lecture 7 (hyperbola and focus directrix equation)
TRANSCRIPT
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Conic Sections Hyperbola and Focus-Directrix Equation
Institute of Mathematics, University of the Philippines Diliman
Mathematics 54Elementary Analysis 2
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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
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
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
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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Equation of a Hyperbola
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Equation of a Hyperbola
Foci : F1 = (c,0) and F2 = (c, 0)
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Equation of a Hyperbola
Foci : F1 = (c,0) and F2 = (c, 0)
Vertices : V1 = (a, 0) and V2 = (a, 0)
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Equation of a Hyperbola
Foci : F1 = (c,0) and F2 = (c, 0)
Vertices : V1 = (a, 0) and V2 = (a, 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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Equation of a Hyperbola
Foci : F1 = (c,0) and F2 = (c, 0)
Vertices : V1 = (a, 0) and V2 = (a, 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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Equation of a Hyperbola
Simplifying, we get
x2
a2 y2
c2a2 = 1
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Equation of a Hyperbola
Simplifying, we get
x2
a2 y2
c2a2 = 1
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Equation of a Hyperbola
Simplifying, we get
x2
a2 y2
c2a2 = 1
Since a< c
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Equation of a Hyperbola
Simplifying, we get
x2
a2 y2
c2a2 = 1
Since a< c, we can let b2= c
2a
2
.
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Equation of a Hyperbola
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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Equation of a Hyperbola
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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Equation of a Hyperbola
If we solve for yin terms ofxwe get
y=x2b2a2
b2
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Equation of a Hyperbola
If we solve for yin terms ofxwe get
y=x2b2a2
b2
Consider the lines y= baxand y=bax.
Hyperbola and Focus-Directrix Equation 5/ 1
f b l
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Equation of a Hyperbola
If we solve for yin terms ofxwe get
y=x2b2a2
b2
Consider the lines y= baxand y=bax. Verify the following:
Hyperbola and Focus-Directrix Equation 5/ 1
E i f H b l
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Equation of a Hyperbola
If we solve for yin terms ofxwe get
y=x2b2a2
b2
Consider the lines y= baxand y=bax. Verify the following:
limx+
x2b2
a2 b2
ba
x
= 1
limx
x2b2
a2b2
ba
x
= 1
Hyperbola and Focus-Directrix Equation 5/ 1
E ti f H b l
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Equation of a Hyperbola
If we solve for yin terms ofxwe get
y=x2b2a2
b2
Consider the lines y= baxand y=bax. Verify the following:
limx+
x2b2
a2 b2
ba
x
= 1
limx
x2b2
a2b2
ba
x
= 1
Hyperbola and Focus-Directrix Equation 5/ 1
E ti f H b l
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Equation of a Hyperbola
If we solve for yin terms ofxwe get
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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Equation of a Hyperbola
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Equation of a Hyperbola
In general a hyperbola centered at the origin has form:
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Equation of a Hyperbola
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Equation of a Hyperbola
In general a hyperbola centered at the origin has form:
x2
a2 y
2
b2= 1
Foci
(c, 0)= (
a2+b2, 0)
Vertices
(a, 0)
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Equation of a Hyperbola
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Equation of a Hyperbola
In general a hyperbola centered at the origin has form:
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)
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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.
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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.
Solution:The hyperbola can be expressed as
x2
36 y
2
4= 1
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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.
Solution:The hyperbola can be expressed as
x2
36 y
2
4= 1
So a
=6 and b
=2.
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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.
Solution:The hyperbola can be expressed as
x2
36 y
2
4= 1
So a
=6 and b
=2.
Then c=a2+b2 =36+4= 210.
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p
Example 1.
Find the foci of the hyperbola given by the equation
x2
9y2 = 4.
Then sketch its graph.
Solution:The hyperbola can be expressed as
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
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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.
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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.
Solution:
b= 3 and c= 4.
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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.
Solution:
b= 3 and c= 4.So a=c2b2
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Examples
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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.
Solution:
b= 3 and c= 4.So a=c2b2 =169=7.
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Examples
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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.
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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Equation of a Hyperbola
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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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Equation of a Hyperbola
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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). Then, the hyperbola has the form
Hyperbola and Focus-Directrix Equation 9/ 1
Equation of a Hyperbola
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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). Then, the hyperbola has the form
(xh)2a2
(yk)2
b2= 1
Foci Vertices
(h
a2+b2, k) (ha, k)
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Equation of a Hyperbola
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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). Then, the hyperbola has the form
(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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Equation of a Hyperbola
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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).
Hyperbola and Focus-Directrix Equation 10/ 1
Equation of a Hyperbola
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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).
Solution:
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Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h
=32 and k
=1.
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Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h
=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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Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h
=
32 and k
=1.
The value ofais the distance from the center to a vertex, which in this case is 32 .
The value ofcis the distance of a focus to the center which is inthis case is 52 .
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Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h=32 and k= 1.
The value ofais the distance from the center to a vertex, which in this case is 32 .
The value ofcis the distance of a focus to the center which is inthis case is 52 .
Now, b2 = c2+a2 = 164 = 4.
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Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h=32 and k= 1.
The value ofais the distance from the center to a vertex, which in this case is 32 .
The value ofcis the distance of a focus to the center which is inthis case is 52 .
Now, b2 = c2+a2 = 164 = 4. Hence, the desired equation is
Hyperbola and Focus-Directrix Equation 10/ 1
Equation of a Hyperbola
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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).
Solution:
The desired hyperbola is centered at ( 32 , 1) so h=32 and k= 1.
The value ofais the distance from the center to a vertex, which in this case is 32 .
The value ofcis the distance of a focus to the center which is inthis case is 52 .
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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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
Hence, we have
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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
Hence, we have
0< e< 1 ellipse
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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
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
e= distance between focidistance between vertices
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.
Hyperbola and Focus-Directrix Equation 12/ 1
Consider the ellipse(xh)2
a2+ (yk)
2
b2= 1.
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a b
Hyperbola and Focus-Directrix Equation 12/ 1
Consider the ellipse(xh)2
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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Consider the ellipse(xh)2
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.
Hyperbola and Focus-Directrix Equation 12/ 1
Consider the ellipse(xh)2
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. 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
Hyperbola and Focus-Directrix Equation 13/ 1
2 2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2 b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
Hyperbola and Focus-Directrix Equation 13/ 1
2 2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2 b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
Hyperbola and Focus-Directrix Equation 13/ 1
PF
( )2 2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2 b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
Hyperbola and Focus-Directrix Equation 13/ 1
PF
( )2 2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2 b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
Hyperbola and Focus-Directrix Equation 13/ 1
PF
( + )2+ 2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2 b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2
Hyperbola and Focus-Directrix Equation 13/ 1
PF
(x+c)2+y2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2= e2x2+2aex+a2
Hyperbola and Focus-Directrix Equation 13/ 1
PF1 =
(x+c)2+y2
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PF1 =
(x+c)2+y2
= (x+ae)2+b2b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2= e2x2+2aex+a2= e
x2+2 a
ex+
ae
2
Hyperbola and Focus-Directrix Equation 13/ 1
PF1 =
(x+c)2+y2
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PF1 =
(x+c) +y
= (x+ae)2+b2b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2= e2x2+2aex+a2= e
x2+2 a
ex+
ae
2= e
x
a
e
2
Hyperbola and Focus-Directrix Equation 13/ 1
PF1 =
(x+c)2+y2
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PF1 =
(x+c) +y
= (x+ae)2+b2b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2= e2x2+2aex+a2= e
x2+2 a
ex+
ae
2= e
x
a
e
2= e
x ae
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PF1 =
(x+c)2+y2
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PF1
(x+c) +y
= (x+ae)2+b2b2
a2 x2
=
a2x2+2a3ex+a4e2+a2b2b2x2
a2
= c2x2+2a3ex+a4e2+a2b2
a2
=
a2e2x2+2a3ex+a4e2+a2b2
a2
=
e2x2+2aex+a2e2+b2
= e2x2+2aex+ c2+b2= e2x2+2aex+a2= e
x2+2 a
ex+
ae
2= e
x
a
e
2= e
x ae
= ePD1
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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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Consider the hyperbola(xh)2
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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Consider the hyperbola(xh)2
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.
Let D1 be the directrix nearest the focus F1 and let D2 be the directrix nearest the
focus F2.
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Consider the hyperbola(xh)2
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.
Let D1 be the directrix nearest the focus F1 and let D2 be the directrix nearest the
focus F2. This pairing of the foci and the directrices will be refered to as the
focus-directrix pairing.
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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
Hyperbola and Focus-Directrix Equation 17/ 1
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Hence, for any focus-directrix pair in an ellipse, hyperbola or parabola we have the
following equation
PF= ePD
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Hence, for any focus-directrix pair in an ellipse, hyperbola or parabola we have the
following equation
PF= ePD
this equation is refered to as the focus-directrix equation.
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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.
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