Conic Sections

Parabola

Conic Sections - Parabola

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

Conic Sections - Parabola

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”.

Conic Sections - Parabola

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

Focus

Directrix

Conic Sections - Parabola

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

Conic Sections - Parabola

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

Conic Sections - 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

Conic Sections - Parabola

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

Conic Sections - Parabola

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

Conic Sections - Parabola

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

Conic Sections - Parabola

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)

Directrix y = -p

p

p( 0, 0)

y = ax2

Conic Sections - Parabola

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

Conic Sections - Parabola

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

Conic Sections - Parabola

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 ( ?, ?)

Conic Sections - Parabola

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)

Conic Sections - Parabola

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 - Parabola 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.

d1 = d2

You finish the rest.

Conic Sections - Parabola 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.

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 px ax p ax p

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

ap

pa

− + − = − + +

+ − = ++ − + = + +

==

=

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

1 1/(4 )4

p or p aa

= =

Conic Sections - Parabola Using 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 1 Make 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.

Parabola – Example 1

( )21 2 38xy + −=

Make a table of values.

x y

-2

-1

0

1

2

3

4

-3 72 8−12 2−71 8−

-1 1811 2

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

Parabola – Example 1

Find the focus and directrix.

Parabola – Example 1

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.

Parabola – Example 1

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

2 Units

Latus Rectum

Parabola

Conic Sections - Parabola The latus rectum is the line segment passing through the focus, perpendicular to the axis of symmetry with endpoints on the parabola.

y = ax2

Focus

Vertex (0, 0)

Latus Rectum

Conic Sections - Parabola In the previous set, we learned that the distance from the vertex to the focus is 1/(4a). Therefore, the focus is

at

y = ax2

Focus

Vertex (0, 0)

Latus Rectum

10,4a

⎛ ⎞⎜ ⎟⎝ ⎠

Conic Sections - Parabola Using the axis of symmetry and the y-value of the focus,

the endpoints of the latus rectum must be

y = ax2

Vertex (0, 0)

Latus Rectum

1,4

xa

⎛ ⎞⎜ ⎟⎝ ⎠

1,4

xa

⎛ ⎞−⎜ ⎟⎝ ⎠

1,4

xa

⎛ ⎞±⎜ ⎟⎝ ⎠

Conic Sections - Parabola Since the equation of the parabola is y = ax2, substitute

for y and solve for x. 14a

2

2

22

1414

12

y ax

axa

xa

xa

=

=

=

=±

Conic Sections - Parabola Replacing x, the endpoints of the latus rectum are

y = ax2

Vertex (0, 0)

Latus Rectum

1 1,2 4a a⎛ ⎞⎜ ⎟⎝ ⎠

1 1,2 4a a

⎛ ⎞−⎜ ⎟⎝ ⎠

1 1,2 4a a

⎛ ⎞−⎜ ⎟⎝ ⎠

and 1 1,2 4a a⎛ ⎞⎜ ⎟⎝ ⎠

Conic Sections - Parabola The length of the latus rectum is

y = ax2

Vertex (0, 0)

Latus Rectum

1 1,2 4a a⎛ ⎞⎜ ⎟⎝ ⎠

1 1,2 4a a

⎛ ⎞−⎜ ⎟⎝ ⎠

1 1 1 1 2 12 2 2 2 2a a a a a a

⎛ ⎞− − = + = =⎜ ⎟⎝ ⎠

Conic Sections - Parabola Given the value of “a” in the quadratic equation

y = a (x – h)2 + k, the length of the latus rectum is

1a

An alternate method to graphing a parabola with the latus rectum is to:

1. Plot the vertex and axis of symmetry 2. Plot the focus and directrix.

3. Use the length of the latus rectum to plot two points on the parabola.

4. Draw the parabola.

Example 2

Graph a parabola using the vertex, focus, axis of symmetry and latus rectum.

Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph.

( )21 1 216xy − +=−

Parabola – Example 2

The vertex is at (1, 2) with the parabola opening down.

( )21 1 216xy − +=−

( )1 414

114 16

p = =−−

=

The focus is 4 units down and the directrix is 4 units up.

The length of the latus rectum is

1 1 16116a

= =−

Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph.

V(1, 2)

Directrix y=6

Focus (1, -2)

Latus Rectum

Axis x=1

Parabola – Example 2 The graph of the parabola ( )

21 1 216xy − +=−

V(1, 2)

Directrix y=6

Focus (1, -2) Latus

Rectum

Axis x=1

x = ay2 Parabola

Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.

Parabola – Graphing x = ay2

Graph x = 2y2 by constructing a table of values.

x y

-3

-2

-1

0

1

2

3

18

8

2

0

2

8

18

Graph x = 2y2 by plotting the points in the table.

Parabola – Graphing x = ay2 Graph the table of values.

Parabola – Graphing x = ay2 One could follow a similar proof to show the distance from the vertex to the focus and directrix to be . 1

4a

Similarly, the length of the latus rectum can be shown to be . 1a

1 1 14 4(2) 8a

= =

1 1 12 2a

= =

Parabola – Graphing x = ay2 Graphing the axis of symmetry, vertex, focus, directrix and latus rectum.

( )1 ,08

18x =

−

Axis y=0

V(0,0)

Directrix

Focus

x = a(y – k)2 + h

Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.

When horizontal and vertical transformations are applied, a vertical shift of k units and a horizontal shift of h units will result in the equation:

x = a(y – k)2 + h

Parabola – x = a(y – k)2 + h We have just seen that a parabola x = ay2 opens to the right when a is positive. When a is negative, the graph will reflect about the y-axis and open to the left.

Note: In both cases of the parabola, the x always goes with h and the y always goes with k.

Example 3

Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.

Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum.

( )21 1 22yx + −=−

What is the vertex? Remember that inside the “function” we always do the opposite. So the graph moves -1 in the y direction and -2 in the x direction. The vertex is (-2, -1)

What is the direction of opening?

The parabola opens to the left since it is x= and “a” is negative.

Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum.

( )21 1 22yx + −=−

What is the distance to the focus and directrix?

The distance is 1 1 1 1

14 2 24( )2a= = =

− −

The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola.

Parabola – Example 3 The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola.

( )21 1 22yx + −=−

Vertex? (-2, -1)

Focus? (-2.5, -1)

Directrix? x = -1.5

Parabola – Example 3

What is the length of the latus rectum?

1 1 2 212a

= = − =−

( )21 1 22yx + −=−

Parabola – Example 3 Construct the latus rectum with a length of 2.

( )21 1 22yx + −=−

Vertex? (-2, -1)

Focus? (-2.5, -1)

Directrix? x = -1.5

Latus Rectum?

2

Construct the parabola.

Parabola – Example 3 The parabola is:

( )21 1 22yx + −=−

Vertex? (-2, -1)

Focus? (-2.5, -1)

Directrix? x = -1.5

Latus Rectum?

2

Building a Table of Rules

Parabola

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

Opens

Vertex

Focus

Axis

Directrix

Latus Rectum

Up Down

(h, k) (h, k)

1,4

h ka

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

+ 1, 4h k

a⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

−

x = h x = h

14

y ka

= − 14y k

a= +

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 + h a > 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

= − 14x h

a= +

1a

1a

(h, k)

(h, k)

1 ,4

h ka

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

+

1 ,4

h ka

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

−

y = k

y = k

14

x ha

= −

14

x ha

= +

Paraboloid Revolution

Parabola

Paraboloid Revolution

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.png GNU Free Documentation License

Paraboloid Revolution

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

Paraboloid Revolution

The 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 Receiver 8 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 + k Since the vertex is (0, 0), h and k are 0. y = ax2

Example 4 – Satellite Receiver 8 ft

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

116a=

Example 4 – Satellite Receiver 8 ft

1 ft

V(0, 0)

(4, 1)

2116y x=The 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 416a= = =

Example 4 – Satellite Receiver 8 ft

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) 112x 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 312a= = =

Sample Problems

Find the length of the latus rectum.

21 ( 3) 112x y += +

1 1 12112a

= =

Sample Problems

b. Sketch the graph given:

21 ( 3) 112x 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 Problems 21 ( 3) 112x y += +

Vertex (1, -3) Opens Right Axis y = -3 Focus (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 yx y

y xy x

− = +

− = +

+ =± −

=− ± −Graph as 2 separate equations in the grapher.

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

3 12( 1)3 12( 1)

y xy x=− + −

=− − −

c. Graph using a grapher.

Sample Problems

2. 2x2 + 8x – 3 + y = 0 a.  Find the vertex, focus, directrix, axis of

symmetry and length of the latus rectum. b. Sketch the graph.

c. Graph using a grapher.

Sample Problems

2. 2x2 + 8x – 3 + y = 0 a.  Find the vertex, focus, directrix, axis of

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

of the latus rectum. 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.

( )18

1 14 4 2a

= =

Sample Problems

2. 2x2 + 8x – 3 + y = 0 or y = -2(x + 2) 2 + 11 a.  Find the vertex, focus, directrix, and length

of the latus rectum.

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 Problems 3. 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 16116

14

a

a

a=

=

=

Sample Problems 3. 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.

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

Sample Problems 3. Write the equation of a parabola with vertex at

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

x = a(y – 3)2 + 2 4

1 16116

1 116 16

14

a

a

a or a

a=

=

−= =

=

Sample Problems 3. Write the equation of a parabola with vertex at

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

x = a(y – 3)2 + 2 1 116 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 of y = a(x – h)2 + k

What is the vertex?

Sample Problems

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

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

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

The vertex is (4, 1).

How can the value of “a” be found? The distance from the focus to the vertex is 1. Therefore

1 144 1

1 14 4

aa

a or a

=

=

−= =

Sample Problems

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

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−= − +