Analytical Geometry
Ellipse
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Ellipse
ļ‚§ An ellipse is the set of all points in a plane such that the
sum of the distances from two points (foci) is a constant.
ļ‚§ When you go from point ā€œFā€ to any point on the
ellipse and hen go on to point "G", you will always
travel the same distance.
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ļ‚§ The two fixed points are called the foci(plural of ā€žfocusā€Ÿ) of
the ellipse .
ļ‚§ The mid point of the line
segment joining the foci is
called the centre of the ellipse.
ļ‚§ The line segment through the
foci of the ellipse is called the
major axis and the line segment
through the centre and perpendicular
to the major axis is called the minor axis.
ļ‚§ The end points of the major axis are called the vertices of
the ellipse
F G
A B
C
D
0
Vertices
Major axis
Minoraxis
Foci
Centre
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ļ‚§ We denote the length of the major axis by 2a, the length of
the minor axis by 2b and the distance between the foci by 2c.
ļ‚§ Thus, the length of the semi major axis is a and semi-minor
axis is b
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Relationship between semi-major axis, semi-minor
axis and the distance of the focus from the centre
of the ellipse
Take a point P at one end of the major axis.
Sum of the distances of the point P to the
foci is
FP + GP = FO + OP +GP
(since FP = FO + OP)
= c + a + a – c
= 2a
Take a point Q at one end of the minor axis.
Sum of the distances from the point Q to the foci is
FQ + GQ =
F G
R P
Q
0
Foci
b
c c
a
a – c
2222
cbcb
22
2 cb
22
cb
22
cb
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Since both P and Q lies on the ellipse.
By the definition of ellipse, we have
or
acb 22
22
i.e.,
22
cba
222
cba i.e., 22
bac
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Eccentricity
ļ‚§ The eccentricity of an ellipse is the ratio of the distances
from the centre of the ellipse to one of the foci and to one of
the vertices of the ellipse (eccentricity is denoted by e)
i.e.,
Then, since the focus is at a distance of c from the centre, in
terms of the eccentricity the focus is at a distance of ae
from the centre.
a
c
e
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Standard equations of an ellipse
ļ‚§ The equation of an ellipse is simplest if the centre of the
ellipse is at the origin and the foci are on the x-axis or
y-axis.
A B
F (-c ,0) G (c ,0)
C
D
P (x, y)
O
X
Y
X
Y
O
(0 ,-c)
(0 , c)
(-b ,0) (b ,0)
a. Foci are on x-axis
b. Foci are on y-axis
12
2
2
2
b
y
a
x
12
2
2
2
a
y
b
x
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ļ‚§ We will derive the equation for the ellipse shown above in
Figure with foci on the x-axis.
Let F and G be the foci and O be the
mid-point of the line segment FG,
Let O be the origin and the line
from O through G be the
positive x-axis and that through F
as the negative x-axis.
Let the coordinates of F be (-c , 0 ) and G be ( c, 0 ).
Let P( x, y ) be any point on the ellipse such that the sum of
the distances from P to the two foci be 2a so given
PF + PG = 2a --------------- (1)
A B
F (-c ,0) G (c ,0)
C
D
P (x, y)
O
X
Y
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Using the distance formula, we have
Squaring both side we get
2222
)(2)( ycxaycx
aycxycx 2)()(
2222
or
2222222
)(4)(4)( ycxaycxaycx
square binomials
2222222
)(4242 ycxaxccxaxccx
222
)(444 ycxaaxc
x
a
c
aycx
22
)(
or
or
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Squaring again and simplifying, we get
122
2
2
2
ca
y
a
x
12
2
2
2
b
y
a
x (Since c2 = a2 – b2)
Hence any point on the ellipse satisfies
12
2
2
2
b
y
a
x
------------------ (2)
Conversely, let P ( x, y) satisfy the equation (2) with 0 < c< a.
Then,
Therefore,
2
2
2
12
a
x
by
22
)( ycxPF
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2
22
22
)(
a
xa
bcx
2
22
222
)()(
a
xa
cacx
(Since b2 = a2 – c2)
2
a
cx
a
a
cx
aPF
Similarly,
a
cx
aPG
Hence,
PF + PG a
a
cx
a
a
cx
a 2 ----------------- (3)
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So, any point that satisfies , satisfies the
geometric condition and so P(x, y) lies on the ellipse.
Hence from (2) and (3), we proved that the equation of an
ellipse with centre of the origin and major axis along the
x-axis is
Similarly, the ellipse lies between the lines y= –b and y = b
and touches these lines.
Similarly, we can derive the equation of the ellipse in Fig
(b) as
These two equations are known as standard equations of the
ellipses.
12
2
2
2
b
y
a
x
12
2
2
2
b
y
a
x
12
2
2
2
a
y
b
x
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• From the standard equations of the ellipses, we
have the following observations:
1. Ellipse is symmetric with respect to both the coordinate
axes since if ( x, y) is a point on the ellipse, then (–x, y),
(x, –y) and (–x, –y) are also points on the ellipse.
2. The foci always lie on the major axis. The major axis can
be determined by finding the intercepts on the axes of
symmetry. That is, major axis is along the x-axis if the
coefficient of x2 has the larger denominator and it is along
the y-axis if the coefficient of y2 has the larger
denominator.
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Ellipse - Equation
2 2
2 2
1
x h y k
a b
The equation of an ellipse centered at (0, 0) is ….
2 2
2 2
1
x y
a b
where c2 = |a2 – b2| and
c is the distance from the
center to the foci.
Shifting the graph over h units and up k units, the
center is at (h, k) and the equation is
where c2 = |a2 – b2| and
c is the distance from the
center to the foci.
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Ellipse - Graphing
2 2
2 2
1
x h y k
a b
where c2 = |a2 – b2| and
c is the distance from the
center to the foci.
Vertices are ā€œaā€ units in
the x direction and ā€œbā€
units in the y direction.
aa
b
b The foci are ā€œcā€ units in
the direction of the
longer (major) axis.
cc
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Graph the ellipse
x
y
a2 = 36
a = ±6
b2 = 16
b = ±4
(x – 2)2
36
(y + 5)2
16
+ = 1
Center = (2,-5)
Vertices: (8,-5) and (-4,-5)
Co-vertices: (2,-1) and (2,-9)
Horizontal Major Axis
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Graph the ellipse
x
y
a2 = 25
a = ±5
b2 = 81
b = ±9
(x + 3)2
25
(y + 1)2
81
+ = 1
Center = (-3,-1)
Vertices: (-3,8) and (-3,-10)
Co-vertices: (-8,-1) and (2,-1)
Vertical Major Axis
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9.4 Ellipses
Horizontal Major Axis:
a2 > b2
a2 – b2 = c2
x2
a2
y2
b2
+ = 1
F1(–c, 0) F2 (c, 0)
y
x
(–a, 0) (a, 0)(0, b)
(0, –b)
O
length of major axis: 2a
length of minor axis: 2b
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9.4 Ellipses
F2(0, –c)
F1 (0, c)
y
x
(0, –b)
(0, b)
(a, 0)(–a, 0)
O
Vertical Major Axis:
b2 > a2
b2 – a2 = c2
x2
a2
y2
b2
+ = 1
length of major axis: 2b
length of minor axis: 2a
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Latus rectum of Ellipse
ļ‚§ Latus rectum of an ellipse is a line segment perpendicular
to the major axis through any of the foci and whose end
points lie on the ellipse.
ļ‚§ To find the length of the latus rectum of the ellipse
A
B
F GO
X
Y
Latus rectum
D
C
12
2
2
2
b
y
a
x
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Let the length of CG is l.
Then the coordinate of the point c are (c , l) i.e., (a e , l)
Since A lies on the ellipse we have ,
But,
Since the ellipse is symmetric with respect to y-axis CG = DG
so the length of the CD (latus rectum)
12
2
2
2
b
y
a
x
1
)(
2
2
2
2
b
l
a
ae 222
1 ebl
2
2
2
22
2
2
2
1
a
b
a
ba
a
c
e
2
2
22
11
a
b
bl
a
b
l
2
a
b
CD
2
2
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Ellipse

  • 1.
  • 2.
    Ellipse ļ‚§ An ellipseis the set of all points in a plane such that the sum of the distances from two points (foci) is a constant. ļ‚§ When you go from point ā€œFā€ to any point on the ellipse and hen go on to point "G", you will always travel the same distance. Ā© iTutor. 2000-2013. All Rights Reserved
  • 3.
    ļ‚§ The twofixed points are called the foci(plural of ā€žfocusā€Ÿ) of the ellipse . ļ‚§ The mid point of the line segment joining the foci is called the centre of the ellipse. ļ‚§ The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. ļ‚§ The end points of the major axis are called the vertices of the ellipse F G A B C D 0 Vertices Major axis Minoraxis Foci Centre Copyright reserved. 2012 The E Tutor
  • 4.
    ļ‚§ We denotethe length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. ļ‚§ Thus, the length of the semi major axis is a and semi-minor axis is b Copyright reserved. 2012 The E Tutor
  • 5.
    Relationship between semi-majoraxis, semi-minor axis and the distance of the focus from the centre of the ellipse Take a point P at one end of the major axis. Sum of the distances of the point P to the foci is FP + GP = FO + OP +GP (since FP = FO + OP) = c + a + a – c = 2a Take a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is FQ + GQ = F G R P Q 0 Foci b c c a a – c 2222 cbcb 22 2 cb 22 cb 22 cb Copyright reserved. 2012 The E Tutor
  • 6.
    Since both Pand Q lies on the ellipse. By the definition of ellipse, we have or acb 22 22 i.e., 22 cba 222 cba i.e., 22 bac Copyright reserved. 2012 The E Tutor
  • 7.
    Eccentricity ļ‚§ The eccentricityof an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., Then, since the focus is at a distance of c from the centre, in terms of the eccentricity the focus is at a distance of ae from the centre. a c e Copyright reserved. 2012 The E Tutor
  • 8.
    Standard equations ofan ellipse ļ‚§ The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis. A B F (-c ,0) G (c ,0) C D P (x, y) O X Y X Y O (0 ,-c) (0 , c) (-b ,0) (b ,0) a. Foci are on x-axis b. Foci are on y-axis 12 2 2 2 b y a x 12 2 2 2 a y b x Copyright reserved. 2012 The E Tutor
  • 9.
    ļ‚§ We willderive the equation for the ellipse shown above in Figure with foci on the x-axis. Let F and G be the foci and O be the mid-point of the line segment FG, Let O be the origin and the line from O through G be the positive x-axis and that through F as the negative x-axis. Let the coordinates of F be (-c , 0 ) and G be ( c, 0 ). Let P( x, y ) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a so given PF + PG = 2a --------------- (1) A B F (-c ,0) G (c ,0) C D P (x, y) O X Y Copyright reserved. 2012 The E Tutor
  • 10.
    Using the distanceformula, we have Squaring both side we get 2222 )(2)( ycxaycx aycxycx 2)()( 2222 or 2222222 )(4)(4)( ycxaycxaycx square binomials 2222222 )(4242 ycxaxccxaxccx 222 )(444 ycxaaxc x a c aycx 22 )( or or Copyright reserved. 2012 The E Tutor
  • 11.
    Squaring again andsimplifying, we get 122 2 2 2 ca y a x 12 2 2 2 b y a x (Since c2 = a2 – b2) Hence any point on the ellipse satisfies 12 2 2 2 b y a x ------------------ (2) Conversely, let P ( x, y) satisfy the equation (2) with 0 < c< a. Then, Therefore, 2 2 2 12 a x by 22 )( ycxPF Copyright reserved. 2012 The E Tutor
  • 12.
    2 22 22 )( a xa bcx 2 22 222 )()( a xa cacx (Since b2 =a2 – c2) 2 a cx a a cx aPF Similarly, a cx aPG Hence, PF + PG a a cx a a cx a 2 ----------------- (3) Copyright reserved. 2012 The E Tutor
  • 13.
    So, any pointthat satisfies , satisfies the geometric condition and so P(x, y) lies on the ellipse. Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is Similarly, the ellipse lies between the lines y= –b and y = b and touches these lines. Similarly, we can derive the equation of the ellipse in Fig (b) as These two equations are known as standard equations of the ellipses. 12 2 2 2 b y a x 12 2 2 2 b y a x 12 2 2 2 a y b x Copyright reserved. 2012 The E Tutor
  • 14.
    • From thestandard equations of the ellipses, we have the following observations: 1. Ellipse is symmetric with respect to both the coordinate axes since if ( x, y) is a point on the ellipse, then (–x, y), (x, –y) and (–x, –y) are also points on the ellipse. 2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator. Copyright reserved. 2012 The E Tutor
  • 15.
    Ellipse - Equation 22 2 2 1 x h y k a b The equation of an ellipse centered at (0, 0) is …. 2 2 2 2 1 x y a b where c2 = |a2 – b2| and c is the distance from the center to the foci. Shifting the graph over h units and up k units, the center is at (h, k) and the equation is where c2 = |a2 – b2| and c is the distance from the center to the foci. Copyright reserved. 2012 The E Tutor
  • 16.
    Ellipse - Graphing 22 2 2 1 x h y k a b where c2 = |a2 – b2| and c is the distance from the center to the foci. Vertices are ā€œaā€ units in the x direction and ā€œbā€ units in the y direction. aa b b The foci are ā€œcā€ units in the direction of the longer (major) axis. cc Copyright reserved. 2012 The E Tutor
  • 17.
    Graph the ellipse x y a2= 36 a = ±6 b2 = 16 b = ±4 (x – 2)2 36 (y + 5)2 16 + = 1 Center = (2,-5) Vertices: (8,-5) and (-4,-5) Co-vertices: (2,-1) and (2,-9) Horizontal Major Axis Copyright reserved. 2012 The E Tutor
  • 18.
    Graph the ellipse x y a2= 25 a = ±5 b2 = 81 b = ±9 (x + 3)2 25 (y + 1)2 81 + = 1 Center = (-3,-1) Vertices: (-3,8) and (-3,-10) Co-vertices: (-8,-1) and (2,-1) Vertical Major Axis Copyright reserved. 2012 The E Tutor
  • 19.
    9.4 Ellipses Horizontal MajorAxis: a2 > b2 a2 – b2 = c2 x2 a2 y2 b2 + = 1 F1(–c, 0) F2 (c, 0) y x (–a, 0) (a, 0)(0, b) (0, –b) O length of major axis: 2a length of minor axis: 2b Copyright reserved. 2012 The E Tutor
  • 20.
    9.4 Ellipses F2(0, –c) F1(0, c) y x (0, –b) (0, b) (a, 0)(–a, 0) O Vertical Major Axis: b2 > a2 b2 – a2 = c2 x2 a2 y2 b2 + = 1 length of major axis: 2b length of minor axis: 2a Copyright reserved. 2012 The E Tutor
  • 21.
    Latus rectum ofEllipse ļ‚§ Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse. ļ‚§ To find the length of the latus rectum of the ellipse A B F GO X Y Latus rectum D C 12 2 2 2 b y a x Copyright reserved. 2012 The E Tutor
  • 22.
    Let the lengthof CG is l. Then the coordinate of the point c are (c , l) i.e., (a e , l) Since A lies on the ellipse we have , But, Since the ellipse is symmetric with respect to y-axis CG = DG so the length of the CD (latus rectum) 12 2 2 2 b y a x 1 )( 2 2 2 2 b l a ae 222 1 ebl 2 2 2 22 2 2 2 1 a b a ba a c e 2 2 22 11 a b bl a b l 2 a b CD 2 2 Copyright reserved. 2012 The E Tutor
  • 23.
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