Page 1
Edurev123
6. Paraboloid and it Properties
6.1 Show that the plane ?? ?? +?? ?? +?? ?? +
?? ?? =?? touches the paraboloid ?? ?? ?? +?? ?? ?? =
???? ?? and find the point of contact.
(2010 : 20 Marks)
Solution:
Given plane
?? =3?? +3?? +7?? +
5
2
=0 (??)
and
Paraboloid =3?? 2
+4?? 2
=10??
Now, let point of contact be (?? 1
,?? 1
,?? 1
) .
? If given plane touches paraboloid at this point of contact, then the plane should be
tangent plane to the paraboloid.
Now, equation of tangent plane at (?? 1
,?? 1
,?? 1
) is
3?? ?? 1
+4?? ?? 1
=
10(?? +?? 1
)
2
? 3?? ?? 1
+4?? ?? 1
-5?? -5?? 1
=0
Comparing (1) and (2), we get
3
3?? 1
=
4
4?? 1
=
7
-5
=
5/2
-521
?
3
3?? 1
=
-7
5
??? 1
=
-5
7
4
4?? 1
=
7
-5
??? 1
=
-5
7
?
7
-5
=
5
-5×221
??? 1
=
5
14
? Point of contact (?? 1
,?? 1
,?? 1
) =(
-5
7
,
-5
7
,
5
14
)
Page 2
Edurev123
6. Paraboloid and it Properties
6.1 Show that the plane ?? ?? +?? ?? +?? ?? +
?? ?? =?? touches the paraboloid ?? ?? ?? +?? ?? ?? =
???? ?? and find the point of contact.
(2010 : 20 Marks)
Solution:
Given plane
?? =3?? +3?? +7?? +
5
2
=0 (??)
and
Paraboloid =3?? 2
+4?? 2
=10??
Now, let point of contact be (?? 1
,?? 1
,?? 1
) .
? If given plane touches paraboloid at this point of contact, then the plane should be
tangent plane to the paraboloid.
Now, equation of tangent plane at (?? 1
,?? 1
,?? 1
) is
3?? ?? 1
+4?? ?? 1
=
10(?? +?? 1
)
2
? 3?? ?? 1
+4?? ?? 1
-5?? -5?? 1
=0
Comparing (1) and (2), we get
3
3?? 1
=
4
4?? 1
=
7
-5
=
5/2
-521
?
3
3?? 1
=
-7
5
??? 1
=
-5
7
4
4?? 1
=
7
-5
??? 1
=
-5
7
?
7
-5
=
5
-5×221
??? 1
=
5
14
? Point of contact (?? 1
,?? 1
,?? 1
) =(
-5
7
,
-5
7
,
5
14
)
6.2. Show that the locus of a point from which the three mutually perpendicular
tangent lines can be drawn to the paraboloid ?? ?? +?? ?? +?? ?? =?? is
?? ?? +?? ?? +?? ?? =??
(2012 : 20 Marks)
Solution:
Let ?? (?? 1
,?? 1
,?? 1
) be the point from which three mutually perpendicular lines can be drawn
to the paraboloid
?? 2
+?? 2
+2?? =0 (??)
Then, the enveloping cone of (i) with the vertex at ??
(?? 1
,?? 1
,?? 1
) is
?? ?? 1
=?? 2
(???? )
?? =?? 2
+?? 2
+2?? ?? 1
=?? 1
2
+?? 1
2
+2?? 1
?? =?? ?? 1
+?? ?? 1
+?? +?? 1
? from (ii), we have
(?? 2
+?? 2
+2?? )(?? 1
2
+?? 1
2
+2?? 1
)=(?? ?? 1
+?? ?? 1
+(?? +?? 1
))
2
For three mutually perpendicular generators, coefficient of ?? 2
+ coefficient of ?? 2
+
coefficient of ?? 2
=0
or
?? 1
2
+?? 1
2
+4?? 1
-1 =0
?? 1
2
+?? 1
2
+4?? 1
=0
? Locus of (?? 1
,?? 1
,?? 1
) is
?? 2
+?? 2
+4?? =1
6.3 Two perpendicular tangent planes to the parabolold ?? ?? +?? ?? =?? ?? intersect in a
straight line in the plane ?? =?? . Obtain the curve to which this straight line
touches.
(2015 : 13 Marks)
Solution:
Let the line of intersection of the two planes be:
???? +???? =?? ,?? =0 (??)
Page 3
Edurev123
6. Paraboloid and it Properties
6.1 Show that the plane ?? ?? +?? ?? +?? ?? +
?? ?? =?? touches the paraboloid ?? ?? ?? +?? ?? ?? =
???? ?? and find the point of contact.
(2010 : 20 Marks)
Solution:
Given plane
?? =3?? +3?? +7?? +
5
2
=0 (??)
and
Paraboloid =3?? 2
+4?? 2
=10??
Now, let point of contact be (?? 1
,?? 1
,?? 1
) .
? If given plane touches paraboloid at this point of contact, then the plane should be
tangent plane to the paraboloid.
Now, equation of tangent plane at (?? 1
,?? 1
,?? 1
) is
3?? ?? 1
+4?? ?? 1
=
10(?? +?? 1
)
2
? 3?? ?? 1
+4?? ?? 1
-5?? -5?? 1
=0
Comparing (1) and (2), we get
3
3?? 1
=
4
4?? 1
=
7
-5
=
5/2
-521
?
3
3?? 1
=
-7
5
??? 1
=
-5
7
4
4?? 1
=
7
-5
??? 1
=
-5
7
?
7
-5
=
5
-5×221
??? 1
=
5
14
? Point of contact (?? 1
,?? 1
,?? 1
) =(
-5
7
,
-5
7
,
5
14
)
6.2. Show that the locus of a point from which the three mutually perpendicular
tangent lines can be drawn to the paraboloid ?? ?? +?? ?? +?? ?? =?? is
?? ?? +?? ?? +?? ?? =??
(2012 : 20 Marks)
Solution:
Let ?? (?? 1
,?? 1
,?? 1
) be the point from which three mutually perpendicular lines can be drawn
to the paraboloid
?? 2
+?? 2
+2?? =0 (??)
Then, the enveloping cone of (i) with the vertex at ??
(?? 1
,?? 1
,?? 1
) is
?? ?? 1
=?? 2
(???? )
?? =?? 2
+?? 2
+2?? ?? 1
=?? 1
2
+?? 1
2
+2?? 1
?? =?? ?? 1
+?? ?? 1
+?? +?? 1
? from (ii), we have
(?? 2
+?? 2
+2?? )(?? 1
2
+?? 1
2
+2?? 1
)=(?? ?? 1
+?? ?? 1
+(?? +?? 1
))
2
For three mutually perpendicular generators, coefficient of ?? 2
+ coefficient of ?? 2
+
coefficient of ?? 2
=0
or
?? 1
2
+?? 1
2
+4?? 1
-1 =0
?? 1
2
+?? 1
2
+4?? 1
=0
? Locus of (?? 1
,?? 1
,?? 1
) is
?? 2
+?? 2
+4?? =1
6.3 Two perpendicular tangent planes to the parabolold ?? ?? +?? ?? =?? ?? intersect in a
straight line in the plane ?? =?? . Obtain the curve to which this straight line
touches.
(2015 : 13 Marks)
Solution:
Let the line of intersection of the two planes be:
???? +???? =?? ,?? =0 (??)
Since this lies on the plane ?? =0 (given).
? Equation of the plane through the line (i) is
(???? +???? -?? )+???? =0
???? +???? +???? =?? (???? )
If the plane (ii) touches the paraboloid, then
?? 2
?? 2
+
?? 2
?? 2
+
2????
?? =0 (condition)
i.e.,
?? 2
+?? 2
+2???? =0 (?????? )
This being quadratic in ?? , gives two values of ?? 1
say ?? 1
and ?? 2
such that
?? 1
·?? 2
=
?? 2
+2????
?? (???? )
Also from (ii), the direction ratio's of the normal to the two tangent planes whose line of
intersection is (ii) are ?? 1
,?? ,?? and ?? 2
,?? ,?? .
Also, as these two tangent planes are perpendicular
? ?? 1
·?? 2
+?? ·?? +?? ·?? =0
? (?? 3
+2???? )+?? 2
+?? 2
=0 from (iv)
? 2?? 2
+?? 2
+2???? =0 (v)
Now, we are to prove that the line (i) touches a parabola (to be found). So, we are to find
the envelope of (i) which satisfies the condition (v).
Eliminating ?? between (i) and (?? ) , the equations of the line of intersection of two tangent
planes is :
2?? 2
+?? 2
+2(???? +???? )?? =0,?? =0
? 2(
?? ?? )
2
+2?? (
?? ?? )+(1+2?? )=0,?? =0
It is quadratic in
?? ?? , so its envelope is given by:
?? 2
-4???? =0,?? =0
? (2?? )
2
-4·2(1+2?? )=0,?? =0
? ?? 2
=2(2?? +1),?? =0
Page 4
Edurev123
6. Paraboloid and it Properties
6.1 Show that the plane ?? ?? +?? ?? +?? ?? +
?? ?? =?? touches the paraboloid ?? ?? ?? +?? ?? ?? =
???? ?? and find the point of contact.
(2010 : 20 Marks)
Solution:
Given plane
?? =3?? +3?? +7?? +
5
2
=0 (??)
and
Paraboloid =3?? 2
+4?? 2
=10??
Now, let point of contact be (?? 1
,?? 1
,?? 1
) .
? If given plane touches paraboloid at this point of contact, then the plane should be
tangent plane to the paraboloid.
Now, equation of tangent plane at (?? 1
,?? 1
,?? 1
) is
3?? ?? 1
+4?? ?? 1
=
10(?? +?? 1
)
2
? 3?? ?? 1
+4?? ?? 1
-5?? -5?? 1
=0
Comparing (1) and (2), we get
3
3?? 1
=
4
4?? 1
=
7
-5
=
5/2
-521
?
3
3?? 1
=
-7
5
??? 1
=
-5
7
4
4?? 1
=
7
-5
??? 1
=
-5
7
?
7
-5
=
5
-5×221
??? 1
=
5
14
? Point of contact (?? 1
,?? 1
,?? 1
) =(
-5
7
,
-5
7
,
5
14
)
6.2. Show that the locus of a point from which the three mutually perpendicular
tangent lines can be drawn to the paraboloid ?? ?? +?? ?? +?? ?? =?? is
?? ?? +?? ?? +?? ?? =??
(2012 : 20 Marks)
Solution:
Let ?? (?? 1
,?? 1
,?? 1
) be the point from which three mutually perpendicular lines can be drawn
to the paraboloid
?? 2
+?? 2
+2?? =0 (??)
Then, the enveloping cone of (i) with the vertex at ??
(?? 1
,?? 1
,?? 1
) is
?? ?? 1
=?? 2
(???? )
?? =?? 2
+?? 2
+2?? ?? 1
=?? 1
2
+?? 1
2
+2?? 1
?? =?? ?? 1
+?? ?? 1
+?? +?? 1
? from (ii), we have
(?? 2
+?? 2
+2?? )(?? 1
2
+?? 1
2
+2?? 1
)=(?? ?? 1
+?? ?? 1
+(?? +?? 1
))
2
For three mutually perpendicular generators, coefficient of ?? 2
+ coefficient of ?? 2
+
coefficient of ?? 2
=0
or
?? 1
2
+?? 1
2
+4?? 1
-1 =0
?? 1
2
+?? 1
2
+4?? 1
=0
? Locus of (?? 1
,?? 1
,?? 1
) is
?? 2
+?? 2
+4?? =1
6.3 Two perpendicular tangent planes to the parabolold ?? ?? +?? ?? =?? ?? intersect in a
straight line in the plane ?? =?? . Obtain the curve to which this straight line
touches.
(2015 : 13 Marks)
Solution:
Let the line of intersection of the two planes be:
???? +???? =?? ,?? =0 (??)
Since this lies on the plane ?? =0 (given).
? Equation of the plane through the line (i) is
(???? +???? -?? )+???? =0
???? +???? +???? =?? (???? )
If the plane (ii) touches the paraboloid, then
?? 2
?? 2
+
?? 2
?? 2
+
2????
?? =0 (condition)
i.e.,
?? 2
+?? 2
+2???? =0 (?????? )
This being quadratic in ?? , gives two values of ?? 1
say ?? 1
and ?? 2
such that
?? 1
·?? 2
=
?? 2
+2????
?? (???? )
Also from (ii), the direction ratio's of the normal to the two tangent planes whose line of
intersection is (ii) are ?? 1
,?? ,?? and ?? 2
,?? ,?? .
Also, as these two tangent planes are perpendicular
? ?? 1
·?? 2
+?? ·?? +?? ·?? =0
? (?? 3
+2???? )+?? 2
+?? 2
=0 from (iv)
? 2?? 2
+?? 2
+2???? =0 (v)
Now, we are to prove that the line (i) touches a parabola (to be found). So, we are to find
the envelope of (i) which satisfies the condition (v).
Eliminating ?? between (i) and (?? ) , the equations of the line of intersection of two tangent
planes is :
2?? 2
+?? 2
+2(???? +???? )?? =0,?? =0
? 2(
?? ?? )
2
+2?? (
?? ?? )+(1+2?? )=0,?? =0
It is quadratic in
?? ?? , so its envelope is given by:
?? 2
-4???? =0,?? =0
? (2?? )
2
-4·2(1+2?? )=0,?? =0
? ?? 2
=2(2?? +1),?? =0
This is the required curve.
6.4 Find the volume of the solid above the ???? -plane and directly below the portion
of the elliptic paraboloid ?? ?? +
?? ?? ?? =?? which is cut off by the plane ?? =?? .
(20i7 : 15 Marks)
Solution:
Equation of ?? surface, cut off cut plane
?? 2
+
?? 2
4
=9;?? =9
i.e.,
?? 2
9
+
?? 2
36
=1;?? =9
Making the transformation,
?? =3?? cos ?? ?? =6?? sin ??
?? :0 to 1;?? :0 to 2??
Page 5
Edurev123
6. Paraboloid and it Properties
6.1 Show that the plane ?? ?? +?? ?? +?? ?? +
?? ?? =?? touches the paraboloid ?? ?? ?? +?? ?? ?? =
???? ?? and find the point of contact.
(2010 : 20 Marks)
Solution:
Given plane
?? =3?? +3?? +7?? +
5
2
=0 (??)
and
Paraboloid =3?? 2
+4?? 2
=10??
Now, let point of contact be (?? 1
,?? 1
,?? 1
) .
? If given plane touches paraboloid at this point of contact, then the plane should be
tangent plane to the paraboloid.
Now, equation of tangent plane at (?? 1
,?? 1
,?? 1
) is
3?? ?? 1
+4?? ?? 1
=
10(?? +?? 1
)
2
? 3?? ?? 1
+4?? ?? 1
-5?? -5?? 1
=0
Comparing (1) and (2), we get
3
3?? 1
=
4
4?? 1
=
7
-5
=
5/2
-521
?
3
3?? 1
=
-7
5
??? 1
=
-5
7
4
4?? 1
=
7
-5
??? 1
=
-5
7
?
7
-5
=
5
-5×221
??? 1
=
5
14
? Point of contact (?? 1
,?? 1
,?? 1
) =(
-5
7
,
-5
7
,
5
14
)
6.2. Show that the locus of a point from which the three mutually perpendicular
tangent lines can be drawn to the paraboloid ?? ?? +?? ?? +?? ?? =?? is
?? ?? +?? ?? +?? ?? =??
(2012 : 20 Marks)
Solution:
Let ?? (?? 1
,?? 1
,?? 1
) be the point from which three mutually perpendicular lines can be drawn
to the paraboloid
?? 2
+?? 2
+2?? =0 (??)
Then, the enveloping cone of (i) with the vertex at ??
(?? 1
,?? 1
,?? 1
) is
?? ?? 1
=?? 2
(???? )
?? =?? 2
+?? 2
+2?? ?? 1
=?? 1
2
+?? 1
2
+2?? 1
?? =?? ?? 1
+?? ?? 1
+?? +?? 1
? from (ii), we have
(?? 2
+?? 2
+2?? )(?? 1
2
+?? 1
2
+2?? 1
)=(?? ?? 1
+?? ?? 1
+(?? +?? 1
))
2
For three mutually perpendicular generators, coefficient of ?? 2
+ coefficient of ?? 2
+
coefficient of ?? 2
=0
or
?? 1
2
+?? 1
2
+4?? 1
-1 =0
?? 1
2
+?? 1
2
+4?? 1
=0
? Locus of (?? 1
,?? 1
,?? 1
) is
?? 2
+?? 2
+4?? =1
6.3 Two perpendicular tangent planes to the parabolold ?? ?? +?? ?? =?? ?? intersect in a
straight line in the plane ?? =?? . Obtain the curve to which this straight line
touches.
(2015 : 13 Marks)
Solution:
Let the line of intersection of the two planes be:
???? +???? =?? ,?? =0 (??)
Since this lies on the plane ?? =0 (given).
? Equation of the plane through the line (i) is
(???? +???? -?? )+???? =0
???? +???? +???? =?? (???? )
If the plane (ii) touches the paraboloid, then
?? 2
?? 2
+
?? 2
?? 2
+
2????
?? =0 (condition)
i.e.,
?? 2
+?? 2
+2???? =0 (?????? )
This being quadratic in ?? , gives two values of ?? 1
say ?? 1
and ?? 2
such that
?? 1
·?? 2
=
?? 2
+2????
?? (???? )
Also from (ii), the direction ratio's of the normal to the two tangent planes whose line of
intersection is (ii) are ?? 1
,?? ,?? and ?? 2
,?? ,?? .
Also, as these two tangent planes are perpendicular
? ?? 1
·?? 2
+?? ·?? +?? ·?? =0
? (?? 3
+2???? )+?? 2
+?? 2
=0 from (iv)
? 2?? 2
+?? 2
+2???? =0 (v)
Now, we are to prove that the line (i) touches a parabola (to be found). So, we are to find
the envelope of (i) which satisfies the condition (v).
Eliminating ?? between (i) and (?? ) , the equations of the line of intersection of two tangent
planes is :
2?? 2
+?? 2
+2(???? +???? )?? =0,?? =0
? 2(
?? ?? )
2
+2?? (
?? ?? )+(1+2?? )=0,?? =0
It is quadratic in
?? ?? , so its envelope is given by:
?? 2
-4???? =0,?? =0
? (2?? )
2
-4·2(1+2?? )=0,?? =0
? ?? 2
=2(2?? +1),?? =0
This is the required curve.
6.4 Find the volume of the solid above the ???? -plane and directly below the portion
of the elliptic paraboloid ?? ?? +
?? ?? ?? =?? which is cut off by the plane ?? =?? .
(20i7 : 15 Marks)
Solution:
Equation of ?? surface, cut off cut plane
?? 2
+
?? 2
4
=9;?? =9
i.e.,
?? 2
9
+
?? 2
36
=1;?? =9
Making the transformation,
?? =3?? cos ?? ?? =6?? sin ??
?? :0 to 1;?? :0 to 2??
?? (?? ,?? )
?? (?? ,?? )
=|
??? ??? ??? ??? ??? ??? ??? ??? |=|
3cos ?? -3?? sin ?? 6sin ?? 6?? cos ?? |
=18?? ?? =??
?? ????? ×???? =? ?
2?? ?? =0
?? ?
1
?? =0
?(9?? 2
)18?????????? (?? =?? 2
+
?? 2
4
)
=9×18? ?
2?? 0
????? ? ?
1
0
??? 3
????
=9×18×2?? ×
1
4
=81??
6.5 Reduce the following equation to the standard form and hence determine the
nature of the coincoid:
?? ?? +?? ?? +?? ?? -???? -???? -???? -?? ?? -?? ?? -?? ?? +???? =??
(2017 ; 15 Marks)
Solution:
Comparing with
?? (?? ,?? ,?? ) =?? ?? 2
+?? ?? 2
+?? ?? 2
+2?????? +2?????? +2h???? +2???? +2???? +2???? +?? =0
The discriminating cubic is :
|
?? -?? h ?? h ?? -?? ?? ?? ?? ?? -?? |=0 or
|
|
1-?? -
1
2
-
1
2
-
1
2
1-?? -
1
2
-
1
2
-
1
2
1-?? |
|
=0
? 4?? 3
-12?? 2
+9?? =0 or ?? (2?? -3)
2
=0
? ?? =
3
2
,
3
2
,0
As this discriminating cube has two roots equal and third root equal to zero, so it is either
a paraboloid of revolution or a right circular cylinder.
The d.r.'s of the axis are given by ???? +h?? +???? =0,h?? +???? +???? =0,???? +???? +???? =0
??.?? ., ?? -
?? 2
-
?? 2
=0,-
?? 2
+?? -
?? 2
=0,-
?? 2
-
?? 2
+?? =0
??.?? ., 2?? -?? -?? =0,-?? +2?? -?? =0,-?? -?? +2?? =0
These gives: ?? =?? =?? =
1
v3
Read More