CHAPTER
3
Cylinder
3.1. DEFINITION
A cylinder is a surface generated by a straight line which is parallel to a fixed line and intersects a given
curve or touches a given surface.
The fixed line is called the axis and the given curve is called the guiding curve of the cylinder.
Any line on the surface of a cylinder is called its generator.
3.2. EQUATION OF A CYLINDER
(a) To find the equation of a cylinder whose generators intersect the conic
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, z = 0
...(1)
and are parallel to the line x/l = y/m = z/n.
...(2)
Suppose P (x1, y1, z1) be any point on the cylinder. The equation of the generator through the
point P and parallel to equation (2) are
x − x1 y − y1 z − z1
=
=
l
m
n
The generator (3) meets the plane z = 0 in the point
Fx
H
...(3)
I
K
lz1
mz
, y1 − 1 , 0
n
n
Since the generator (3) meets the given conic (1),
1
F
H
I
K
2
−
F
H
lz1
n
+ 2h x1 −
lz1
n
I F y − mz I + b F y
KH nK H
I
K
F
H
2
1
1
1
−
mz1
n
+ 2g x1 −
F
H
or
I
K
lz1
n
mz1
+ 2f y1 −
+c=0
n
a (nx1 – lz1)2 + 2h (nx1 – lz1) (ny1 – mz1) + b (ny1 – mz1)2 + 2gn (nx1 – lz1)
a x1 −
I
K
+ 2fn (ny1 – mz1) + cn2 = 0
80
81
CYLINDER
Thus the locus of (x1, y1, z1) is
a (nx – lz)2 + 2h (nx – lz) (ny – mz) + b (ny – mz)2 + 2gn (nx – lz) + 2fn (ny – mz) + cn2 = 0
This is the required equation of the cylinder.
Corollary 1: The equation of the cylinder whose generators are parallel to the z-axis, then
(l = 0, m = 0, n = 0) put in equation (4), we get
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0
Corollary 2: The equation of the form f(x, y) = 0 represents a cylinder whose generator are
parallel to z-axis.
Corollary 3: The equation of the cylinder whose axis is z-axis and whose generators intersection the circle x2 + y2 = a2, z = 0, is given by x2 + y2 = a2.
3.3. RIGHT CIRCULAR CYLINDER
A right circular cylinder is a surface generated by a straight line passing through the point on a fixed
circle and is perpendicular to its plane.
The normal to the plane of the circle through its centre is called the axis of the cylinder and the
section by a plane which is perpendicular to the axis is called the normal section i.e., a circle.
The radius of the normal section is also called the radius of the cylinder.
The length of the perpendicular from any point on a right circular cylinder to its axis is equal to
its radius.
3.4. EQUATION OF A RIGHT CIRCULAR CYLINDER
To find the equation of a right circular cylinder whose axis is the line
x − x1 y − y1 z − z1
and whose radius is r.
=
=
l
m
n
Suppose P (x, y, z) be any point on the cylinder and PN be the length of the point perpendicular
from the point P on a given line and the given line
passes through the point A (x1, y1, z1).
Let
P (x, y, z)
M
PN = r
AN =
( x − x1 ) l + ( y − y1 ) m + ( z − z1 ) n
l2 + m2 + n2
AP = distance between the point A and P
= ( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2
r
N
A (x1, y1, z1)
82
ENGINEERING MATHEMATICS—II
Now, we have
PN2 = AP2 – AN2
= {(x – x1)2 + (y – y1)2 + (z – z1)2} –
{l ( x − x1 ) + m ( y − y1 ) + n ( z − z1 )}2
l 2 + m 2 + n2
PN2 = p2 = {(x – x1)2 + (y – y1)2 + (z – z1)2} –
Hence
{l ( x − x1 ) + m ( y − y1 ) + n ( z − z1 )}2
l 2 + m2 + n 2
Since, the radius of the given cylinder is r, so by definition of right circular cylinder we have
p = r. i.e.,
r2 (l2 + m2 + n2) = {(x – x1)2 + (y – y1)2 + (z – z1)2} (λ2 + m2 + n2)
– {l(x – x1) + m (y – y1) + n (z – z1)}2
or
or
r2 (l2 + m2 + n2) = [(y – y1) n – (z – z1) m]2 + [(z – z1) l – (x – x1) n]2 + [(x – x1) m – (y – y1) l]2
r2 (l2 + m2 + n2) = Σ[(y – y1)n – (z – z1)m]2
This is required equation of right circular cylinder.
3.5. ENVELOPING CYLINDER
A cylinder whose generator touches a given surface and is directed in a given direction is called an
enveloping cylinder.
3.6. EQUATION OF AN ENVELOPING CYLINDER
To find the equation of the enveloping cylinder whose generator touch the sphere ax2 + by2 + cz2 = 1,
x y z
and are parallel to line = = .
l m n
Let P (x1, y1, z1) be any point on the given enveloping cylinder. The equation of the generator of
the cylinder through the point P and parallel to the line
x y z
= = is
l m n
x − x1 y − y1 z − z1
=
=
= r (say)
l
m
n
The coordinates if any point on the generator (1) are (lr + x1, mr + y1, nr + z1)
...(1)
Suppose the given sphere meets the point (lr + x1, mr + y1, nr + z1), then we have
a (lr + x1)2 + b (mr + y1)2 + c (nr + z1)2 =1
or
2
r2 (al2 + bm2 + cn2) + 2r (alx1 + bmy1 + cnz1) + ax12 + by12 + cz1 = 1
The line (1) will touch given sphere if the equation (2) has equal roots. Therefore we get
(alx1 + bmy1 + cnz1)2 = (al2 + bm2 + cn2) ( ax12 + by12 + cz12 − 1 )
...(2)
83
CYLINDER
Hence the locus of (x1, y1, z1) is
(alx1 + bmy1 + cnz1)2 = (al2 + bm2 + cn2) ( ax12 + by12 + cz12 − 1 )
This is required equation of enveloping cylinder.
3.7. EQUATION OF A TANGENT PLANE TO THE CYLINDER
To find the equation of a tangent plane to the cylinder whose equation is ax2 + 2hxy + by2 + 2gx + 2fy
+ c = 0 at the point P (x1, y1, z1).
Suppose the given equation of the cylinder is
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
...(1)
Since, the point P (x1, y1, z1) line on (1), then
ax12 + 2hx1y1 + by12 + 2gx1 + 2fy1 + c = 0
...(2)
Let the equation of a line which passes through the point P (x1, y1, z1) and whose direction
cosine are l, m, n be
x − x1 y − y1 z − z1
=
=
=r
l
m
n
...(3)
any point on the line (3) is (lr + x1, mr + y1, nr + z1) and the given cylinder (1) are given by
a (lr + x1)2 + 2h (lr + x1) (mr + y1) + b (mr + y1)2 + 2g (lr + x1) + 2f (mr + y1) + c = 0
r2 (al2 + 2hlm + bm2) + 2r[l(ax1 + hy1 + g) + m (hx1 + by1 + f)]
+ (ax21 + 2hx1y1 + by12 + 2gx1 + 2fy1 + c) = 0
Using equation (2), we get
r2 (al2 + 2hlm + bm2) + 2r [l(ax1 + hy1 + g) + m(hx1 + by1 + f)] = 0
...(4)
One root of this equation is zero. This given line (3) will be a tangent line at (x1, y1, z1). If the
other root is also zero. Equation (4) other root is zero if
l [hx1 + by1 + g] + m [hx1 + by1 + f}] = 0
Eliminating l, m, n between equation (3) and (5), we get
(x – x1)[hx1 + by1 + g] + (y – y1)[hx1 + by1 + f] = 0
or
x (ax1 + hy1 + g) + y (hx1 + by1 + f) + gx1 + fy1 + c
= ax12 + 2hx1y1 + by12 + 2gx1 + 2fy1 + c = 0
Using equation (2), we get
x (ax1 + hy1 + g) + y (hx1 + by1 + f) + gx1 + fy1 + c = 0
This is the required equation of tangent plane to the cylinder.
...(5)
84
ENGINEERING MATHEMATICS—II
Corollary 1: The tangent plane at the point (x1, y1, z1) to the cylinder
ax12 + 2hx1y1 + by12 + 2gx1 + 2fy1 + c = 0
...(1)
x (ax1 + hy1 + g) + y (hx1 + by1 + f) + (gx1 + fy1 + c) = 0
or
axx1 + h (xy1 + x1y) + byy1 + g (x + x1) + f (y + y1) + c = 0
this equation is obtained by replacing x2 by xx1, y2 by yy1, 2x by x + x1, 2y by y + y1 and 2xy by xy1 + yx1
in (1).
SOLVED EXAMPLES
Example 1. Find the equation of a cylinder whose generators are parallel to the line
x = y/2 = – z and passing through the curve is 3x2 + 2y2 = 1, z = 0.
Sol. The equation of the giving curve is
3x2 + 2y2 = 1, z = 0
...(1)
The equation of the giving line is
x y
z
= =
...(2)
1 2 −1
Let us consider a point P (x1, y1, z1) on the cylinder. The equation of generator through the point
P (x1, y1, z1) which is a line parallel to the given line (2) are
x − x1 y − y1 z − z1
=
=
...(3)
1
2
−1
The generator (3) meets the plane z = 0 in the point given by
x − x1 y − y1 z − z1
=
=
i.e., (x1 + z1, y1 + 2z1, 0)
1
2
−1
Since the generator (3) meets the conic (1). Hence the point (x1 + z1, y1 + 2z1, 0) will satisfy the
equation of the conic given by (1), we have
3 (x1 + z1)2 + 2(y1 + 2z1)2 = 1
or
3 ( x12 + 2x1z1 + z12 ) + 2( y12 + 4y1z1 + 4 z12 ) = 1
x12 + 6x1z1 + 11 z12 + 2 y12 + 8y1z1 – 1 = 0
The locus of P (x1, y1, z1) is
3x2 + 6xz + 11z2 + 2y2 + 8yz – 1 = 0
This is required equation of the cylinder.
Example 2. Find the equation of the circular cylinder whose generating lines have the direction
cosines l, m, n and which pass through the fixed circle x2 + y2 = a2 in ZOX plane.
Sol. The equation of the guiding curve (circle) are
x2 + y2 = a2, ZOX lane i.e., y = 0
...(1)
85
CYLINDER
let us consider a point P (x1, y1, z1) on the cylinder. The equation of generator through the point P (x1,
y1, z1) and with direction cosine l, m, n are
x − x1 y − y1 z − z1
=
=
l
m
n
the generator (2) meet the plane y = 0 in point given by
x − x1 y − y1 z − z1
=
=
l
m
n
i.e.
...(2)
Fx
H
1
−
ly1
ny
, 0, z1 − 1
m
m
F
H
Since, the generator (2) meets the curve (1). Hence, the point x1 −
I
K
ly1
ny
, 0, z1 − 1
m
m
I will satisfy
K
the equation of the curve given by (1), we have
Fx
H
or
2
I F
K H
I
K
ly1
ny1 2
z
−
+ 1
= a2
1
m
m
(mx1 – ly1)2 + (mz1 – ny1)2 = a2m2
−
The locus of P (x1, y1, z1) is
(mx1 – ly)2 + (mz1 – ny1)2 = a2m2
This is the required equation of cylinder.
Example 3. Find the equation of the cylinder which intersects the curve ax2 + by2 + cz2 = 1,
lx + my + nz = p and whose generators are parallel to x-axis.
Sol. The given equation of the guiding curve are
and
ax2 + by2 + cz2 = 1
...(1)
lx + my + nz = p
...(2)
Since, the generators of the cylinder are parallel to x-axis, so the equation of the cylinder will
not contain terms of x. Thus the equation of the cylinder will be obtained by eliminates x between
equation(1) and (2), we get
a
(p – my – nz)2 + by2 + cz2 = 1
l2
a(p – my – nz)2 + bl2y2 + cl2z2 = l2
or
a (am2 + bl2)y2 + (an2 + cl2)z2 + 2amnyz – 2amby – 2anbz + (ap2 – l2) = 0
This is the required equation of the cylinder.
Example 4. Find the equation of a right circular cylinder described on the circle through the
three points (1, 0, 0) , (0, 1, 0) , (0, 0, 1) are guiding circle.
Sol. Let the given three points A (1, 0, 0), B (0, 1, 0), C (0, 0, 1).
The equation of the sphere OABC is x2 + y2 + z2 – x – y – z = 0
and the equation of the plan ABC is
x+y+z=1
86
ENGINEERING MATHEMATICS—II
Therefore, the equation of the circle ABC is
x2 + y2 + z2 – x – y – z = 0
and
x+y+z=1
...(1)
Since, the cylinder is a right circular cylinder, then the axis of the given cylinder is perpendicular to the plane x + y + z. So direction ratio of the axis are (1, 1, 1).
The generator through (x1, y1, z1) and parallel to the axis has equation
x − x1 y − y1 z − z1
=
=
=r
1
1
1
Any point on this line (r + x1, r + y1, r + z1)lies on the circle (1), if
r + x1, r + y1, r + z1 = 1
or
and
3r = 1 – (x1 + y1 + z1)
(r + x1)2 + (r + y1)2 + (r + z1)2 – (r + x1 + r + y1 + r + z1) = 0
...(2)
...(3)
Multiply by 3 in equation (3) and using (2), we get
2
2
2
[1 – (x1 + y1 + z1)]2 + 2 (x1 + y1 + z1)[1 – (x1 + y1 + z1)] + 3[ x1 + y1 + z1 − 1] = 0
or
3 ( x12 + y12 + z12 − 1 ] – (x1 + y1 + z1)2 + 1 = 0
or
x12 + y12 + z12 – x1 y1 + y1z1 – x1z1 = 1
The locus of P (x1, y1, z1) is or x2 + y2 + z2 – xy – yz – zx = 1.
This is required equation of the right circular cylinder.
Example 5. Find the tangent plane to the cylinder 3x2 + 8xy + 5y2 + x + 7y + 6 = 0 at the point
(1, –1, 2).
Sol. The tangent plane at the point (x1, y1, z1) to the cylinder
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
is
...(1)
axx1 + h (xy1 + x1y) + byy1 + g (x + x1) + f (y + y1) + c = 0
The tangent plane the given cylinder at (1, – 1, 2) is
3x (1) + 4[x (– 1) + 1(y)] + 5y (– 1) +
or
1
7
(x + 1) + (y – 1) + 6 = 0
2
2
x – 5y + 6 = 0
This is the required equation of the tangent plane.
Example 6. Find the equation of the sphere enveloping cylinder of the sphere x2 + y2 + z2 – 2x
+ 4y – 1 = 0 having its generation parallel to the line x = y = z.
Sol. The equation of the given sphere is
x2 + y2 + z2 – 2x + 4y – 1 = 0
...(1)
the generators of the enveloping cylinder are parallel to the line
x=y= z
...(2)
87
CYLINDER
let us consider a point P (x1, y1, z1) on the given enveloping cylinder. The equation of the generator
through the point P (x1, y1, z1) and parallel to the line
x = y = z is
x − x1 y − y1 z − z1
=r
=
=
1
1
1
...(3)
the point of intersection of the line (3) and the given (1) are given by
(r + x1)2 + (r + y1)2 + (r + z1)2 – 2(r + x1) + 4(r + y1) – 1 = 0
or
2
2
2
3r2 + 2 (x1 + y1 + z1 + 1)r + ( x1 + y1 + z1 – 2x1 + 4y1 – 1) = 0
...(4)
For enveloping cylinder, the equation (4) must have equal roots. This requires
(x1 + y1 + z1 + 1)2 = 3( x12 + y12 + z12 – 2x1 + 4y1 – 1).
or
x12 + y12 + z12 – y1z1 – x1y1 – 4x1 + 5y1 – z1 – 2 = 0
The locus of P (x1, y1, z1) is
x2 + y2 + z2 – yz – xy – 4x + 5y – z – 2 = 0
This is the required equation of the enveloping cylinder.
Example 7. Find the equation of the right circular cylinder whose axis is x – 2 = z, y = 0 and
which passes through the point (3, 0, 0).
Sol. The given equation of the axis of the cylinder is
x−2 y−0 z−0
=
=
1
0
1
...(1)
We know r = the length of the perpendicular for a point (3, 0, 0) on the cylinder to the axis (1)
=
1
(1) + 0 + 1
2
2
2
[( 0 − 0 . 0) 2 + {1. 0 − 1(3 − 2)}2 + {0 .(3 − 2) − 1. 0}2 ] =
1
2
Let us consider a point P (x1, y1, z1) on the cylinder. The length of the perpendicular from the
point P to the given axis (1) is equal to the radius of the cylinder. i.e.,
{1 . y – 0 . z}2 + {1 . z – 1 . (x – 2)}2 + {0 . (x – 2) – 1 . y}2 =
FG 1 IJ
H 2K
2
(1 + 0 + 1)
y2 + (z – x + 2)2 + y2 = 1
x2 + 2y2 + z2 – 2zx – 4x + 4z + 3 = 0
This is the required equation of the right circular cylinder.
EXERCISE 3.1
1. Find the equation of the cylinder whose generators are parallel to the line x/1 = y/ – 2 = z/3 and whose
guiding curve is the ellipse x2 + 2y2 = 1, z = 0.
88
ENGINEERING MATHEMATICS—II
2. Find the equation of the cylinder whose generators are parallel to the line x/1 = y/ – 2 = z/3 and whose
guiding curve is the ellipse x2 + 2y2 = 1, z = 3.
3. Find the equation of the cylinder whose generatoring lines have the direction cosines l, m, n and which
passes through the fixed circle is the ellipse x2 + z2 = 1 in the ZOX plane.
4. Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and passes
through the curve x2 + y2 = 16 and z = 0.
5. Find the equation of the cylinder whose generators are parallel to the line x/4 = y/ – 2 = z/3 and which
intersects the ellipse 4x2 + 2y2 = 1, z = 0.
6. Find the equation of the surface generated by a straight line which is parallel to the line y = mx, z = nx and
intersects the ellipse x2/a2 + y2/b2 = 1, z = 0.
7. Find the equation of the cylinder with generators parallel to z-axis and passes through the curve ax2 + by2
= 2cz, lx + my + nz = p.
8. Find the equation of the cylinder with generators parallel to x-axis and passing through the curve ax2 + by2
= 2cz, lx + my + nz = p.
9. Find the equation of the right circular cylinder of radius 2 whose axis is the line (x – 1)/2 = y/3
= (z – 3)/1.
10. Find the equation of the right circular cylinder whose axis is (x – 2)/2 = (y – 1) = z/3 and which passes
through (0, 0, 1).
11. Find the equation of the right circular cylinder of radius 4 whose axis is the line x =2y = – z.
12. Find the equation of the right circular cylinder of radius 3 whose axis is the line (x – 1)/2 = (y – 3)/2
= (z – 5) / – 7.
13. Find the equation of the right circular cylinder whose guiding circle is x2 + y2 + z2 = 9, x – 2y + 2z = 3.
14. Show that the coordinate of the foot of perpendicular from a point P (x1, y1, z1) on the line x = y = – z are
1
1
1
(x1 + y1 – z1), (x1 + y1 – z1), (x1 + y1 – z1).
3
3
3
15. Find the equation of the right circular cylinder whose guiding circle passes through the points (a, 0, 0),
(0, b, 0), (0, 0, c).
16. Find the equation of the right circular cylinder with generators parallel to z-axis and intersect the surfaces
ax2 + by2 + cz2 = 1, lx + my + nz = p.
17. Find the equation of the right circular cylinder of radius 2 whose axis is the line (x – 1)/2 = (y – 2) =
(z – 3)/2.
18. Find the equation of the right circular cylinder whose one section is the circle x2 + y2 + z2 – x – y – z = 0,
x + y + z = 1 is x2 + y2 + z2 – yz – zx – xy = 1.
19. Find the equation of the right circular cylinder of radius 2 whose axis passes through (1, 2, 3) and has
direction cosines proportional to 2, 3, – 6.
20. Find the equation of the cylinder whose generating line are parallel to the line x/l = y/m = z/n and which
touches the sphere x2 + y2 + z2 = a2.
21. Find the enveloping cylinder of the sphere x2 + y 2 + z 2 = 1 having its generators parallel to the line
x/1 = y/2 = z/3.
22. Show that the enveloping cylinder of the conicoid ax2 + by2 + cz2 = 1 with generators perpendicular to the
z-axis meets the plane z = 0 in parabolas.
89
CYLINDER
23. Find the equation of a right circular cylinder which envelopes a sphere of centre (a, b, c) and radius r, and
has its generator parallel to a line with direction ratio l, m, n.
24.
Find the equation of the right circular cylinder of radius 5 and having for its axis the line x/2 = y/3 = z/6.
25. Find the equation of a cylinder whose generator touches the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0
whose generators are parallel to the line x/l = y/m = z/n.
26. Prove that the enveloping cylinder of ellipsoid
x 2 y2 z 2
=
=
= 1,whose generator are parallel to line
a2 b2 c2
x
y
z
=
= meet the plane z = 0 in circles.
2
2
0 ± a −b
c
ANSWERS
1. 9 (x2 + 2y2 + z2) – 6xz + 24yz = 9.
2. 3x2 + 6y2 + 3z2 + 8yz – 2zx + 6x – 24y – 18z + 24 = 0.
3. (mx – ly)2 + (mz – ny)2 = m2.
4. 9x2 + 9y2 + 5z2 – 6zx – 12yz – 144 = 0.
5. 4 (3x – 4z)2 + 2 (3y + 2z)2 = 9.
6. b2 (nx – z)2 + a2 (ny – mz)2 = a2b2n2.
7. anx2 + bny2 + 2c(lx + my) – 2pc = 0.
8. (am2 + bl2) y2 + an2z2 + 2amnyz – 2apmy – 2 (apn + cl2)z = 0.
9. 10x2 + 5y2 + 13z2 – 12xy – 6yz – 4zx – 8x + 30y – 74z + 59 = 0.
10. 10x2 + 13y2 + 5z2 – 4xy – 6yz – 12zx – 36x – 18y + 30z – 35 = 0.
11. 5x2 + 8y2 + 5z2 – 4xy + 4yz + 8zx – 144 = 0.
12. 5x2 + 5y2 + 8z2 – 8xy + 4yz + 4zx – 6x – 42y – 96z + 225 = 0.
13. 8x2 + 5y2 + 5z2 + 4xy + 8yz – 4zx – 72 = 0.
F1
Ha
I
K
F
H
IF
KH
x y z
x y z
1
1
+ + −1
+ + −2
+
(x2 + y2 + z2 – ax – by – cz) =
a b c
a b c
b2 c2
16. (an2 + cl2) x2 + (bn2 + cm2) y2 + 2clmny – 2cplx – 2cpmy + cp2 – n2 = 0.
15.
2
+
I
K
17. 5x2 + 8y2 + 5z2 – 4xy – 4yz – 8zx + 22x – 16y – 14z – 10 = 0.
19. 45x2 + 40y2 + 13z2 – 36yz – 24zx + 12xy – 42x – 280y – 126z + 294 = 0.
20. (lx + my + nz)2 = (l2 + m2 + n2) (x2 + y2 + z2 – a2).
21. (x + 2y + 3z)2 = 14 (x2 + y2 + z2 – 1).
22. ab (mx – ly)2 = (al2 + bm2), z = 0.
23. {l (x – a) + m (y – b) + n (z – c)}2 = (l2 + m2 + n2){(x – a)2 + (y – b)2 + (z – c)2 – r2}.
24. 45x2 + 40y2 + 13z2 – 12xy – 36yz – 24zx – 1225 = 0.
25. {l (x + u) + m (y + v) + n (z + w)}2 = (l2 + m2 + n2) ( x2 + y2 + z2 + 2ux + 2vy + 2wz + d).
26. x2 + y2 = a2, z = 0.