Analytic Geometry Ellipse Hyperbola

8/9/2019 Analytic Geometry Ellipse Hyperbola

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Analytic GeometryAnalytic Geometry


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What are Conics?What are Conics?

Conics are curves formed by the intersection of a

plane and a pair of circular cones. It is also

known as conics sections.

Kinds of Conics:

1. Circle

2. Parabola

3. Ellipse

4. Hyperbola


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Circle Ellipse

Parabola Hyperbola

Conic SectionsConic Sections


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Ellipses

Set of all points in a plane such that the sum of the

distances from two given points in a plane, called thefoci, is constant.

fociSum of the distances:

12 units

vertex vertex

co-vertex

co-vertex

The major axis is the line segment joining the vertices

(through the foci)

The minor axis is the line segment joining the co-vertices

(perpendicular to the major axis)


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Equation of the Ellipse

The equations are derived using the distance

formula. Consider the point P(x,y) the two foci

and the center at the origin.

)0,(and)0,( 21 cFcF

axis-yon theFoci,1)2(

axis-xon theFoci,1)1(

2

2

2

2

2

2

2

2

!

!

b

x

a

yor

b

y

a

x

See the graph for each equation


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)0,( a )0,(a

(1)

(2)),( bo

),0( b

1Fy 2Fy

),0( b

),0( b

)0,(a)0,( a1Fy

2Fy

1V 2V1B

2B 1V

2V

2B


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Remember the following:

10,)(tyeccentrici

axisminortheoflength2

axismajortheofength2

222222

21

21

!

!!

!!

!!

ea

ce

baccab

bBB

la

Where c is the distance from the centre of the ellipse to Foci


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Example: Write an equation of the ellipse whose vertices are

(0, 3) and (0, 3) and whose co-vertices are (2, 0) and (2, 0). Find

the foci of the ellipse.

x2

+ y2

= 1

b2

a2

Since the major axis is

vertical, the equation isthe following:

Use c2

= a2 b

2to find

c. c2

= 32 2

2

c2

= 9 4 = 5

(0, c)

(0, c)

(0, 3)

(0, 3)

(2, 0)(2, 0)

Since a = 3 b = 2

The equation isx

2+ y

2= 1

4 9

The foci are 0, and 0, 5


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Examples

ind the vertices, foci , eccentricity, major and minor

axes of the ellipse and sketch the graph.

8.05

35

62therefore3

102therefore5,135

)0,0(1

925

2222

2

2

2

2

22

!

!

!!

!!

!!!

"!

a

ba

a

ce

bb

aayx

baCyx


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Sketch of the graph

)0,5(2V1)0,5( V

)3,0(1B

)30(2 B

y)0,4( 1F2)0,4( Fy C


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Example: Write the equation in standard form of

9x2

+ 16y2

= 144. Find the foci and vertices of the ellipse.

Get the equation in standard form (make it equal to 1):9x

2+ 16y

2= 144

144 144 144

Use c2 = a2 b2 to find

c.

c2

= 42 3

2

c2

= 16 9 = 7

c =(c, 0)(c,0)

(4,0) (4, 0)

(0, 3)

(0,-3)

That means a = 4 b = 3

Vertices:

Foci:

4, and 4,

Simplify...

x2

+ y2

= 1

16 9

7,0 and 7,0


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Eccentricity Ellipses can be very flat, or nearly circular.

We use the eccentricity of an ellipse as a

measure of how far from an ellipse is from

being circular:


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Illustrations of Eccentricity


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Example

Halleys comet has an elliptical orbit witheccentricity e = 0.967 .

The closest that Halleys comet comes to the

sun is 0.587 AU. One astronomical unit(AU) is the average

distance from Earth to the sun, roughly 93,000,000

mi.

What is the maximum distance of the comet

from the sun, to the nearest 0.1 AU?


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Example (contd)


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ExerciseExercise

A 4 v 8 foot elliptical tabletop is to be cut out of a 4 v 8 foot

rectangular sheet of a teak plywood

Determine:

1. the distance of the foci from the edge of the table top alongthe major axis

2. the total length of the string required to draw the ellipse,

with both ends being fastened at the foci.

4

8


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HyperbolaHyperbola


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Definition and Equation

We can define hyperbolas as follows:

This leads to a simple equation for a

hyperbola; we choose the

x-axis as the line through the foci F and F ; origin to be the centerof the hyperbola, that is,

the midpoint of line segment FF.


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Equation (contd) Thus the foci are located

at ( c, 0) .

By definition ofhyperbola, either

d(P,F) d(P,F) = 2a

or

d(P,F) d(P,F) = 2a .


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Equation (contd) After algebra this becomes

where b denotes

The resulting hyperbola is shown on thenext slide:

22

2 21

b !

2 2 .c a


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Equation (contd)


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Equation (contd)

Some terminology:

The x-intercepts (a, 0) are the vertices;

The line segment is the transverseaxis;

The line segment WW is the conjugateaxis.

The lines y = (b/a)x are asymptotes for the

hyperbola.

To draw the asymptotes, first draw the auxiliary

rectangle shown by the dotted line on the preceding

slide.


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oci on they-axis

Similarly, taking thefoci on they-axis leads

to

Now the

vertices are (0, a) ; asymptotes are

y = (a/b)x .

2 2

2 21

y x

a b !


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Summary Here is a summary of the standard equations

of a hyperbola with center at the origin:


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Remember the following when sketching aRemember the following when sketching a

hyperbolahyperbola ::

1. The given equation must be expressed in the

standard form. Know the center.

2. Plot the vertices. (y-intercepts and x-intercepts)

3. Draw the central box.(Length =2a, witdh =2b)using thevertices.

4. On the central box draw the diagonals and extend these

two beyond the box. These are the asymptotes.

5. Sketch the hyperbola starting on the vertices approaching

the asymptotes.


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Example


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Example (contd)


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Example (contd)


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Sketch the following1. ind the vertices, foci, and asymptotes and sketch the

graph.

47.442

22

4:

)0,0(

4,2:142

1164

).

2222

2

2

2

2

22

!!!

s!s!s!

!!!

!

bac

xxxa

byAsy ptotes

C

bawhereyx

yxa


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Sketch:

)4,0()4,0(

)0,2()0,2(

)0,5.4()0,5.4(

11

21

21

BB

VV

FF

V1V2

B1

B2


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Example


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Example (contd)

The equations of

the asymptotes are

2

2

a y x x

b! s ! s


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Example


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Example (contd)


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Example (contd)


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Analytic Geometry Ellipse Hyperbola