UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Hyperboloid of One and Two Sheets and its Properties
Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Edurev123
8. Hyperboloid of One and Two Sheets and
its Properties
8.1 Find the vertices of the skew quadrilateral formed by the four generators of the
hyperboloid
?? ?? ?? +?? ?? -?? ?? =????
passing through (???? ,?? ,?? ) and (???? ,?? ,-?? ) .
(2010 : 20 Marks)
Solution:
Given, the equation of hyperboloid is
?? 2
4
+?? 2
-?? 2
=49
It can be rewritten as
(
?? 2
-?? )(
?? 2
+?? )=(7-?? )(7+?? )
? The equation of two systems of generating lines are :
(
?? 2
-?? )=?? (7-?? ),?? (
?? 2
+?? )=(7+?? ) (1)
(
?? 2
-?? )=?? (7+?? ),?? (
?? 2
+?? )=(7-?? ) (2)
(1) and (2) pass through (10,5,1) and (14,2,-2) for ?? =2,?? =
1
3
and ?? =
9
5
,?? =1.
So, the two systems of generating lines are
(
?? 2
-?? )=2(7-?? ), (
?? 2
+?? )=7+?? (3)
(
?? 2
-?? )=
1
3
(7+?? ),
1
3
(
?? 2
+?? )=7-?? (4)
(
?? 2
-?? )=
9
5
(7-?? ),
9
5
(
?? 2
+2)=7+?? (5)
(
?? 2
-?? )=1(7+?? ),
?? 2
+?? =7-?? (6)
Solving (3) and (6), we get the vertices as (14,
7
3
,-
7
3
) .
Page 2
Edurev123
8. Hyperboloid of One and Two Sheets and
its Properties
8.1 Find the vertices of the skew quadrilateral formed by the four generators of the
hyperboloid
?? ?? ?? +?? ?? -?? ?? =????
passing through (???? ,?? ,?? ) and (???? ,?? ,-?? ) .
(2010 : 20 Marks)
Solution:
Given, the equation of hyperboloid is
?? 2
4
+?? 2
-?? 2
=49
It can be rewritten as
(
?? 2
-?? )(
?? 2
+?? )=(7-?? )(7+?? )
? The equation of two systems of generating lines are :
(
?? 2
-?? )=?? (7-?? ),?? (
?? 2
+?? )=(7+?? ) (1)
(
?? 2
-?? )=?? (7+?? ),?? (
?? 2
+?? )=(7-?? ) (2)
(1) and (2) pass through (10,5,1) and (14,2,-2) for ?? =2,?? =
1
3
and ?? =
9
5
,?? =1.
So, the two systems of generating lines are
(
?? 2
-?? )=2(7-?? ), (
?? 2
+?? )=7+?? (3)
(
?? 2
-?? )=
1
3
(7+?? ),
1
3
(
?? 2
+?? )=7-?? (4)
(
?? 2
-?? )=
9
5
(7-?? ),
9
5
(
?? 2
+2)=7+?? (5)
(
?? 2
-?? )=1(7+?? ),
?? 2
+?? =7-?? (6)
Solving (3) and (6), we get the vertices as (14,
7
3
,-
7
3
) .
Solving (4) and (5), we get the vertices as (
21
2
,
77
16
,
21
16
) .
? Other two vertices are (14,
7
3
,-
7
3
) and (
21
2
,
77
16
,
21
16
) .
8.2 Show that the generators through any one of the ends of an equiconjugate
diameter of the principal elliptic section of the hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? are
inclined to each other at an angle of ????
°
if ?? ?? +?? ?? =?? ?? ?? . Find also the condition
for the generators to be perpendicular to each other.
(2011 : 20 Marks)
Solution:
Let ( ?? cos ?? ,?? sin ?? ,0 ) be the given point on the diameter.
The equations of the two generators through this point are
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±??
The direction ratios of two generators are (?? sin ?? ,-?? cos ?? ,?? ) and (?? sin ?? ,-?? cos ?? ,-?? )
respectively.
Let ?? be the angle between two generators and let ?? be the parameter of the end points
of conjugate diameters.
cos ?? =
?? 1
?? 2
+?? 1
?? 2
+?? 1
?? 2
v?? 1
2
:?? 1
2
+?? 1
2
v?? 2
2
+?? 2
2
+?? 2
2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
Putting ?? =60
°
and ?? =45
°
( ? equi-conjugal diameters means equal length of conjugal
diameters, l.e., v?? 2
cos
2
?? +?? 2
sin
2
?? which is equal to v?? 2
sin
2
?? +?? 2
cos
2
?? and is
possible for ?? =45
°
).
? from (i), we get
cos 60
°
=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
?
1
2
=
?? 2
+?? 2
-2?? 2
?? 2
+?? 2
+2?? 2
? ?? 2
+?? 2
+2?? 2
=2?? 2
+2?? 2
-4?? 2
? ?? 2
+?? 2
=6?? 2
Page 3
Edurev123
8. Hyperboloid of One and Two Sheets and
its Properties
8.1 Find the vertices of the skew quadrilateral formed by the four generators of the
hyperboloid
?? ?? ?? +?? ?? -?? ?? =????
passing through (???? ,?? ,?? ) and (???? ,?? ,-?? ) .
(2010 : 20 Marks)
Solution:
Given, the equation of hyperboloid is
?? 2
4
+?? 2
-?? 2
=49
It can be rewritten as
(
?? 2
-?? )(
?? 2
+?? )=(7-?? )(7+?? )
? The equation of two systems of generating lines are :
(
?? 2
-?? )=?? (7-?? ),?? (
?? 2
+?? )=(7+?? ) (1)
(
?? 2
-?? )=?? (7+?? ),?? (
?? 2
+?? )=(7-?? ) (2)
(1) and (2) pass through (10,5,1) and (14,2,-2) for ?? =2,?? =
1
3
and ?? =
9
5
,?? =1.
So, the two systems of generating lines are
(
?? 2
-?? )=2(7-?? ), (
?? 2
+?? )=7+?? (3)
(
?? 2
-?? )=
1
3
(7+?? ),
1
3
(
?? 2
+?? )=7-?? (4)
(
?? 2
-?? )=
9
5
(7-?? ),
9
5
(
?? 2
+2)=7+?? (5)
(
?? 2
-?? )=1(7+?? ),
?? 2
+?? =7-?? (6)
Solving (3) and (6), we get the vertices as (14,
7
3
,-
7
3
) .
Solving (4) and (5), we get the vertices as (
21
2
,
77
16
,
21
16
) .
? Other two vertices are (14,
7
3
,-
7
3
) and (
21
2
,
77
16
,
21
16
) .
8.2 Show that the generators through any one of the ends of an equiconjugate
diameter of the principal elliptic section of the hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? are
inclined to each other at an angle of ????
°
if ?? ?? +?? ?? =?? ?? ?? . Find also the condition
for the generators to be perpendicular to each other.
(2011 : 20 Marks)
Solution:
Let ( ?? cos ?? ,?? sin ?? ,0 ) be the given point on the diameter.
The equations of the two generators through this point are
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±??
The direction ratios of two generators are (?? sin ?? ,-?? cos ?? ,?? ) and (?? sin ?? ,-?? cos ?? ,-?? )
respectively.
Let ?? be the angle between two generators and let ?? be the parameter of the end points
of conjugate diameters.
cos ?? =
?? 1
?? 2
+?? 1
?? 2
+?? 1
?? 2
v?? 1
2
:?? 1
2
+?? 1
2
v?? 2
2
+?? 2
2
+?? 2
2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
Putting ?? =60
°
and ?? =45
°
( ? equi-conjugal diameters means equal length of conjugal
diameters, l.e., v?? 2
cos
2
?? +?? 2
sin
2
?? which is equal to v?? 2
sin
2
?? +?? 2
cos
2
?? and is
possible for ?? =45
°
).
? from (i), we get
cos 60
°
=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
?
1
2
=
?? 2
+?? 2
-2?? 2
?? 2
+?? 2
+2?? 2
? ?? 2
+?? 2
+2?? 2
=2?? 2
+2?? 2
-4?? 2
? ?? 2
+?? 2
=6?? 2
Again, put ?? =90
°
and ?? =45
°
in (i), we get
0=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
? ?? 2
+?? 2
=2?? 2
,
which is the required condition for the generators to be perpendicular to each other.
8.3 A variable generator meets two generators of the same system through the
extremities ?? and ?? '
of the minor axis of the principal elliptic section of the
hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? in ?? and ?? '
. Prove that ?? ?? '
·?? '
?? '
=?? ?? +?? ?? .
(Note: There is minor error in actual question. It must be
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? or ???? ·
?? '
?? =?? ?? +
?? ?? ?? ).
(2012 : 20 Marks)
Solution:
The generator through any general point (?? cos ?? ,?? sin ?? ) on the principal elliptic section is
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±?? (??)(?????? h ?????????????? )
Taking the positive system for the extremity of minor axis ?? =
?? 2
and
3?? 2
i.e.,
?? -?? ?? =
?? -?? 0
=
?? ?? (ii)
and
?? -?? =
?? +?? 0
=
?? ?? (?????? )
Any point on these lines is
?? ?? =
?? -?? ?? =
?? ?? =?? and
?? -?? =
?? +?? 0
=
?? ?? =?? 2
?? =???? ,?? =?? ,?? =?? ?? 1
and ?? =-???? ,?? =-?? ,?? =?? ?? 2
Note that the distance of such a point from ?? is v2
2
+?? 2
+?? 2
?? 1
=v?? 2
+?? 2
?? 1
.
For, ?? ,?? this general point must bs on the variable generaior whose equation is given by
(i) (taking the other system).
Page 4
Edurev123
8. Hyperboloid of One and Two Sheets and
its Properties
8.1 Find the vertices of the skew quadrilateral formed by the four generators of the
hyperboloid
?? ?? ?? +?? ?? -?? ?? =????
passing through (???? ,?? ,?? ) and (???? ,?? ,-?? ) .
(2010 : 20 Marks)
Solution:
Given, the equation of hyperboloid is
?? 2
4
+?? 2
-?? 2
=49
It can be rewritten as
(
?? 2
-?? )(
?? 2
+?? )=(7-?? )(7+?? )
? The equation of two systems of generating lines are :
(
?? 2
-?? )=?? (7-?? ),?? (
?? 2
+?? )=(7+?? ) (1)
(
?? 2
-?? )=?? (7+?? ),?? (
?? 2
+?? )=(7-?? ) (2)
(1) and (2) pass through (10,5,1) and (14,2,-2) for ?? =2,?? =
1
3
and ?? =
9
5
,?? =1.
So, the two systems of generating lines are
(
?? 2
-?? )=2(7-?? ), (
?? 2
+?? )=7+?? (3)
(
?? 2
-?? )=
1
3
(7+?? ),
1
3
(
?? 2
+?? )=7-?? (4)
(
?? 2
-?? )=
9
5
(7-?? ),
9
5
(
?? 2
+2)=7+?? (5)
(
?? 2
-?? )=1(7+?? ),
?? 2
+?? =7-?? (6)
Solving (3) and (6), we get the vertices as (14,
7
3
,-
7
3
) .
Solving (4) and (5), we get the vertices as (
21
2
,
77
16
,
21
16
) .
? Other two vertices are (14,
7
3
,-
7
3
) and (
21
2
,
77
16
,
21
16
) .
8.2 Show that the generators through any one of the ends of an equiconjugate
diameter of the principal elliptic section of the hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? are
inclined to each other at an angle of ????
°
if ?? ?? +?? ?? =?? ?? ?? . Find also the condition
for the generators to be perpendicular to each other.
(2011 : 20 Marks)
Solution:
Let ( ?? cos ?? ,?? sin ?? ,0 ) be the given point on the diameter.
The equations of the two generators through this point are
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±??
The direction ratios of two generators are (?? sin ?? ,-?? cos ?? ,?? ) and (?? sin ?? ,-?? cos ?? ,-?? )
respectively.
Let ?? be the angle between two generators and let ?? be the parameter of the end points
of conjugate diameters.
cos ?? =
?? 1
?? 2
+?? 1
?? 2
+?? 1
?? 2
v?? 1
2
:?? 1
2
+?? 1
2
v?? 2
2
+?? 2
2
+?? 2
2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
Putting ?? =60
°
and ?? =45
°
( ? equi-conjugal diameters means equal length of conjugal
diameters, l.e., v?? 2
cos
2
?? +?? 2
sin
2
?? which is equal to v?? 2
sin
2
?? +?? 2
cos
2
?? and is
possible for ?? =45
°
).
? from (i), we get
cos 60
°
=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
?
1
2
=
?? 2
+?? 2
-2?? 2
?? 2
+?? 2
+2?? 2
? ?? 2
+?? 2
+2?? 2
=2?? 2
+2?? 2
-4?? 2
? ?? 2
+?? 2
=6?? 2
Again, put ?? =90
°
and ?? =45
°
in (i), we get
0=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
? ?? 2
+?? 2
=2?? 2
,
which is the required condition for the generators to be perpendicular to each other.
8.3 A variable generator meets two generators of the same system through the
extremities ?? and ?? '
of the minor axis of the principal elliptic section of the
hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? in ?? and ?? '
. Prove that ?? ?? '
·?? '
?? '
=?? ?? +?? ?? .
(Note: There is minor error in actual question. It must be
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? or ???? ·
?? '
?? =?? ?? +
?? ?? ?? ).
(2012 : 20 Marks)
Solution:
The generator through any general point (?? cos ?? ,?? sin ?? ) on the principal elliptic section is
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±?? (??)(?????? h ?????????????? )
Taking the positive system for the extremity of minor axis ?? =
?? 2
and
3?? 2
i.e.,
?? -?? ?? =
?? -?? 0
=
?? ?? (ii)
and
?? -?? =
?? +?? 0
=
?? ?? (?????? )
Any point on these lines is
?? ?? =
?? -?? ?? =
?? ?? =?? and
?? -?? =
?? +?? 0
=
?? ?? =?? 2
?? =???? ,?? =?? ,?? =?? ?? 1
and ?? =-???? ,?? =-?? ,?? =?? ?? 2
Note that the distance of such a point from ?? is v2
2
+?? 2
+?? 2
?? 1
=v?? 2
+?? 2
?? 1
.
For, ?? ,?? this general point must bs on the variable generaior whose equation is given by
(i) (taking the other system).
?
?? ?? 1
-?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =-?? 1
and
-?? ?? 1
-?? cos ?? ?? sin ?? =
-?? -?? sin ?? -?? cos ?? =-?? 2
? ?? 1
=
cos ?? 1+sin ?? ?? 2
=-(
1+sin ?? cos ?? )
? ???? =v?? 2
+?? 2
|?? 1
|
?? '
?? '
=v?? 2
+?? 2
|?? 2
|
? ???? ·?? '
?? '
=(?? 2
+?? 2
)
cos ?? 1+sin ?? ×
1+sin ?? cos ?? =?? 2
+?? 2
8.4 Find the equations of the two generating lines through any point
(?? ?????? ?? ,?? ?????? ?? ,?? ) , of the principal elliptic section
?? ?? ?? ?? +
?? ?? ?? ?? =?? ,?? =?? , of the
hyperboloid by the plane ?? =?? .
(2014 : 15 Marks)
Solution:
Any point on the elliptic section of the hyperboloid is (?? cos ?? ,?? sin ?? ,0) .
? Equations of any line through this point is
?? -?? cos ?? ?? =
?? -?? sin ?? ?? =
?? -0
?? =?? ( say ) (??)
Any point on this line is (???? +?? cos ?? ,???? +?? sin ?? , and it lies on given hyperboloid, if
(???? +?? cos ?? )
2
?? 2
+
(???? +?? sin ?? )
2
?? 2
-
?? 2
?? 2
?? 2
=1
or (
?? 2
?? 2
+
?? 2
?? 2
-
?? 2
?? 2
)?? 2
+2(
??cos ?? ?? +
?? sin ?? ?? )?? =0 (???? )
If the line (i) generator of given hyperboloid, then (i) lies wholly on the hyperboloid and
the condition for which from (ii) are
(
?? 2
?? 2
)+(
?? 2
?? 2
)-(
?? 2
?? 2
) =0 (?????? )
??cos ?? ?? +
?? sin ?? ?? =0 (???? )
from (iv) we get,
Page 5
Edurev123
8. Hyperboloid of One and Two Sheets and
its Properties
8.1 Find the vertices of the skew quadrilateral formed by the four generators of the
hyperboloid
?? ?? ?? +?? ?? -?? ?? =????
passing through (???? ,?? ,?? ) and (???? ,?? ,-?? ) .
(2010 : 20 Marks)
Solution:
Given, the equation of hyperboloid is
?? 2
4
+?? 2
-?? 2
=49
It can be rewritten as
(
?? 2
-?? )(
?? 2
+?? )=(7-?? )(7+?? )
? The equation of two systems of generating lines are :
(
?? 2
-?? )=?? (7-?? ),?? (
?? 2
+?? )=(7+?? ) (1)
(
?? 2
-?? )=?? (7+?? ),?? (
?? 2
+?? )=(7-?? ) (2)
(1) and (2) pass through (10,5,1) and (14,2,-2) for ?? =2,?? =
1
3
and ?? =
9
5
,?? =1.
So, the two systems of generating lines are
(
?? 2
-?? )=2(7-?? ), (
?? 2
+?? )=7+?? (3)
(
?? 2
-?? )=
1
3
(7+?? ),
1
3
(
?? 2
+?? )=7-?? (4)
(
?? 2
-?? )=
9
5
(7-?? ),
9
5
(
?? 2
+2)=7+?? (5)
(
?? 2
-?? )=1(7+?? ),
?? 2
+?? =7-?? (6)
Solving (3) and (6), we get the vertices as (14,
7
3
,-
7
3
) .
Solving (4) and (5), we get the vertices as (
21
2
,
77
16
,
21
16
) .
? Other two vertices are (14,
7
3
,-
7
3
) and (
21
2
,
77
16
,
21
16
) .
8.2 Show that the generators through any one of the ends of an equiconjugate
diameter of the principal elliptic section of the hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? are
inclined to each other at an angle of ????
°
if ?? ?? +?? ?? =?? ?? ?? . Find also the condition
for the generators to be perpendicular to each other.
(2011 : 20 Marks)
Solution:
Let ( ?? cos ?? ,?? sin ?? ,0 ) be the given point on the diameter.
The equations of the two generators through this point are
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±??
The direction ratios of two generators are (?? sin ?? ,-?? cos ?? ,?? ) and (?? sin ?? ,-?? cos ?? ,-?? )
respectively.
Let ?? be the angle between two generators and let ?? be the parameter of the end points
of conjugate diameters.
cos ?? =
?? 1
?? 2
+?? 1
?? 2
+?? 1
?? 2
v?? 1
2
:?? 1
2
+?? 1
2
v?? 2
2
+?? 2
2
+?? 2
2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
v?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
=
?? 2
sin
2
?? +?? 2
cos
2
?? -?? 2
?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
Putting ?? =60
°
and ?? =45
°
( ? equi-conjugal diameters means equal length of conjugal
diameters, l.e., v?? 2
cos
2
?? +?? 2
sin
2
?? which is equal to v?? 2
sin
2
?? +?? 2
cos
2
?? and is
possible for ?? =45
°
).
? from (i), we get
cos 60
°
=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
?
1
2
=
?? 2
+?? 2
-2?? 2
?? 2
+?? 2
+2?? 2
? ?? 2
+?? 2
+2?? 2
=2?? 2
+2?? 2
-4?? 2
? ?? 2
+?? 2
=6?? 2
Again, put ?? =90
°
and ?? =45
°
in (i), we get
0=
?? 2
2
+
?? 2
2
-?? 2
?? 2
2
+
?? 2
2
+?? 2
? ?? 2
+?? 2
=2?? 2
,
which is the required condition for the generators to be perpendicular to each other.
8.3 A variable generator meets two generators of the same system through the
extremities ?? and ?? '
of the minor axis of the principal elliptic section of the
hyperboloid
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? in ?? and ?? '
. Prove that ?? ?? '
·?? '
?? '
=?? ?? +?? ?? .
(Note: There is minor error in actual question. It must be
?? ?? ?? ?? +
?? ?? ?? ?? -
?? ?? ?? ?? =?? or ???? ·
?? '
?? =?? ?? +
?? ?? ?? ).
(2012 : 20 Marks)
Solution:
The generator through any general point (?? cos ?? ,?? sin ?? ) on the principal elliptic section is
?? -?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±?? (??)(?????? h ?????????????? )
Taking the positive system for the extremity of minor axis ?? =
?? 2
and
3?? 2
i.e.,
?? -?? ?? =
?? -?? 0
=
?? ?? (ii)
and
?? -?? =
?? +?? 0
=
?? ?? (?????? )
Any point on these lines is
?? ?? =
?? -?? ?? =
?? ?? =?? and
?? -?? =
?? +?? 0
=
?? ?? =?? 2
?? =???? ,?? =?? ,?? =?? ?? 1
and ?? =-???? ,?? =-?? ,?? =?? ?? 2
Note that the distance of such a point from ?? is v2
2
+?? 2
+?? 2
?? 1
=v?? 2
+?? 2
?? 1
.
For, ?? ,?? this general point must bs on the variable generaior whose equation is given by
(i) (taking the other system).
?
?? ?? 1
-?? cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =-?? 1
and
-?? ?? 1
-?? cos ?? ?? sin ?? =
-?? -?? sin ?? -?? cos ?? =-?? 2
? ?? 1
=
cos ?? 1+sin ?? ?? 2
=-(
1+sin ?? cos ?? )
? ???? =v?? 2
+?? 2
|?? 1
|
?? '
?? '
=v?? 2
+?? 2
|?? 2
|
? ???? ·?? '
?? '
=(?? 2
+?? 2
)
cos ?? 1+sin ?? ×
1+sin ?? cos ?? =?? 2
+?? 2
8.4 Find the equations of the two generating lines through any point
(?? ?????? ?? ,?? ?????? ?? ,?? ) , of the principal elliptic section
?? ?? ?? ?? +
?? ?? ?? ?? =?? ,?? =?? , of the
hyperboloid by the plane ?? =?? .
(2014 : 15 Marks)
Solution:
Any point on the elliptic section of the hyperboloid is (?? cos ?? ,?? sin ?? ,0) .
? Equations of any line through this point is
?? -?? cos ?? ?? =
?? -?? sin ?? ?? =
?? -0
?? =?? ( say ) (??)
Any point on this line is (???? +?? cos ?? ,???? +?? sin ?? , and it lies on given hyperboloid, if
(???? +?? cos ?? )
2
?? 2
+
(???? +?? sin ?? )
2
?? 2
-
?? 2
?? 2
?? 2
=1
or (
?? 2
?? 2
+
?? 2
?? 2
-
?? 2
?? 2
)?? 2
+2(
??cos ?? ?? +
?? sin ?? ?? )?? =0 (???? )
If the line (i) generator of given hyperboloid, then (i) lies wholly on the hyperboloid and
the condition for which from (ii) are
(
?? 2
?? 2
)+(
?? 2
?? 2
)-(
?? 2
?? 2
) =0 (?????? )
??cos ?? ?? +
?? sin ?? ?? =0 (???? )
from (iv) we get,
?? asin ?? =
?? -?? cos ?? or
(??/?? )
sin ?? =
(?? /-?? )
cos ?? ?
(??/?? )
sin ?? =
(?? /-?? )
cos ?? =v
(?? 2
/?? 2
)+(?? 2
/?? 2
)
vsin
2
?? +cos
2
?? =
v?? 2
/?? 2
1
?
?? ?? sin ?? =
?? -?? cos ?? =
?? ±??
? The equation to the required generated from (i) are
?? -cos ?? ?? sin ?? =
?? -?? sin ?? -?? cos ?? =
?? ±??
8.5. Find all the asymptotes of the curve (?? ?? +?? )?? =(?? -?? )
?? .
(2020: 10 marks)
Solution:
Here the given curve (2?? -3)?? =(?? -1)
2
2???? -3?? =?? 2
-2?? +1
?? 2
-2???? -2?? -3?? +1 =0
Putting ?? =???? +??
?? 2
-[2?? (???? +?? )]-2?? -3(???? +?? )+1=0
?? 2
-2?? 2
?? -2???? -2?? -3???? -3?? +1=0
(1-2?? )?? 2
-?? (2?? +2+3?? )-3?? +1=0
Putting coefficient of ?? 2
=0 and ?? =0
1-2?? =0
?? =1/2
2?? +2+3?? =0
2?? +2+3×
1
2
=0
2?? +
4+3
2
=0
2?? +
7
2
=0
?? =-
7
4
Thus ?? =
?? 2
-
7
2
is one asymptote. Equating highest power equal to zero. We get
asymptotes parallel to ?? -axis.
Read More
|
387 videos|203 docs
|
Related Exams
About this Document
|
Dec 18, 2024 Last updated |
Document Description: Hyperboloid of One and Two Sheets and its Properties for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Hyperboloid of One and Two Sheets and its Properties have been prepared according to the UPSC exam syllabus. Information about Hyperboloid of One and Two Sheets and its Properties covers topics
like and Hyperboloid of One and Two Sheets and its Properties Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Hyperboloid of One and Two Sheets and its Properties.
Introduction of Hyperboloid of One and Two Sheets and its Properties in English is available as part of our Mathematics Optional Notes for UPSC
for UPSC & Hyperboloid of One and Two Sheets and its Properties in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Hyperboloid of One and Two Sheets and its Properties
In this doc you can find the meaning of Hyperboloid of One and Two Sheets and its Properties defined & explained in the simplest way possible. Besides explaining types of
Hyperboloid of One and Two Sheets and its Properties theory, EduRev gives you an ample number of questions to practice Hyperboloid of One and Two Sheets and its Properties tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Exam
,study material
,practice quizzes
,Viva Questions
,video lectures
,MCQs
,Summary
,Objective type Questions
,Free
,Sample Paper
,Semester Notes
,Extra Questions
,mock tests for examination
,shortcuts and tricks
,Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC
,Important questions
,Previous Year Questions with Solutions
,Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC
,ppt
,Hyperboloid of One and Two Sheets and its Properties | Mathematics Optional Notes for UPSC
,past year papers
;
Additional Information about Hyperboloid of One and Two Sheets and its Properties for UPSC Preparation
Hyperboloid of One and Two Sheets and its Properties Free PDF Download
The Hyperboloid of One and Two Sheets and its Properties is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Hyperboloid of One and Two Sheets and its Properties now and kickstart your journey towards success in the UPSC exam.
Importance of Hyperboloid of One and Two Sheets and its Properties
The importance of Hyperboloid of One and Two Sheets and its Properties cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Hyperboloid of One and Two Sheets and its Properties Notes
Hyperboloid of One and Two Sheets and its Properties Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Hyperboloid of One and Two Sheets and its Properties.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Hyperboloid of One and Two Sheets and its Properties Notes on EduRev are your ultimate resource for success.
Hyperboloid of One and Two Sheets and its Properties UPSC Questions
The "Hyperboloid of One and Two Sheets and its Properties UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Hyperboloid of One and Two Sheets and its Properties on the App
Students of UPSC can study Hyperboloid of One and Two Sheets and its Properties alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Hyperboloid of One and Two Sheets and its Properties,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Hyperboloid of One and Two Sheets and its Properties is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup