the value of p is 3 1or 2. Because p is positive, the graph opens up.

focus (7, 10), directrix x = 1

Since the directrix is x = 1, the graph opens horizontally. Since the directrix is to the left of the focus, the graph opens to the right.

Write an equation for each parabola.

The focus and vertex share the same x coordinate.

The focus is (h, k + p ), so the value of p is 4 5 or1.


4p (y k) = (x h)2

4( 1)(y 5) = (x 3)2

4(y 5) = (x 3)2


7-1 Parabolas

The vertex is (0, 4) and the directrix is x = 2.

Since the directrix is x = h p or 2 = 0 p ,p is 2.Write the equation for the parabola in standard form using the values of h,p , and k .

4p (x h) = (y k)2

4(2)(x 0) = [y ( 4)]2

8x = (y + 4)2

The focus is ( 9, 1) and the directrix is x = 1. The

directrix is x = h p or h p = 1 and the x

coordinate of the focus is h + p = 9.

Solve the system of equations.

Substitute h = 5 into the equation for the directrix

to find that p is 4.Write the equation for the parabola in standard form using the values of h,p , and k .

h p

h +p

= 1

= 9

2h = 10

h = 5

4p (x h) = (y k)2

4( 4)[x ( 5)] = (y 1)2

16(x + 5) = (y 1)2

The focus is (2, 2) and the directrix is y = k p =

12. The y coordinate of the focus is k + p = 2.Solve the system of equations.

Substitute k = 7 into the equation for the directrix to find that p = 5.Write the equation for the parabola in standard form using the values of h,p , and k .

k p

k +p

= 12

= 2

2k = 14

k = 7

4p (y k) = (x h)2

4(5)[y ( 7)] = (x 2)2

20(y + 7) = (x 2)2

BRIDGES

below is in the shape of a parabola. The two main support towers are 208 meters apart and 80 meterstall. The distance from the top of the parabola to the water below is 60 meters.

a. Write an equation that models the shape of the arch. Let the railroad track represent the x axis.

b. Two vertical supports attached to the arch are equidistant from the center of the parabola as shown in the diagram. Find their lengths if they are 86.4 meters apart.

a. If the railroad track represents the x axis and we assume that the vertex lies on the y axis, then

the vertex is at the point (0, 20). The arch meets each support tower 104 meters to the left and to the right of the vertex and 80 feet below the railroad or the x axis. Thus, two points on the

parabola are at ( 104, 80) and (104, 80). Find peSolutions Manual - Powered by Cognero

7-1 Parabolas

railroad or the x axis. Thus, two points on the

parabola are at ( 104, 80) and (104, 80). Find pusing the vertex and a point on the parabola.


b. If the supports are 86.4 meters apart and are equidistant from the center of the parabola, then they are located 43.2 meters from the center of theparabola. Therefore, the supports meet the parabola when x = 43.2. Find y when x = 43.2.

Therefore, the supports meets the parabola at the

points (43.2, 30.35) and ( 43.2, 30.35) and the supports are 30.35 meters in length.

4p (y k) = (x h)2

4p [ 80 ( 20)] = (104 0)2

240p = 10816

p = 45.066

4p (y k) = (x h)2

4( 45.066)[y ( 20)] = (x 0)2

180.27(y + 20) = x2

180.27(y + 20) = x2

180.27(y + 20) = 43.22

180.27(y + 20) = 1866.24

y + 20 = 10.35

y = 30.35


with each set of characteristics.


with each set of characteristics.

vertex (1, 8), contains (11, 13), opens vertically

If the graph opens vertically, then the standard

form is 4p (y k) = (x h)2. Find p using the

vertex and the point on the parabola.


4p (y k) = (x h)2

4p (13 8) = (11 1)2

20p = 100p = 5

4p (y k) = (x h)2

4(5)(y 8) = (x 1)2

20(y 8) = (x 1)2

vertex ( 6, 4), contains ( 10, 8), opens horizontally


7-1 Parabolas

vertex ( 6, 4), contains ( 10, 8), opens horizontally

If the graph opens horizontally, then the standard

form is 4p (x h) = (y k)2. Find p using the

vertex and the point on the parabola.


4p (x h) = (y k)2

4p [ 10 ( 6)] = (8 4)2

16p = 16

p = 1

4p (x h) = (y k)2

4( 1)[x ( 6)] = (y 4)2

4(x + 6) = (y 4)2

= (y 4)

opens vertically, passes through points ( 12, 14),

(0, 2), and (6, 5)

If the graph opens vertically, then the standard

form is 4p (y k) = (x h)2. Find an equation for

the parabola for each set of points.Equation 1:

Equation 2:

Equation 3:

4p (y k) = (x h)2

4p ( 14 k) = ( 12 h)2

56p 4kp = 144 + 24h + h2

4p (y k) = (x h)2

4p ( 2 k) = (0 h)2

8p 4kp = h2

4p (y k) = (x h)2

4p ( 5 k) = (6 h)2

1, 1),

Parabolic reflectors with microphones

Determine the point of tangency for each

In a searchlight, the bulb is

Use the standard form of the equation of

of a parabola is the line segment

A solar furnace in France s

Write a possible equation for a parabola with

such that the line given is tangent to

In this

Abigail and Jaden are

parabolic sector

Which point on a parabola is

Without graphing, determine in

of a function s

Write a letter outlining and

Find the maximum and minimum values of the

) and for what values of

Talia is surveying a rectangular lot

Find the value of each expression using the

Locate the vertical asymptotes, and sketch

What is the solution set for 3(4 + 1)2

is a positive number, then

Which is the parent function of the graph shown

Solve this system of equations. Subtract Equation 2 from Equation 1.

Subtract Equation 2 from Equation 3.

Solve the system consisting of the two new equations.

Substitute h = 0 into one of the equations and solve forp .

Substitutep = 3 and h = 0 into one of the original equations and solve for k .


4p (y k) = (x h)

4p ( 5 k) = (6 h)2

20p 4kp = 36 12h + h2

56p 4kp = 144 + 24h + h2

8p 4kp = h2

48p = 144 + 24h

20p 4kp = 36 12h + h2

8p 4kp = h2

12p = 36 12h

48p12p

= 144 + 24h

= 36 12h

48p48p

= 144 + 24h

= 144 48h0 = 24h0 = h

12p = 36 12h

12p = 36 12(0)12p = 36

p = 3

8p 4kp = h2

8( 3) 4( 3)k = 02

24 + 12k = 0

12k = 24

k = 2

4p (y k) = (x h)2

4( 3)[y ( 2)] = (x 0)2

12(y + 2) = x2

1, 1),









In this


parabolic sector



of a function s











7-1 Parabolas

opens horizontally, passes through points ( 1, 1),(5, 3), and (15, 7)

If the graph opens horizontally, then the standard

form is 4p (x h) = (y k)2. Find an equation for

the parabola for each set of points.Equation 1:

Equation 2:

Equation 3:

Solve this system of equations. Subtract Equation 2 from Equation 1.

Subtract Equation 2 from Equation 3.

Solve the system consisting of the two new equations.

4p (x h) = (y k)2

4p ( 1 h) = ( 1 k)2

4p 4hp = 1 + 2k + k2

4p (x h) = (y k)2

4p (5 h) = (3 k)2

20p 4hp = 9 6k + k2

4p (x h) = (y k)2

4p (15 h) = (7 k)2

60p 4hp = 49 14k + k2

4p 4hp = 1 + 2k + k2

20p 4hp = 9 6k + k2

24p = 8 + 8k

60p 4hp = 49 14k + k2

20p 4hp = 9 6k + k2

40p = 40 8k

24p40p

= 8 + 8k

= 40 8k16p = 32

p = 2









In this


parabolic sector



of a function s










Substitutep = 2 into one of the equations and solve for k .

Substitutep = 2 and k = 5 into one of the original equations and solve for h.


16p = 32p = 2

40p = 40 8k

40(2) = 40 8k

80 = 40 8k

40 = 8k5 = k

4p 4hp = 1 + 2k + k2

4(2) 4(2)h = 1 + 2( 5) + ( 5)2

8 8h = 16

8h = 24

h = 3

4p (x h) = (y k)2

4(2)[x ( 3)] = [y ( 5)]2

8(x + 3) = (y + 5)2

Parabolic reflectors with microphones located at the focus are used to capture sounds

reflector and are concentrated toward the microphone.

a. How far from the reflector should a microphone be placed if the reflector has a width of 3 feet and a depth of 1.25 feet?








In this


parabolic sector



of a function s











7-1 Parabolas

a. How far from the reflector should a microphone be placed if the reflector has a width of 3 feet and a depth of 1.25 feet?

b. Write an equation to model a different parabolic reflector that is 4 feet wide and 2 feet deep, if the vertex of the reflector is located at (3, 5) and the parabola opens to the left.

c. Graph the equation. Specify the domain and range.

a. The vertex and focus of a parabola in standard form are (h, k) and (h + p , k) respectively. Find pto determine the distance the microphone should befrom the reflector. Sketch a coordinate plane on topof the reflector with the vertex located at the origin.The width represents the distance the reflector spans vertically. Since the reflector has a width of 3 feet, 1.5 feet of the reflector will exist on each side of the x axis. The depth represents the distance the reflector spans horizontally along the x axis. Using a depth of 1.25 feet and a width of

1.5 feet, the points (1.25, 1.5) and (1.25, 1.5) mustlie on the reflector. Since the parabola opens to the

right, the standard form is 4p (x h) = (y k)2.

Substitute the values of x, y , h, and k to solve for p .

The focus of the parabola is at (h + p , k) or (0.45, 0). Thus, the microphone should be placed at the point (0.45, 0). This is 0.45 feet from the vertex of the reflector.

b. Since the parabola opens to the left, the standard

form is 4p (x h) = (y k)2. The vertex is (3, 5).

The depth represents the distance the reflector spans horizontally and the width represents the distance the reflector spans vertically. Points that are 2 feet to the left of the vertex and 2 feet above and below it will lie on the reflector. Thus, the points (1, 7) and (1, 3) lie on the parabola.Substitute the values of x, y , h, and k to solve for p .

Substitute values for h, k , and p to write an equation to model the reflector.

4p (x h) = (y k)2

4p (1.25 0) = (1.5 0)2

5p = 2.25p = 0.45

4p (x h) = (y k)2

4p (1 3) = (7 5)2

8p = 4p = 0.5








In this


parabolic sector



of a function s











c. Use a table of values to graph the parabola. Since the reflector has a depth of 2 feet, graph the

x

4p (x h) = (y k)2

4( 0.5)(x3)

= (y 5)2

2(x 3) = (y 5)2


equation and line.

(x + 2)2 = 2y

y = 4x

The point of tangency occurs at the solution for the system of equations.Use substitution to solve the system.

The solution to the system occurs when x = 2. To find the y coordinate, substitute x = 2 in one of the original equations.

x + 2)2 = 2y

(x + 2)2 = 2(4x)

x2

+ 4x + 4 = 8x

x2

4x + 4 = 0

(x 2)2 = 0

y = 4x

y = 4(2)

y = 8

The point of tangency occurs at (2, 8).

(y 8)2 = 12x







In this


parabolic sector



of a function s











7-1 Parabolas

(y 8)2 = 12x

y = x + 11


The solution to the system occurs when x = 3. To find the y coordinate, substitute x = 3 into one of the original equations.

y = x + 11

y = 3 + 11

y = 14


(y 8)2 = 12x

[(x + 11) 8]2

= 12x

(x + 3)2 = 12x

x2

+ 6x + 9 = 12x

x2

6x + 9 = 0

(x 3)2 = 0

(y + 3)2 = x + 4







In this


parabolic sector



of a function s










(y + 3)2 = x + 4


The solution to the system occurs when x = 0. To find the y coordinate, substitute x = 0 into one of the original equations.

y = 1


(y + 3)2 = x + 4

= x + 4

= x + 4

= x + 4

= 0

(x + 5)2 = 4(y + 1)







In this


parabolic sector



of a function s











7-1 Parabolas

(x + 5)2 = 4(y + 1)

y = 2x + 13


The solution to the system occurs when x = 9. To

find the y coordinate, substitute x = 9 into one of the original equations.

y = 2x + 13

y = 2( 9) + 13

y = 5

The point of tangency occurs at ( 9, 5).

x + 5)2 = 4(y + 1)

(x + 5)2 = 4[(2x + 13) + 1]

(x + 5)2 = 4(2x + 14)

x2

+ 10x + 25 = 8x 56

x2

+ 18x + 81 = 0

(x + 9)2 = 0

ILLUMINATION In a searchlight, the bulb is placed at the focus of a parabolic mirror 1.5 feet from the vertex. This causes the light rays from thebulb to bounce off the mirror as parallel rays, thus providing a concentrated beam of light.

a.Write an equation for the parabola if the focal diameter of the bulb is 2 feet, as shown in the diagram.b.Suppose the focal diameter is increased to 3 feet.If the depth of both searchlights is 3.5 feet, how much greater is the width of the opening of the larger light? Round to the nearest hundredth.

a. Sketch a coordinate plane on top of the searchlight with the vertex at the origin. The focus is 1.5 feet from the vertex located at the point (1.5, 0). Since the focal diameter of the bulb is 2 feet, 1 foot will lie above and below the x axis. Therefore,

the points (1.5, 1) and (1.5, 1) lie on the parabola.






In this


parabolic sector



of a function s










foot will lie above and below the x axis. Therefore,

the points (1.5, 1) and (1.5, 1) lie on the parabola. Substitute the values of x, y , h, and k to solve for p .


b. Write an equation in standard form for a parabola with a focal diameter of 3 feet. Sketch a coordinate plane on top of the new searchlight with the vertex at the origin. The focus is still 1.5 feet from the vertex located at the point (1.5, 0). Since the focal diameter of the bulb is 3 feet, 1.5 feet will lie above and below the x axis. Therefore, the points (1.5, 1.5) and (1.5, 1.5) lie on the parabola. Substitute the values of x, y , h, and k to solve for p .


The depth of both searchlights is 3.5 feet. The depth represents the distance the reflector spans horizontally and the width represents the distance the reflector spans vertically. To find the width of the first searchlight, substitute x = 3.5 into the equation found in part a and solve for y .

4p (x h) = (y k)2

4p (1.5 0) = (1 0)2

6p = 1

p =

4p (x h) = (y k)2

4 (x 0) = (y 0)2

x = y2

4p (x h) = (y k)2

4p (1.5 0) = (1.5 0)2

6p = 2.25p = 0.375

4p (x h) = (y k)2

4(0.375)(x 0) = (y 0)2

1.5x = y2

x = y2

(3.5) = y2

= y2






In this


parabolic sector



of a function s











7-1 Parabolas

At a depth of 3.5 feet, the reflector will span 1.53 feet above and below the x-axis for a total width of2(1.53) or 3.06 feet. Repeat the same process to find the width of the second reflector.

At a depth of 3.5 feet, the reflector will span 2.29 feet above and below the x-axis for a total width of2(2.29) or 4.58 feet. The difference between the two widths is 4.58 3.06 or 1.52 feet.

= y2

= y

or 1.53 = y

1.5x = y2

1.5(3.5) = y2

5.25 = y2

or 2.29 = y






In this


parabolic sector



of a function s










Use the standard form of the equation ofa parabola to prove the general form of the equation of a parabola.

The general form of equations for conic sections is

Ax2 + Bxy + Cy

2 + Dx + Ey + F = 0, where A, B,

C, D, E, and F are constants and A, B, and Ccannot all be zero. Expand each standard form of equations for a parabola.

Once values are given for p , k , and h, these variables become constants. Thus, they may be represented by constants D, E, and F. Repeat the process using the other standard form of equation for a parabola.

y k)2

= 4p (x h)

y2 2ky + k2

= 4px 4ph

y2 4px 2ky + k2

+ 4ph= 0

y2 + ( 4p )x + ( 2k)y

+ (k2 + 4ph )

= 0

y2 + Dx + Ey + F = 0

(x h)2

= 4p (y k)

x2

2hx + h2

= 4py 4pk

x2

2hx 4py + h2 + 4pk = 0

x2

+ ( 2h)x + ( 4p )y + (h2 +

4pk )= 0

x2

+ Dx + Ey + F = 0

The latus rectum of a parabola is the line segmentthat passes through the focus, is perpendicular to the axis of the parabola, and has endpoints on the

parabola. The length of the latus rectum is |4p |units, where p is the distance from the vertex to thefocus.

a.Write an equation for the parabola with vertex at ( 3, 2), axis y = 2, and latus rectum 8 units long.b.Prove that the endpoints of the latus rectum and point of intersection of the axis and directrix are thevertices of a right isosceles triangle.




In this


parabolic sector



of a function s











7-1 Parabolas

b.Prove that the endpoints of the latus rectum and point of intersection of the axis and directrix are thevertices of a right isosceles triangle.

a. Since the parabola has an axis of y = 2, the parabola will open horizontally and will have the

standard form of (y k)2 = 4p (x h). The latus

rectum is 8 units long. This can be used to solve for

p.|4p | = 8, so p = Substitute values for h, k , and p to write an equationfor the parabola.

b.

To prove that the endpoints of the latus rectum Xand W and the point of intersection of the axis and directrix D are the vertices of a right isosceles

triangle XDW, we need to show that XDW is a

right angle and that . Since ,

4p (x h) = (y k)2

4( 2)[x ( 3)] = (y 2)2

8(x + 3) = (y 2)2

, and XFD WFD, XFD

FWD by SAS. Thus, . To prove that

XDW is a right angle, we can first prove that

XDF and FDW are both

XFD is a right angle, XFD is an

isosceles right triangle. Therefore, XDF is .

This also means that FDW is . Resultantly,

XDW is a right angle. Thus, XDW is an isosceles right triangle.

SOLAR ENERGY A solar furnace in France s



In this


parabolic sector



of a function s










SOLAR ENERGY A solar furnace in France sEastern Pyrenees uses a parabolic mirror that is illuminated by sunlight reflected off a field of heliostats, which are devices that track and

performed in the focal zone part of a tower. If the parabolic mirror is 6.25 meters deep, how many meters in front of the parabola is the focal zone?

The focal zone is located at the focus of the parabolic mirror. The vertex and focus of a parabola in standard form are (h, k) and (h + p , k)respectively. Find p to determine the distance the focal zone is from the mirror. Sketch a coordinate plane on top of the diagram with the vertex located at the origin. The width represents the distance the reflector spans vertically and is shown to be 40 meters. Since the reflector has a width of 40 meters, 20 meters of the reflector will exist on eachside of the x axis. The depth represents the distance the reflector spans horizontally along the x axis. Using a depth of 6.25 meters and a width

of 20 meters, the points (6.25, 20) and (6.25, 20)must lie on the mirror. Since the parabola opens to

the right, the standard form is 4p (x h) = (y k)2.

Substitute the values of x, y , h, and k to solve for p .

The focus of the parabola is at (h + p , k) or (16, 0).Thus, the focal zone is at the point (16, 0). This is 16 meters from the vertex of the mirror.

4p (x h) = (y k)2

4p (6.25 0) = (20 0)2

25p = 400p = 16


focus F such that the line given is tangent to

the parabola.

In this


parabolic sector



of a function s











7-1 Parabolas

the parabola.

Find point A, the point of intersection of the axis and the line tangent to the parabola.

The equation for the tangent line is y = 2x 2.Since point A lies on the y axis, the x coordinate is0. Substitute x = 0 to find y .

y = 2x 2

y = 2(0) 2

y = 2

A is at (0, 2). The segment formed by FArepresents one of the legs of the isosceles triangle formed by F, A, and the point of tangency. Find the length of this leg.

d = 5

The lengths of the legs of the isosceles triangle are 5 units. This is the distance from F to the point of tangency (x, y). Use the distance formula and the tangent line to solve for the point of tangency.

The solutions to this equation are x = 0 and 4. SinceA has an x coordinate of 0, the point of tangency must have an x coordinate of 4. Substitute x = 4 to solve for y .

y = 2x 2

y = 2(4) 2

y = 6

d

5

25 = x2 + (y 3)

2

25 = x2 + y

26y + 9

25 = x2 + (2x 2)

26(2x 2) + 9

25 = x2 + 4x

28x + 4 12x + 12 + 9

25 = 5x2

20x + 25

0 = 5x2

20x

0 = 5x(x 4)

In this


parabolic sector



of a function s










y = 2(4) 2

y = 6

The point of tangency is located at (4, 6). Since the parabola opens upward, the standard form of the

equation is (x h)2 = 4p (y k) and the focus is

located at (h, k + p ). The focus is at the point (0,

3); thus, h = 0 and k + p = 3 or k = 3 p . Use, h,the standard form of the equation, the point of tangency, and the equation for k to solve for p .

The solutions to this equation are p = 4 and 1. Since the parabola opens upward, p is positive;

thus,p = 1. Use p to find k . k = 3 (1) or 2. The vertex of the parabola is (0, 2). Use the vertex and p to write the standard form of the equation.

x h)2 = 4p (y k)

(4 0)2 = 4p (6 k)

16 = 24p 4pk16 = 24p 4p (3 p )16 = 24p 12p + 4p

2

0 = 4p2 + 12p 16

0 = 4(p + 4)(p 1)

x h)2 = 4p (y k)

(x 0)2 = 4(1)(y 2)

x2 = 4(y 2)


The equation for the tangent line is y = x + 2.

Since point A lies on the x axis, the y coordinate is0. Substitute y = 0 to find x.

y = x + 2

0 = x + 2

In this


parabolic sector



of a function s











7-1 Parabolas

0 = x + 2

2 = x

4 = x

A is at ( 4, 0). The segment formed by FArepresents one of the legs of the isosceles triangle formed by F, A, and the point of tangency. Find the length of this leg.

d =

d =

d = 5


The solutions to this equation are x = 4. Since A

has an x coordinate of 4, the point of tangency must have an x coordinate of 4. Substitute x = 4 to solve for y .

y = x + 2

y = (4) + 2

y = 4

The point of tangency is located at (4, 4). Since the parabola opens to the right, the standard form of

the equation is 4p (x h) = (y k)2 and the focus is

d=

5 =

25 = (x 1)2 + y

2

25 = x2

2x + 1 + y2

25 = x2

2x + 1 +

25 = x2

2x + 1 + x2 + 2x + 4

25 = x2 + 5

20 = x2

16 = x2

4 = x

In this


parabolic sector



of a function s










parabola opens to the right, the standard form of

the equation is 4p (x h) = (y k)2 and the focus is

located at (h + p , k). The focus is at the point (1,

0); thus, k = 0 and h + p = 1 or h = 1 p . Use, k ,the standard form of the equation, the point of tangency, and the equation for h to solve for p .

The solutions to this equation are p = 4 and 1. Since the parabola opens to the right, p is positive;

thus,p = 1. Use p to find h. h = 1 (1) or 0. The vertex of the parabola is (0, 0). Use the vertex and p to write the standard form of the equation.

4p (x h) = (y k)2

4p (4 h) = (4 0)2

16p 4ph = 16

16p 4p (1 p ) = 16

16p 4p + 4p2 = 16

4p2 + 12p 16 = 0

4(p + 4)(p 1) = 0

4p (x h) = (y k)2

4(1)(x 0) = (y 0)2

4x = y2


The equation for the tangent line is y = x + 6.

Since point A lies on the x axis, the y coordinate is0. Substitute y = 0 to find x.

y = x + 6

0 = x + 6

6 = x

12 = x

A is at ( 12, 0). The segment formed by FArepresents one of the legs of the isosceles triangle

In this


parabolic sector



of a function s











7-1 Parabolas

A is at ( 12, 0). The segment formed by FArepresents one of the legs of the isosceles triangle formed by F, A, and the point of tangency. Find the length of this leg.

d =

d =

d = 15


The solutions to this equation are x = 12. Since A

has an x coordinate of 12, the point of tangency must have an x coordinate of 12. Substitute x = 12 to solve for y .

y = x + 6

y = (12) + 6

y = 12

The point of tangency is located at (12, 12). Since the parabola opens to the right, the standard form

of the equation is 4p (x h) = (y k)2 and the

focus is located at (h + p , k). The focus is at the

point (3, 0); thus, k = 0 and h + p = 3 or h = 3 p .Use, k , the standard form of the equation, the point of tangency, and the equation for h to solve for p .

d =

15 =

225 = (x 3)2 + y

2

225 = x2

6x + 9 + y2

225 = x2

6x + 9 +

225 = x2

6x + 9 + x2 + 6x + 36

225 = x2 + 45

180 = x2

144 = x2

12 = x

4p (x h) = (y k)2

4p (12 h) = (12 0)2

In this


parabolic sector



of a function s










The solutions to this equation are p = 12 and 3. Since the parabola opens to the right, p is positive;

thus,p = 3. Use p to find h. h = 3 (3) or 0. The vertex of the parabola is (0, 0). Use the vertex and p to write the standard form of the equation.

4p (x h) = (y k)

4p (12 h) = (12 0)2

48p 4ph = 144

48p 4p (3 p ) = 144

48p 12p + 4p2 = 144

4p2 + 36p 144 = 0

4(p + 12)(p 3) = 0

4p (x h) = (y k)2

4(3)(x 0) = (y 0)2

12x = y2

Find point A, the point of intersection of the axis and the line tangent to the parabola.The equation for the tangent line is y = 4x + 3. Since point A lies on the y axis, the x coordinate is0. Substitute x = 0 to find y .

y = 4x + 3

y = 4(0) + 3

y = 3

A is at (0, 3). The segment formed by FArepresents one of the legs of the isosceles triangle formed by F, A, and the point of tangency. Find the length of this leg.

The lengths of the legs of the isosceles triangle are

F to

d=

d=

d =

In this


parabolic sector



of a function s











7-1 Parabolas

The lengths of the legs of the isosceles triangle are

F to

the point of tangency (x, y). Use the distance formula and the tangent line to solve for the point oftangency.

The solutions to this equation are x = 0 and 1.Since A has an x coordinate of 0, the point of

tangency must have an x coordinate of 1.

Substitute x = 1 to solve for y .

y = 4x + 3

y = 4( 1) + 3

y = 1

The point of tangency is located at ( 1, 1). Since the parabola opens down, the standard form of the

equation is (x h)2 = 4p (y k) and the focus is

located at (h, k + p ). The focus is at the point

; thus, h = 0 and k + p = k = p .

Use, h, the standard form of the equation, the point of tangency, and the equation for k to solve for p .

d=

=

= x2 +

= x2 + y

2

= x2 + (4x + 3)

2

= x2 + 16x

2 + 24x + 9 7x

= 17x2 + 17x +

0 = 17x2 + 17x

0 = 17x(x + 1)

x h)2 = 4p (y k)

( 1 0)2 = 4p ( 1 k)

1 = 4p 4pk

1 =

1 = 4p 4p2

In this


parabolic sector



of a function s










The solutions to this equation are p = 2 and .

Since the parabola opens down, p is negative; thus,

p = . Use p to find k .

vertex of the parabola is (0, 1). Use the vertex and p to write the standard form of the equation.

1 = 4p 4p2

0 = 4p2

p 1

0 = 8p2

15p 2

0 = (8p + 1)(p 2)

x h)2 = 4p (y k)

(x 0)2 =

x2 =

In this problem, you will investigate how the shape of a parabola changes as the position of the focus changes.a. GEOMETRIC

vertex and the focus of each parabola.

i. y2

= 4(x 2)

ii. y2

= 8(x 2)

iii. y2

= 16(x 2)b. GRAPHICAL a

using a different color for each. Label each focus.c. VERBAL aparabola's shape and the distance between its vertex and focus.d. ANALYTICAL

parabola that has the same vertex as (x + 1)2 = 20

(y + 7), but is narrower.e. ANALYTICAL

graphs of x2 = 2(y + 1), x

2 = 12(y + 1), and x

2 =

5(y + 1). Check your conjecture by graphing the parabolas.

a.

i. Written in standard form, (y k)2 = 4p (x h),p

is the distance between the vertex and the focus.

y2 = 4(x 2) can be written as y

2 = 4(1)(x 2),


parabolic sector



of a function s











7-1 Parabolas

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7-1 parabolas - montville.net · the parabola. (x ±3) 2 = 12(y ±7) 62/87,21 the equation is in...

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