Lecture 7 (Hyperbola and Focus Directrix Equation)

7/31/2019 Lecture 7 (Hyperbola and Focus Directrix Equation)

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Conic Sections Hyperbola and Focus-Directrix Equation

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54Elementary Analysis 2

Hyperbola and Focus-Directrix Equation 1/ 1


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Hyperbola

The set of all points in a plane whose distances from the two fixed points

(focuses/foci) have a constant difference.



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Hyperbola





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Hyperbola





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Hyperbola





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Equation of a Hyperbola



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Foci : F1 = (c,0) and F2 = (c, 0)



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Foci : F1 = (c,0) and F2 = (c, 0)

Vertices : V1 = (a, 0) and V2 = (a, 0)



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Foci : F1 = (c,0) and F2 = (c, 0)


From the above figure and from the definition of the hyperbola we have(x+ c)2+y2

(xc)2+y2 = constant



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Foci : F1 = (c,0) and F2 = (c, 0)


From the above figure and from the definition of the hyperbola we have(x+ c)2+y2

(xc)2+y2 = 2a



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Simplifying, we get

x2

a2 y2

c2a2 = 1



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Simplifying, we get

x2

a2 y2

c2a2 = 1



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Simplifying, we get

x2

a2 y2

c2a2 = 1

Since a< c



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Simplifying, we get

x2

a2 y2

c2a2 = 1

Since a< c, we can let b2= c

2a

2

.



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Simplifying, we get

x2

a2 y2

c2a2 = 1

Since a< c, we can let b2= c

2a

2

. Thus

x2

a2 y

2

b2= 1



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Simplifying, we get

x2

a2 y2

c2a2 = 1

Since a< c, we can let b2

= c2

a2

. Thus

x2

a2 y

2

b2= 1

Note that the rectangle formed above is called the auxillary rectangle of the

hyperbola.Hyperbola and Focus-Directrix Equation 4/ 1


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If we solve for yin terms ofxwe get

y=x2b2a2

b2



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y=x2b2a2

b2

Consider the lines y= baxand y=bax.


f b l


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y=x2b2a2

b2

Consider the lines y= baxand y=bax. Verify the following:


E i f H b l


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y=x2b2a2

b2


limx+

x2b2

a2 b2

ba

x

= 1

limx

x2b2

a2b2

ba

x

= 1


E ti f H b l


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y=x2b2a2

b2


limx+

x2b2

a2 b2

ba

x

= 1

limx

x2b2

a2b2

ba

x

= 1


E ti f H b l


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y=x2b2a2 b2Consider the lines y= baxand y=

bax. Verify the following:

limx+

x2b2

a2 b2

ba

x

= 1

limx

x2b2

a2b2

ba

x

= 1

Hence, the lines y

=baxserve as asymptotes of the hyperbola.




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In general a hyperbola centered at the origin has form:




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x2

a2 y

2

b2= 1

Foci

(c, 0)= (

a2+b2, 0)

Vertices

(a, 0)




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x2

a2 y

2

b2= 1

Foci

(c, 0)= (

a2+b2, 0)

Vertices

(a, 0)

y2

b2 x2

a2= 1

Foci

(0,c)= (0,

a2+b2)

Vertices

(0,b)


Examples


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Examples

Example 1.

Find the foci of the hyperbola given by the equation

x2

9y2 = 4.

Then sketch its graph.


Examples


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Examples

Example 1.


x2

9y2 = 4.


Solution:The hyperbola can be expressed as

x2

36 y

2

4= 1


Examples


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Examples

Example 1.


x2

9y2 = 4.



x2

36 y

2

4= 1

So a

=6 and b

=2.


Examples


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Examples

Example 1.


x2

9y2 = 4.



x2

36 y

2

4= 1

So a

=6 and b

=2.

Then c=a2+b2 =36+4= 210.


Examples


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p

Example 1.


x2

9y2 = 4.



x2

36 y

2

4= 1

So a

=6 and b

=2.

Then c=a2+b2 =36+4= 210.Hence, F: (2

10,0)

insert graph here


Examples


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p

Example 2.

Find the equation of the vertical hyperbola centered at the origin with a vertex at

(0,3) and a focus at (0,4). Then sketch its graph.


Examples


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p

Example 2.



Solution:

b= 3 and c= 4.


Examples


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p

Example 2.



Solution:

b= 3 and c= 4.So a=c2b2


Examples


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Example 2.



Solution:

b= 3 and c= 4.So a=c2b2 =169=7.


Examples


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Example 2.



Solution:

b= 3 and c= 4.So a=c2b2 =169=7.Thus, the hyperbola has the equation

y2

9 x

2

7= 1

insert graph here




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In general, we may consider a hyperbola with either a vertical or a horizontal

transverse axis centered at the point (h, k).




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transverse axis centered at the point (h, k). Then, the hyperbola has the form




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(xh)2a2

(yk)2

b2= 1

Foci Vertices

(h

a2+b2, k) (ha, k)




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(xh)2a2

(yk)2

b2= 1

Foci Vertices

(h

a2+b2, k) (ha, k)

(yk)2b2

(xh)2a2

= 1

a< bFoci

(h, kc)= (h, k

a2+b2)

Vertices

(h, k

b)




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Example

Find the equation of the hyperbola with foci (1,1) and (4,1) and vertices (0,1) and(3,1).




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Example


Solution:




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Example


Solution:

The desired hyperbola is centered at ( 32 , 1) so h

=32 and k

=1.




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Example


Solution:


=32 and k

=1.

The value ofais the distance from the center to a vertex, which in this case is 32 .




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Example


Solution:


=

32 and k

=1.


The value ofcis the distance of a focus to the center which is inthis case is 52 .




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Example


Solution:

The desired hyperbola is centered at ( 32 , 1) so h=32 and k= 1.



Now, b2 = c2+a2 = 164 = 4.




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Example


Solution:




Now, b2 = c2+a2 = 164 = 4. Hence, the desired equation is




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Example


Solution:




Now, b2 = c2+a2 = 164 = 4. Hence, the desired equation is

(x 32 )232

2 (y1)2(2)2 = 1



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For the parabola, we define the eccentricityeto be e= 1.



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For the parabola, we define the eccentricityeto be e= 1.For the ellipse and hyperbola, we define eccentricity to be

e= distance between focidistance between vertices



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Hence, we have



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Hence, we have

0< e< 1 ellipse



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Hence, we have

0< e< 1 ellipse

e= 1 parabola



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For the parabola, we define the eccentricityeto be e= 1.

For the ellipse and hyperbola, we define eccentricity to be


Hence, we have

0< e< 1 ellipse

e= 1 parabola

e> 1 hyperbola



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Consider the ellipse(xh)2

a2+ (yk)

2

b2= 1.



a2+ (yk)

2

b2= 1.


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a b



a2+ (yk)

2

b2= 1.


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a b

The lines perpendicular to the major axis of this ellipse at distances ae from thecenter are the directrices of this ellipse.



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a2+ (yk)

2

b2= 1.


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The lines perpendicular to the major axis of this ellipse at distances ae from the

center are the directrices of this ellipse.

Let D1 be the directrix nearest the focus F1 and let D2 be the directrix nearest the

focus F2.



a2+ (yk)

2

b2= 1.


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The lines perpendicular to the major axis of this ellipse at distances ae from the

center are the directrices of this ellipse.


focus F2. This pairing of the foci and the directrices will be refered to as the

focus-directrix pairing.



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PF1 =

(x+c)2+y2



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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2


2 2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2

=

a2x2+2a3ex+a4e2+a2b2b2x2

a2


2 2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2

=


a2

= c2x2+2a3ex+a4e2+a2b2

a2


PF

( )2 2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2

=


a2


a2

=

a2e2x2+2a3ex+a4e2+a2b2

a2


PF

( )2 2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2


PF

( + )2+ 2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2 b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2

= e2x2+2aex+ c2+b2


PF

(x+c)2+y2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2

= e2x2+2aex+ c2+b2= e2x2+2aex+a2


PF1 =

(x+c)2+y2


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PF1 =

(x+c)2+y2

= (x+ae)2+b2b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2

= e2x2+2aex+ c2+b2= e2x2+2aex+a2= e

x2+2 a

ex+

ae

2


PF1 =

(x+c)2+y2


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PF1 =

(x+c) +y

= (x+ae)2+b2b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2


x2+2 a

ex+

ae

2= e

x

a

e

2


PF1 =

(x+c)2+y2


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PF1 =

(x+c) +y

= (x+ae)2+b2b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2


x2+2 a

ex+

ae

2= e

x

a

e

2= e

x ae


PF1 =

(x+c)2+y2


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PF1

(x+c) +y

= (x+ae)2+b2b2

a2 x2

=


a2


a2

=


a2

=

e2x2+2aex+a2e2+b2


x2+2 a

ex+

ae

2= e

x

a

e

2= e

x ae

= ePD1


Thus, ifPis a point on the ellipse


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Thus, ifPis a point on the ellipse then

PF1 = ePD1


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Thus, ifPis a point on the ellipse then

PF1 = ePD1 PF2 = ePD2


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Consider the hyperbola(xh)2

a2 (yk)

2

b2= 1.



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a2 (yk)

2

b2= 1.

The lines perpendicular to the transverse axis of this hyperbola at distances aefrom the center are the directrices of this ellipse.



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a2 (yk)

2

b2= 1.



focus F2.



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a2 (yk)

2

b2= 1.



focus F2. This pairing of the foci and the directrices will be refered to as the

focus-directrix pairing.


Similarly, ifPis a point on the ellipse


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Similarly, ifPis a point on the ellipse then

PF1 = ePD1 PF2 = ePD2


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Hence, for any focus-directrix pair in an ellipse, hyperbola or parabola we have the

following equation



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following equation

PF= ePD



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following equation

PF= ePD

this equation is refered to as the focus-directrix equation.


Exercises


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1 Identify the following conic sections and determine their eccentricity:

a. 2x23y2+4x+6y1= 0b. 2x2+3y2+16x18y53= 0c. 9x+y2+4y5= 0d. 4x2x=y2+1e. 7yy2x= 0

2 Determine the equation of the parabola whose focus and vertex are the vertexand focus, respectively of the parabola with equation x+4y2y= 0.

3 Let M= 3. Determine the equations of the hyperbola and ellipse having(2,1) as the foci and M as the length of the conjugate axis and minor axis,respectively.


Documents

Lecture 7 (Hyperbola and Focus Directrix Equation)