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Conic Sections

Parabola

Conic Sections - Parabola

The intersection of a plane with one nappe of the cone is a parabola.

OBJECTIVES:-

Paraboloid Revolution

They are commonly used today in satellite technology as well as lighting in motor vehicle headlights and flashlights.


The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.


The line is called the directrix and the point is called the focus.

Focus

Directrix


The line perpendicular to the directrix passing through the focus is the axis of symmetry. The vertex is the point of intersection of the axis of symmetry with the parabola.

Focus

Directrix

Axis of Symmetry

Vertex


The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d1 = d2 for any point (x, y) on the parabola.

Focus

Directrix

d1

d2

Finding the Focus and Directrix

Parabola


We know that a parabola has a basic equation y = ax2. The vertex is at (0, 0). The distance from the vertex to the focus and directrix is the same. Let’s call it p.

Focus

Directrix

p

p

y = ax2


Find the point for the focus and the equation of the directrix if the vertex is at (0, 0).

Focus( ?, ?)

Directrix ???

p

p( 0, 0)

y = ax2


The focus is p units up from (0, 0), so the focus is at the point (0, p).

Focus( 0, p)

Directrix ???

p

p( 0, 0)

y = ax2


The directrix is a horizontal line p units below the origin. Find the equation of the directrix.

Focus( 0, p)

Directrix ???

p

p( 0, 0)

y = ax2


The directrix is a horizontal line p units below the origin or a horizontal line through the point (0, -p). The equation is y = -p.

Focus( 0, p)

Directrixy = -p

p

p( 0, 0)

y = ax2


The definition of the parabola indicates the distance d1 from any point (x, y) on the curve to the focus and the distance d2 from the point to the directrix must be equal.

Focus( 0, p)

Directrix y = -p

( 0, 0)

( x, y)

y = ax2

d1

d2


However, the parabola is y = ax2. We can substitute for y in the point (x, y). The point on the curve is (x, ax2).

Focus( 0, p)

Directrix y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2


What is the coordinates of the point on the directrix immediately below the point (x, ax2)?

Focus( 0, p)

Directrix y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2( ?, ?)


The x value is the same as the point (x, ax2) and the y value is on the line y = -p, so the point must be (x, -p).

Focus( 0, p)

Directrix y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2( x, -p)


d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p.

Focus( 0, p)

Directrix y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2( x, -p)

Conic Sections - Parabolad1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p.

d1 = d2

You finish the rest.

Conic Sections - Parabolad1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p.

d1 = d2

2 2 4 2 2 2 4 2 2

2 2

2 2 2 2 2 2( 0) ( ) ( ) ( )2 2 2 2 2( ) ( ) ( )2 2

41 4

14

x ax p x x ax p

x ax p ax px a x ax p p a x ax p p

x ax pap

pa

Conic Sections - ParabolaTherefore, the distance p from the vertex to the focus and the vertex to the directrix is given by the formula

11/(4 )

4p or p a

a

Conic Sections - ParabolaUsing transformations, we can shift the parabola y=ax2 horizontally and vertically. If the parabola is shifted h units right and k units up, the equation would be

2( )y a x h k The vertex is shifted from (0, 0) to (h, k). Recall that when “a” is positive, the graph opens up. When “a” is negative, the graph reflects about the x-axis and opens down.

Example 1

Graph a parabola.

Find the vertex, focus and directrix.

Parabola – Example 1Make a table of values. Graph the function. Find the vertex, focus, and directrix.

21 2 38xy

Parabola – Example 1

21 2 38xy

The vertex is (-2, -3). Since the parabola opens up and the axis of symmetry passes through the vertex, the axis of symmetry is x = -2.


21 2 38xy

Make a table of values.

x y

-2

-1

0

1

2

3

4

-372 812 271 8

-1

1811 2

Plot the points on the graph!Use the line of symmetry to plot the other side of the graph.


Find the focus and directrix.


14

pa

The focus and directrix are “p” units from the vertex

where

21 2 38xy

1 1

2114 28

p

The focus and directrix are 2 units from the vertex.


Focus: (-2, -1) Directrix: y = -5

2 Units

Building a Table of Rules

Parabola

Table of Rules - y = a(x - h)2 + ka > 0 a < 0

Opens

Vertex

Focus

Axis

Directrix

Latus Rectum

Up Down

(h, k) (h, k)

1,4

h ka

1,4

h ka

x = h x = h

14

y ka

14

y ka

1a

1a

(h, k)

(h, k)

1,4

h ka

1,4

h ka

x = h

x = h

14

y ka

14

y ka

Table of Rules - x = a(y - k)2 + ha > 0 a < 0

Opens

Vertex

Focus

Axis

Directrix

Latus Rectum

Right Left

(h, k) (h, k)

1 ,4

h ka

1 ,4

h ka

y = k y = k

14

x ha

14

x ha

1a

1a

(h, k)

(h, k)

1 ,4

h ka

1 ,4

h ka

y = k

y = k

14

x ha

14

x ha


Parabola


A paraboloid revolution results from rotating a parabola around its axis of symmetry as shown at the right.

http://commons.wikimedia.org/wiki/Image:ParaboloidOfRevolution.pngGNU Free Documentation License

Paraboloid RevolutionThe focus becomes an important point. As waves approach a properly positioned parabolic reflector, they reflect back toward the focus. Since the distance traveled by all of the waves is the same, the wave is concentrated at the focus where the receiver is positioned.

Example 4 – Satellite Receiver

A satellite dish has a diameter of 8 feet. The depth of the dish is 1 foot at the center of the dish. Where should the receiver be placed?

8 ft

1 ft

Let the vertex be at (0, 0). What are the coordinates of a point at the diameter of the dish?

V(0, 0)

(?, ?)

Example 4 – Satellite Receiver8 ft

1 ft

With a vertex of (0, 0), the point on the diameter would be (4, 1). Fit a parabolic equation passing through these two points.

V(0, 0)

(4, 1)

y = a(x – h)2 + kSince the vertex is (0, 0), h and k are 0.y = ax2


1 ft

V(0, 0)

(4, 1)

y = ax2

The parabola must pass through the point (4, 1).

1 = a(4)2 Solve for a.

1 = 16a

116

a


1 ft

V(0, 0)

(4, 1)

2116

y xThe model for the parabola is:

The receiver should be placed at the focus. Locate the focus of the parabola.

Distance to the focus is:

1 1 1

4114 4 416

a


1 ft

V(0, 0)

(4, 1)

The receiver should be placed 4 ft. above the vertex.

Sample Problems

Parabola

Sample Problems

1. (y + 3)2 = 12(x -1)

a. Find the vertex, focus, directrix, and length of the latus rectum.

b. Sketch the graph.

c. Graph using a grapher.

Sample Problems

1. (y + 3)2 = 12(x -1)

a. Find the vertex, focus, directrix, axis of symmetry and length of the latus rectum.

Since the y term is squared, solve for x.

2

2

1 ( 3) 1121 ( 3) 112

y x

x y

Sample Problems

Find the direction of opening and vertex.

21 ( 3) 112

x y

The parabola opens to the right with a vertex at (1, -3).

Find the distance from the vertex to the focus.

1 1 1

3114 4 312

a

Sample Problems

Find the length of the latus rectum.

21 ( 3) 112

x y

1 112

112

a

Sample Problems

b. Sketch the graph given:

21 ( 3) 112

x y

• The parabola opens to the right.

• The vertex is (1, -3)

• The distance to the focus and directrix is 3.

• The length of the latus rectum is 12.

Sample Problems21 ( 3) 1

12x y

Vertex (1, -3)Opens RightAxis y = -3Focus (4, -3)Directrix x = -2

Sample Problems

1. (y + 3)2 = 12(x -1)c. Graph using a grapher.

Solve the equation for y.

2

2

12( 1) ( 3)

12( 1) ( 3)

3 12( 1)

3 12( 1)

x y

x y

y x

y x

Graph as 2 separate equations in the grapher.

Sample Problems1. (y + 3)2 = 12(x -1)

3 12( 1)

3 12( 1)

y x

y x


Sample Problems

2. 2x2 + 8x – 3 + y = 0


b. Sketch the graph.


Sample Problems

2. 2x2 + 8x – 3 + y = 0


Solve for y since x is squared.

y = -2x2 - 8x + 3

Complete the square.

y = -2(x2 + 4x ) + 3

y = -2(x2 + 4x + 4 ) + 3 + 8 (-2*4) is -8. To balance the side, we must add 8.

y = -2(x + 2) 2 + 11

Sample Problems

2. 2x2 + 8x – 3 + y = 0


y = -2(x + 2) 2 + 11

The parabola opens down with a vertex at (-2, 11).

Find the direction of opening and the vertex.

Find the distance to the focus and directrix.

1

8

1 14 4 2a

Sample Problems

2. 2x2 + 8x – 3 + y = 0 or y = -2(x + 2) 2 + 11


Graph the table of values and use the axis of symmetry to plot the other side of the parabola.

Since the latus rectum is quite small, make a table of values to graph.

x y

-2 11-1 9 0 3

1 -7

Sample Problems

2. 2x2 + 8x – 3 + y = 0 or y = -2(x + 2) 2 + 11

b. Sketch the graph using the axis of symmetry.

x y

-2 11-1 9 0 3

1 -7

Sample Problems

2. 2x2 + 8x – 3 + y = 0 or y = -2(x + 2) 2 + 11

c. Graph with a grapher.

Solve for y.

y = -2x2 - 8x + 3

Sample Problems

3. Write the equation of a parabola with vertex at (3, 2) and focus at (-1, 2).

Plot the known points.

What can be determined from these points?

Sample Problems

3. Write the equation of a parabola with vertex at (3, 2) and focus at (-1, 2).

The parabola opens the the left and has a model of x = a(y – k)2 + h.

Can you determine any of the values a, h, or k in the model?

The vertex is (3, 2) so h is 3 and k is 2.

x = a(y – 3)2 + 2

Sample Problems3. Write the equation of a parabola with vertex at

(3, 2) and focus at (-1, 2).

How can we find the value of “a”?

x = a(y – 3)2 + 2

The distance from the vertex to the focus is 4.

4

1 16

116

14

a

a

a


(3, 2) and focus at (-1, 2).

How can we find the value of “a”?

x = a(y – 3)2 + 2

The distance from the vertex to the focus is 4.

How can this be used to solve for “a”?


(3, 2) and focus at (-1, 2).

x = a(y – 3)2 + 2

4

1 16

116

1 116 16

14

a

a

a or a

a


(3, 2) and focus at (-1, 2).

x = a(y – 3)2 + 21 116 16a or a

Which is the correct value of “a”?

Since the parabola opens to the left, a must be negative.

21 316x y

Sample Problems

4. Write the equation of a parabola with focus at (4, 0) and directrix y = 2.

Graph the known values.

What can be determined from the graph?

The parabola opens down and has a model ofy = a(x – h)2 + k

What is the vertex?

Sample Problems


The vertex must be on the axis of symmetry, the same distance from the focus and directrix. The vertex must be the midpoint of the focus and the intersection of the axis and directrix.

The vertex is (4, 1)

Sample Problems


The vertex is (4, 1).

How can the value of “a” be found?

The distance from the focus to the vertex is 1. Therefore

1 14

4 1

1 14 4

a

a

a or a

Sample Problems


1 14 4a or a

Since the parabola opens down, a must be negative and the vertex is (4, 1). Write the model.

Which value of a?

21 4 14y x

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