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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.
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