UNIT 6 CONES AND CYL13'!'23ERS

Structure

6.1 Inuoduction Objectives

6.2 Cones

6.3 Tangent Planes

6.4 Cylinders

6.5 Summary

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6.1 PNTRQDUCTHON

In ~ b c previous unit we discusscd a very commonly round thrce-dimensional object. In lhis tinit we look at two more common1 y found three-dimensional objects, namdy, a cone and a cylinder. But, what you will see in this unit may surprise you-whal people usually call a cone or a cylinder are only portions of very particular cases of what mathematicians rcfcr to as a cone or a cylinder.

We shall start our discussion on cones by defining them, and deriving their equations. Then wc shall concentrate on cones whose vertices are the origin. In particular, we will obtain the langent planes of such cones.

t

The other surface that we will discuss in lhis unit is a cylindcr. We shall define a general cylindcr, and then focus on a right circular cylinder.

'The contcnts in this unit are of malhematicnl intercst, of coursc. But, they arc also of intercst to astronomers, physicists, engineers and architects, among otl~crs. This is because of the many applications that cones and cylinders have in various ficlds of science and enginecr- 1ng.

The surfaces that you will study in this unit are particular cascs of conicoids, which you will study in thc ncxl block. So if you go through Lhis unit carefully and ensure ha t you achieve llie I'ollowing objectives, you will find the next block msicr lo undcrslanti.

Objectives

Altcr sludying this unit you should be able Lo

e obtain h e equation of a conc if you know its vertex and base curve;

prove and use the fact that a sccond degree equation in 3 variables represents a cone with vertex at the origin i1 it is homogeneous;

a obtain the tangent planes to a conc;

0 oblain the equation of a right circular cylinder if you know ils axis and base curve.

Let us now see what a cone is.

The Sphere, Cone and Wintter 6.2 CONES

When you see an ice-cream cone, do you ever think that it is a set of lines? That is exactly what it is, as you will see in this section.

Definition: A cone is a set of lines hat intersect a given curve and pass through a fixed point which is not in ~ h c plane of the curve. The fixed point is called the vertex of the cone, and the curve is called the base curve (or directrix) of the cone.

Each line that makes up a cone is called a generator of the cone.

Thus, we can also define a cone in the following way.

Definition: A cone is a surface generated by a line that intersects a given curve and passes through a fixed point which doesn't lie in the plane of the curve.

For example, in Fig. I (a), we give the cone generated by a line passing through the point P, and intersecting the circlc C. The basc curves of the concs in Fig. 1 (b) and Fig. 1 (c) are an ellipse and a parabola, rcspcctively.

Fig.! : (a) A circular coac, (b) an dliptic cone, (c) a parvbnlic cone.

At this point we would like to make an important rcmark.

Remark 1: A cone is a.set of lines passing through its vertex ind base curve. Thus, it cxtcnds beyond the vertex and the base curve. So the concs in our diagrams arc only a portion ,of the irctual concs.

Now let us inlroduce some new terms. ,

Definitions: A cone who'sc base curve is a circle is called a circular cone. The line joining I the vemx of a circular cone to the centre of ils base curve is called the axis of the cone. If

the axis of a circular cone is perpendicular ~o the plane of the basc curve, then the cone is I

called a right circular cone,

Thus, the cone in Fig. 1 (a) is a right circular cone, while the one in Fig. 2 is not.

i:ig. 2: A circular cunc which is not Cones were given great imporlance by the ancient Greeks who were studying the problem of right ctrcular. doubling the cube. A teacher of Alexander the Great, Menaechrnus, is supposed to have

geomeuically proved the following result. This result is the reason for the continuing importance of cones.

Theorem 1: Every planar section of a cone is a conic.

This theorem is tlle reason for an ellipse, parabola or hyperbola to be called a conic section Cones a n d Cylinders

(see Fig. 1 of Unit 3). It was proved by the Greek astronoiner Appolonius (approximately 200 B.C.). We will not give the proof here.

NOW, according to Theorem 1 would you call a pair of intersecting lines a conic? If you cut a right circular cone by a plane that contains its axis, what will the resulting curve be? Sec Fig. 3.

LCL us now see how we can represent a cone algebraically. We shall first talk about a right circular cone, which we'shall refer to as an r.c. cone.

So, let us mke an r.c. cone. Let us assume that its vemx is at the origin, and is axis ishe -.

z-axis (see Fig. 4). Then the base cuke, which is.a circle of radius r (say), lies in.a . plane .. . that . . ,

is parallel to the XY-plane. Let this plane be z = k, where k is a constant. Then, any gcnerator will intersect this curve in a point (a, b, k), for.somc a, b E. R. So the angle'

1'1~. 3: A pair of lnterstctlng lin& between the generator and the axis of the cone will be 9 = tan-' , which is a conslant. Is n conic scctlon.

This is true for any generator of the conc.

Thus, cvery line that makes up the cone makes a fixed angle 0 with the axis of the cone. l l i s angle is called the semi-vertical angle (or generating angle) of the cone.

Wc can now define an r.c. cone in thc following way.

Definition: A right circular cone is a,surface generated by a line which passes through a fixed point (its vcrtex), and makes a constant angle with a fixcd line through the fixed point. i

Lct us obtain the equation of an r.c. cone in terns sf iu scmi-vcrtical angle. Let us assume [hat thc vertex of the cone is 0 (0,0,0) and axis is the z-axis. (We can always choose ow coordinate system in this manner.) Now take any point P (x, y, z) on the cone (see Fig. 5). Then, the direction ratios of OP are x, y, z, and of the conc's axis are 0,0, 1. Thus, from Equation (9) of Unit 4, we get

z cos 0 - ------- I(: J7Tps Pig. 4: A right rircvlar conc wllh

~ h u s , x2+y2+~Z - i? S ~ C ~ 9 vertex st the origin and brzw c w r v c ~ ~ + ~ ' = r ~ , z=k. + x ~ + ~ ~ = z2 tan2 e. ......( 1)

(1) is called the standard form of the equation ofa right circular cone.

Now, why don't you try the following cxercist?s?

x- a y - b El ) Show that the equation of thc r.c, cot~e with vertex at (a, b, c), axis - = -

01 P

- 2 - C - ---- and semi-vertical angle 8 is Y

[a (x-a) +P(y-11) + y (z-c)] "a2i-B2-ky2) ((x-a)" (y-b)? + (z-c)" ccos28 ..... . (2)

E2) Can you deduce (1) from (2) ?

E3) Find :lie equation of the r.c. cone whose axis is the x-axis, vertex is the origin and

R semi-vertical angle is , . ' 3 Y*

Let us now look at a cone whose vertex is the origin. In this situation we have the following Iiip. 5: x~ + y~ = z1

rcsult.

The Sphere, Cone and Theorem 2: The cquatibn of a cone whose base curve is a conic and whose vertex is (Iylintler (O,O,O) is a homogencous cquation of degrcc 2 in 3 variablcs.

An equation is h o m o g e ~ ~ c o u s of Proof: Lct us assumc that thc basc curvc is lhc conic llcgrcc 2 i f cnch 01 ils Lcrrns i s of tlcgrce 2. ax2+ 2hxy + by2 + 2gx + 2fy + c = 0, z = k.

Any gcncralor of the cone passes through (0, 0,b). Thus, it is or the rorm

x y z A = - = - ..... P Y'

This line intcrsccts thc plane z = k at the point (7, 5. t). This poinl should lic on the conic. Thus,

Eliminating a, P, yfro~n this quation and (3), we gct

This is the cquation of thc conc. As you can scc, i t is homogeneous or dcgrec 2 in the 3 variablcs x, y and z.

x2 y2

For cxamplc, thc equation of the conc whose base curve is Llic cllipse - + - = 1 in the 4 9

plane z = 5 , and whosc vcrlcx is tlic origin, is

Do you sce apattcm in thc way we obtain thc ccjuation of the cone rrom the equation of the base curve? The following remark is about his.

Remark 2: To find thc cquation or thc cone with vcrtcx al(O,O, 0) and basc curve in the planc Ax + By + Cz = D', D f 0, wc simply Iiomogenise the equation or the curve, as

v Ax+By+Cz follows. Wc multiply ~ h c lincar terms by , and thc cons~n t tcrni by

D

[Ax + + )"; and wc lwve the quadratic terms as thcy are. The equation that wc get

by his process is al~o~nogeneouseqi~ation of degrec 2, and is the equation of the conc.

Let us look at a fcw exarnplcs or concs with their vcrticcs at Ulc origin.

Example 1: Show that thc crluatiori ol the cone with the axes as generators is fyz + gzx + hxy = 0, where T, g, h E R.

Solution: By Thcorcm 2, thc equation of the conc is

ax2+by2+cz2+2hy + 2gzx + 2hxy = 0, for solnc a,b, c, f, g, h E R .

Since the x-axis is a gcncrator, (1,0,O) lies on it. Therefore, a = 0. Similarly, as it passcs through (O,1,0) and (0,(!, I), b = c = 0. So the equation becomes Tyz + gzx + hxy = 0.

Example 2: Find thc cquation of the cone with vertex at the origin, and whose base curve is the c i r c ~ e ' x ~ + ~ ~ + z ~ = Id, x+2y+2z = 9.

I

Sulution : On holnogenising the cquation 01' thc sphere, we gct Cones and Cylinders

This is a second degree homogeneous equation in x, y, z and passes Lhrough the circle. Hcnce, it is the required equation of the cone.

Thc following excrciscs will give you somc practice in homogenising equations.

E4) Find Lhe equation of the cone with vertex at thc origin and base curve

a) the parabola y2 = 4ax, z = 3,

Y2 z2 b) the ellipse - + - = 1, x = -2. 3 5

E5) Find the equation of the conc passing Lhrough 2x2 + 3y2 + 4z2 = 1 and x + y + z = 1. and with vertex at the origin.

Let us go back to Theorem 2 now Do you think its converse is true? Consider the following rcsull.

l'heorem 3: A homogeneous equation of the second dcgrec in 3 variables represents a cone whose vertex is at thc origin.

Prook Lct the given cquation be

ax2+ by2 + cz2 + 2tyz + 2gzx + 2hxy = 0. .. . . . .(4)

Lct P (a , p, y) be a point on this surface and 0 the origin. Thcn OP is given by the equations

x y z -=-=-= r (say). a P Y

So any poiat on OP is (ra, rp, ry). Sincc P lics on (4),

acx2+ bp2+ c f+ 2fpy + 2gy a + 2hap = 0. .. . . ..(5) Mul~iplying (5) lhroughout by r2, we get

a(m)' + b (rp12 + c(ry12 + 2f(rp) (ry) + 2g (ry) (ra) + 2h (m) (rp) = 0.

Thus, (m, rp, ry) also lies on (4), for any r E R. In particular, 0 lics on (4). So, the line OP lies on lhc surface given by (4). In othcr words, OP is a gcncrator of (4). Thus, thc surface (4) is generated by lines through thc origin. Eqch of hese lines will also pass through any curvc obtained by intersecting (4) by a plane, and any' of thcse cuwcs can be treatcd as a base curve. Thus, (4) represents a conc with the origin as vertex.

So, from what you have seen so far in this scclion; whcnevcr you come across a homogeneous cquation in 3 variables ~f degree2, you know that it represents a cone.

expressions. Thus, in his casc (4) reprrscnls a pair of planes containing the origin. We shall consider this case as a degenerate cone, and any point on h c linc of intcrscction ofthe two planes can be considered as its vertcx.

Remark3: If

Using Thcorem 3, we can show Lhat if a, P, y E R, then q hornogcneous equation in x - a, Y - p, z -y represcnis a cone wilh vcrtcx at (a, p, y). (We shall discuss this kind of shifting in dctail in Unit 7.)

1

Why don't you uy some exercises now?

a h g h b 1:

6 f C

= 0, then (4) can bc writtcn as a product of two lincar

'I'he Sphere, Cone and Cylinder X Y Z

E6) If - = - = - is a generator of the cone given by the homogeneous equation (4), then u P - Y

show that (a, p, y) lies on (4).

E7) Which of the following equations represents a cone?

3x+4y+5z=o;x2+y2+z2=9;3(x2+y2+z2)=xy;xyz= Y Z + Z X + X ~ .

uL v L w L E8) If - + - + - = d, show that ax2+ by2+ cz2+ 2ux + 2vy + 2wz + d = 0 represents

a b c a cone.

Let us now go back to Example 1. This is an example of a cone with three mutually perpendicular generators. Itc equation has no term containing x2, or z'. Does this mean i h a ~ whenever a cone has thrce mutually perpendicular generators, its equation must have no term with x2, Y2 or z2? The following theorem helps us to answcr this question. I

Theorem 4: If the cone ax2+ by2+ cz2+ 2fyx + 2gzx + 2hxy = 0 has 3 mutually perpendicular generalors, hen a + b + c = 0.

ProoT: Let the direction cosines of Lhe three. mutually perpendicular generators be li, mi, Q,

where i = 1,2,3. Since they are mutually pcrpcndicular, we can rotate our coordinate system so that these lines become the new coordinate axes.

Then he direction cosines of the previous coordinate axes with respect to the ncw axes are 1,. 12, 13; ml, m2, m3 and n,, n2, n3, respectively.

So Unit 4 (Equations (3).and (10)) lell us that

lI2 + 122 + 1; = 1

r n 1 2 + m ~ + m 3 2 = 1 1

Further, since the perpendicular lines arc generators of the cone, using E6 we get

Adding these e~uations, and using (6), we gel a + b + c = 0.

If the lines of interseaion of a cone and a plane through thc cone's vertex are imaginary, the intersection rcduccs to a single point, namely, the . cone's vertex (as in Fig. l{e) of Unit 3).

Actually, the converse of ibis rcsult is also true. The proof uses a fact that you have already scen in Fig. 3 in the case of an r.c. cone, namcly,

any plane through the vertex of a cone intersects the cone in two lines, which may or may not be distinct.

The following result, which wc shall not prove, tells us about the angle bctween the lines of intersection.

Theorem 5: The angle between the two lines in which the plane ux + vy + wz = 0 intersects the cone C(x, y, z) = ax2 + by2 + cz2+2fyz + 2gzx + 2hxy = 0 is

Cones and Cylinders

Looking at (7), can you give the condition under which the angle will be n,2?

whcrc p2 =

Thc lincs of intersection of the plane and the cone will be perpendicular iff

a h g u

h b f v

& f C W

U v w 0

Lct us usc (8) LO solvc h e following example, which includes the converse of Theorem 4.

Example 3: Show chat i l a + b + c = 0, lhen [he conc

C(x, y, Z) =ax 2 + byZ+ cz2+ 2fyz + 2gzx + 2hxy = O

has infinitely many s a s of three mutually perpendicular gcncralors.

x y z Solution: Let - = - = - be any generator of [lie conc. Thcn, by E6, we know that

a P Y C (a, (1, y) = 0. Therefore, using thc fact that a+b+c = 0 and (8), we sce [hat he plane ax + $y + y z = 0 intcrscc~~ thc cone in two mutually pcrpendicular gencralors, say Land L'.

Now X = Y = 5 is normal to h c planc ax + py + y z = 0. Thus, it is perpendicular to both L a P Y

and I.'. 'Thus, thcsc thrcc lines form a scl ol'lhrcc mulually pcrpendicular gencralors of he conc.

\

x y z Nolc that wc chosc - = - = - arbitrarily. 'fhus, for each generiltor chosen we get n set of U P Y

lliree nnutually pcrpendicular generalors. Hcnce, thc conc has infinitely many such sets of gcncralors.

Why don'l you try somc cxcrciscs now?

EV) Find the anglc bctwccn thc lincs of intzrscction of 3 x + y + 5 z = O a n d 6 y z - 2 z x + 5 x y = O .

ElO) Prove that ax + by + cz = 0, whcrc abc ;t 0 , cuts the conc yz + zx + xy = 0 in

1 1 1 pcrpcndicular lincs ifl' - + - + - = O. a 13 c

El 1 Provc chat i f a right circular cone has threc mutually pcq~endiculx generators, its semi-vertical anglc is tan-' \IZ

Lct us now discuss thc intcrscclion of a line and a cone.

I 6.3 'FANGENT PLANES

In thc prcvious unit you saw h a i a line can intersect a sphere in at mosl two poinls. What do you C X ~ ? C L ' ~ in ~11c case of he inlcrscctiors of a line and a cone? Let's see.

Let thc cquadon of Lhe cone be (4), that is, ax2 + by% cc2 + 2fyz + 2gzx + 2hxy = 0.

'fhe Sphere, Cone and Cylinder

Note that, I)y shifting thc origin i f ncccssnry, wc can always assunx this cqualion as thc conc's ccluation.

For convcnicncc, we will wrilc C (x, y, z).= ax2+ byZ+ cz2+ 2fyz + 2gzx + 2hxy.

Now consider Lhc line ?!.! = X'= 3. Any point on Lhis lint is given by ri. P Y

(x, + ra, yl + rp, zl + I-/), for some r E R. Thus, the line will intersect Lhe conc, if this point lics on thc conc for some r E R.

This happens i f

a (xl+ ra)'+b (y + rp)'+c (zI + r y12+2f (Y1 + rp) ( L ~ + r y) + 2g (zl + r y) (xl + ra) + 2h (xl + ra) (y + rp) = 0.

H ?C (a, P, y) + 2 r ( a (ax, + hyl + gzl) + (hxl+ by1 + fzl) + y ( g x ~ + f y ~ + c ~ I ) )

Now, if(xl, yl, z,) docsn't lie on the conc, thcn (9) is a quadratic in r, and hcnce has two roots. Each root corresponds to a point of interscction of the line and the cone. Thus, wc havc just proved the following result.

Theorem 6: A straight line, passing thhugh a point not on cone, mecls thc cone in at most two points.

x-Xl - y-y1 2 - 2 1 Now suppose that ~ h c linc - - -- = - is a kngcnt to the cone (4) a1 (xl, yl, zl). CY P Y

Thcn, since (xl , y I , zl) lies on Lhc conc, C (xl, y zl) = 0. So (9) bc'comes

This cquation must havc coincitlcnt roots, sincc'thc line is a tangcnt to thc conc at

( X I , y l l zl). Thc condition Ijr thisis

a (axl + hyl + gzl)+ P (hx, + byl + fq) + y (gxl + fy, + cxl) =.0. ......( 10)

X - X z - %I So, (10) is Lllc condition for 1 = Y = -- LO be langcnt to the cone a P Y

N o t that (1 0) is satisfied by infinitely Inany values of a , P, y. Thus,

at each point of a cone we can draw inl'initely mimy tangents to tlie cone.

Now, from S e c . 4.3.3, you know ha1 (I 0) ~clls us hat etch of thcsc lincs is perpcndicular to thc linc with direction ratios

axl+ hyl+gzl, hxl+ byl+l 'zl ,gxl+ fyl+czl.

Thus, thc sct of all thc Langcnt lines at (x,, y z,) is thc plwc #

.(x - XI) (axl + hyl + gxl) + (y- yl) (hxl + byl+ Szl) + (z-zl) (gxl + fyl -t czl) = 0

* x (axl + hyl+ gzl)+ p (hx~ + by1 + fzl) + I , (gxl + fyl + cz,) = 0, ......( 1 1)

since C(x I , y 1, zl) = 0.

This plane is defined to bc thc tilngent plane to the cone at (xl, y l , z,).

Thus, (I 1) is the cquation of thc langcnt planc at (xl, y zl) to thc cone

ax2+ by2+ cz2+ 2 f y ~ + 2gzx + 2hxy = 0.

Thcrc is a very simple working rulc for writing (1 1).

Rule of thun~b: To write [he equation of the tangent planearl any point (a, P, y) on Lhe cone

(41, rcplacc x2 by ax, y2 by py, z2 by yz, 2yz by yy + PL, 2zx by c c ~ + y x and 2xy by px+*y . For cxamplc, thc tangcnt plane to the cone 2x2 + y2 - 2x2 = o at (I , O , I ) is 2x (I)*+ y (01 - (x + z) = 0, that is, x = z.

So far wc havc only found the tangent plane lo a cone whose vertcx is at the origin. What about a gcneral cone? The following remark is about this.

Remark 4: We can find h e Langcnl plane lo a cone with vemx at (a, b, c) in the sanhe manner as above. All wc nccd lo do is Lo shift Ihe origin Lo (a, b, c) and find the tangent planc in thc ncw coordinate system. And then wc can shift back to h e old coordinate systcm. making the rcquircd transformations in thc cquation of the plane. This will give us thc rcquircd equation.

Now, if y5u look closcly at (1 I), you will notice that (0.0.0) satisfies it. Thus, the tangent plane to a cone passes through the vertex of the cane. -

Tllerefore, the tangent plane at I-'(x,, J.,, z,) contains P as well as the veltex 0 of the cone. I-Ience, it contains the generator OP of the cone. Thus,

thc tangcnt plane to a conc touches [he conc along thc gcncrator passing through thc point of contact.

This gcncrator is callcd h e generator of contact of thc plane. In Fig. 6 OP is the gcncrator of conlac1 of thc rangent planc T.

You can try somc exercises now.

El?) Find thc equation of the wngcnt plsnc at the point (1, 1, 1) to thc cone 7 4

5yz - 8zx - 3xy = 0.

E13) Usc Theorem 5 to obtain thc condition untlcr which a given plane is langent to a conc.

If,you'vc solved El 3, you would havc secn that thc condition for uxcvycwz = 0 to bc tangcnt to the cone (4). that is, ax2 + by2 + ca2+ 2hxy + 2fyz + 2gzx = 0 is

X Y Z Thus, the line -.= - = -, which is h e normal at (O,O,O) Lo thc tangent planc, is a genera- u v w

lor of the cone Ax2 + B~~ + ez2 + 2Hxy + 2Fyz + 2Gzx = 0. .,....( 12)

a h g u

h b f v

g f c w

u v w o

Thus, (12) is the cone generated by tho nonnals to the langcnt plailles at the vcrlex (0,0,0) of the cone (4). Since it is homogeneous, its vertex is also (0.0.0).

= 0, that is,

Fig. 6: T Is the tangent plane to the cone at P.

Note that (12) is nothing but the de~rrninant equation

'Fhe Sphere, Cone and Cylinder

Now, if we consider tt!e surface generated by the normals at (0, 0, 0) to the tangent planes to (121, what do we get? On calculating, you will get a surprise! The cone is (4), because BC-F~=~~,CA-G~=~A,AB-H~=CA,GH-AF=~A,HF-BG=~A,FG-CH=M, where

a h g x

h b f y

g f c z

x y z o

= 0.

Actually, the following example shows why the name is appropriate.

x2 yZ z2 Exampie 4: Show that the cones axZ+ by2 + cz2 = 0 and - + - + - = 0 are reciprocal.

a b c

/ A = h

(Here abc # 0 .)

Because of this relationship between (4) and (12) we call them reciprocal cones.

a h' g

b f

g f c

Solution: The reciprocal cone of ax2+ by2+ cz2 = O is given by the deeminant equation

a O O x

O b O Y = O O O C Z

x y z o

x2 . Y2 z2 a - + - + - = 0, dividing throughout by abc,

a b c This is the required equation.

Now you can do h e following exercises. This will help you to undersmnd reciprocal cones better.

- -- - - - -

E14) Find the cone on which the perpendiculars drawn from the origin to rangent planes to thecone 19x2+ 1 ly2+ 3z2+ 6yz - lOzx - 26xy = 0 lie.

E15) Prove that the cone (4) has three mutually perpendicular tangent planes iff b ~ + c a + a b = f ' + ~ ~ + h2.

And now let us shift our attention to another surface that is gcneratcd by lines.

6.4 CYLINDERS

You must have come across several examples of the surface that we are going to discuss in this section, For instance, a drain pipe is cylindrical in shape, and so is a pencil. Bu\ for us, the pencil will not be a cylinder, only its surface will, according to the following definition.

Definition: A cylinder is the set of all lines which intersect a given curve and which are to a fixed line which does not lie in the plane of the curve. The fixed line is called

he axis of the cylinder and the curve is called the base curve (or directrix) of the cylinder.

~ l l the figures in Fig. 7 represenl portions of cylinders.

Cones and Cylinders

Fig. 7: a) A elrcular cylinder, b) an clliptic cylinder, c) a parabolic cylinder. T. In Fig. 7 (b) the cylinder's base curve is an ellipse, while il is a parabola in Fig. 7 (c). Fig. 7 (a) is an example of a right circular cylinder according to the following definilion.

Ilefinitiun: A cylinder whose base curve is a circle, and whose axis passos through the cenlrc of the base curve and is perpendicular LO he plane of he base curve, i.s called a right circular cylinder.

As you can see, in common parlance when people lalk of a cylinder, hey mean a portion of a right circular cylinder.

Henceforth, in this section, by a cylinder we shall mean a right circular cylinder.

Lcl us now find the equation of a cybnder. We shall first assume that its axis is the z-axis, ' and its basc curve is the circle xZ + Y2 = ?, z = O (see Fig. 8).

Let P (xl, yl, 21) be any point on the cylinder. Lei the generator ~11rough P intersect the plane of the base curve (that is, the XY -plane) in M. Then the perpendicular distance of P from the / axis is OM =dm. dx Bul this is also r. Thus,

r Z = x f c y:.

Fig. 8: The cylinder x2 + yi = ra.

This equation is true for every point P (x,, y,, zl) on the cylinder. Thus, the equation 6f the cylinder is

xZ t yZ = 2. ......( 13)

You may wonder why z is not figuring in the equation. This is because whatever value of z you take, the equation of the cylinder remains x2 + y2 = 8. What does this mean georneirically? It says that whatever plane parallel to the XY-plane you take, say z = t, and take its intersection with the cylinder, you will always get the circle x 2 t y 2 = P .

Thus, in a sense, the cylinder is made up of infinitely many cii-cles, each piled up on the The radius of a cylinder i s h e

other! radius of its base cslrve.

I

'I'he Sphere, Cone and Cylinder

Note hat thc plane z = t is perpendicular to Ihe axis of Ihe cylindcr x2+g = 2.

Also note that the length of the perpendicular from any point on a cylinder to its axis is equal to its radius.

Using this fact, let us find the equation of a cylinder of radius rand whose axis is x- a y- b z - c - = - = - (see Fig. 9). a P u

Fig. 9: A right clrmlrr cyllnder with r x b AM.

Let P (x,, yl, 2,) be any point on the cylinder. Let A bc the paint (a, b, c), which lies on the axis, and M be the foot of tlie perpendicular fro111 P onto the axis. Then PM = r.

Also, AM = AP cos 8, where 8 is the angle between the lines AM and AP.

Since AMP is a right-angled Lriangle, we AP' = AM' + MP~. Thus,

1 This equation holds for any point (xl, yl, zl) on Ihecylindcr.

x- a y - b z- c - Thus, the equalion of the right circular cylindcr with radius rand axis - = - - - a P r

i s ( ( x - a ) 2 + 0 , - b ) Z + ( z - ~ ~ 2 - ~ ) ( a 2 + ~ 2 + y 2 ) = ( ( x - a ) a + 0 , - b ) ~ + ( z - ~ ) y ) 2 . ...( 14)

Let us look at an cxample.

Example 5: Find Ihe equation of he cylinder having for i!.s base the circle X'+~~+Z' = 9, X-y+z=3.

Solution : The centre of the sphere is (0, Q, 0). and radius 3. The distance between (O,O, 0 ) and Ihe p l a ~ ~ : x - y + z = 3 is a. So the radius of h e base circle is .a = 6 (see Fig. 10).

The axis of Ihe cylindcr is perpendicular tp the planc x - y + z = 3 and passes through (O,0,0).

Flg. 10. x y z So ils quations are - = - = - 50 I - I 1 '

Cones and Cylinders

Thus, using (14), we get the required equation as

~ ( X ~ + ~ ~ + Z ~ - ~ ) = ( X - ~ + Z ) ~

*X2+ y2+z2+xy + Y ~ - % x - 9 = o .

Why don't you try an exercisc now'?

E16) Find the equation of thc cylinder

a) whose axis is x = 2y = -z and radius is 4,

x - 1 - y 2-3 b) whose axis is - -, = - and radius is 2. 2 3 1

We shall end our discussion on cylinders here. Let us now briefly review what wc have covered in this unit

6.5 SUMMARY

In this unit we have discussed thc following pints.

1) A cone is a surface gencrateci by a line passing through a f ixed point (its vertex) and , intersecting a given curve (its base curve), such that the vertex docs not lie in the plane of the base curve.

2) A cone whose basc curve is a circle, and for which the linc joining ils vertex to the centre of thc base curve is perpendicular to the planc of the base curve, is called u right cccular cone. -

3) A planar sccrion of a conc is a conic.

4) The equation ofa right circnlar conc with scmi-vcrtical angle 8 is r2+g a x2 tan2 0.

5 ) A second dcgrce cquation in x, y, s! rcpresenls a conc with vcrtex at LhC origin if it is homogeneous. .

6) The cone w2+hy2+cx2+2fyx+2gzx+2hxy = 0 has 3 mumally pcrpcgdicubr gcncratars iFa+b+c = 0,

- * 7). Any plane through the vertex of a cow intcrrrects ihc cone in two divlincl w coincident

lincs. Thc angle betwcen ~ h c lincs oblaincd by intersecting ux+vy+wz = O wilh

C (x, y, z) = axZ + by2 + ci2 + 21'yz+ 2gzx + 2hxy = 0 is

Thus, the plane is tangent to the cone iff p2 = 0.

where P' =

8) The equation of the langent planc to thc cone ax2.+ by2 + cz2+ 2fyz + 2gzx + 2hxy = 0 at the point P (xl, yl, z,) is

(a1 + hy, + gz,) x + (hxl + byl + lzI) y + ( p l + fyI + C Z ~ ) z = 0.

This contains the linc OP, wherc 0 is the vertex of thc cone.

h b f v

g f c w

u v w o '

The Sphere, Cane and Cylinder

9) The cone formed by the normals to the tangent planes to a given cone at its vertex is the reciprocal of the given cone. The reciprocal cone of the cone ax2+ by2+ cz2+ 2fyz + 2gm + 2hxy = 0 is given by

h b f y I f c z l = o . l x Y 2 01

10) A cylinder is a surface generated by a line which is parallel to a fixed line (its axis) and which cuts a given curve (its base curve), such that the line and curve are not in the same plane.

11) A right circular cylinder is a cylinder whose base curve is a circle and axis is the line through the centre of the circle and perpendicular to its plane.

12) The equation of a right circular cylinder wilh base c w e a circle of radius r and centre (0,0,O) in the plane z = 0 is x2 e y2 = r2.

x - a y- b z- c 13) The equation of a right circular cylinder of radius r and axis - = - = - is

a P Y

And, now, you may like to go back $0 Sec. 6.1 to see if you have achieved the objectives listed there. You must have solved the exercises as you came to them in the unit. In the next section wc have given our answers to the exercises. You may like to h ~ v e a look at them.

El) Let P (x, y, z) be any point on h e cone. Since V (a, b, c) is its vertex, the direction ratios of PV are x -a, y - b, z - c. Also, the direction ratios of the axis of the cone are a , P, y.

Hence, we get (2).

E2) Yes. Just take a = 0 = P, y = 1, a = b = c = 0 in (2), and you will get (1).

E3) The direction ratios of h e axis are 1,0,0, and the vertex is (0, 0,O). If (x, y, z) is any point on the cone, ttien

a x.l+y.O+z.O cos 7= ,/- * x2+ y 2 + z 2 = 4 ~ , which is the required equation.

E5) 2 ~ ~ + 3 ~ + 4 z ~ = ( x + ~ + z ) ~ H x2+ 2y2f 3z2-2xy - 2yz -2zx = 0.

x y '2 U) Let-=-=-= a P v '1 say. Then putting x = m, y = I$, z = ry in (4), and dividing

throughwt by 3, we get

3

aa2 + bp2 + C* + 2fpy + 2 g p + 2h txp = 0. :. (pl, p, y) lies on (4).

E7) Only 3 (x2 + y2 + z2) = xy. .

E8) Subsliluling the value of d in lhe cquillion, we can write it as

U v W which is a hornogcneous ctlualion of dcgrcc 2 iq x + -, y + -, 'z + -.

a b c

Thus, il is a conc will1 vertcx a1 a ' b ' c

E9) The required anglc is

0

5 - where P' = 2

-1

3

EIO) I n lhis silualion (8) tells us Lhat Llic lincs will bc pcrl)cndicular iff bc + ca + ab = 0.

1 1 1 @ - + - + - = ( I . a h c

El I) Lcl ils scmi-vcrlical anglc bc 0. Thcn ~ h c qualion ol' ~l ic conc is ( I ) , Lllat is, x2 + y2= z2 lani 0. Sirlcc this has lhrcc rn~~lually pcrpcndicular gcncralors, Tllcorcm 4 lclls us Lllal ] + I - t a n " = O = , ~ = t a n ' ~ & .

E12) Tllc rcquircd cquaiion is

E13) A tnngcnl planc musl touch L I I ~ conc along n gcncralor. Thus, the two lincs rrl' inlcfscclior?, o l Ll~e plane and Lhe conc musl coincidc. Thus, Ihc angle bclwccn ~hese 1i11cs must be 0 . Thus, Iiom Thcorcm 5, wc sw lhal ux+vy+wz = 0 is a ungcnl lo ~lic conc C (x, y, z) = 0 e P = 0 (since u2+ v2 + w2 ;t 0.)

*

Cones and Cylinders

e

a h g u

h b f v

g f c w

u v w o

= 0.

'The Sphere, Cone and Cylinder

E15) The cone will have three mutually perpendicular tangent planes iff the reciprocal '

cone has three mutually perpendicular generators. Using Thcorem 4 and its converse, we see that this happens iff in (12): A + B + C = 0, that is, iff (bc - f2) + (ca - g2) + (ab - h2) = 0, that is i f fbc+ca+ab= c ~ + ~ ~ + h2.

E14) The required cone is Ihc reciprocal of the given cone. Thus its equation is

1 E16) a) The equation is (x2 + y2 + z2.- 16) (1+ -+I) = (x + 1 - z12 4 2

19 -13 -5 x

-13 11 3 y

-5 3 . 3 z

x y z o

b) The required equation is 14 ((x-112+ )2+(~-33)~-4] = ( 2 ( ~ - 1 ) + 3 y + ( z - 3 ) } ~

10x2+5y2+ 13z2-~',xy-6yz-4xz-8x+ 30y-71z+ 59=0.

= 0

e 3x2+4y2+ 5z2+ 2yz + 4 z x + 6xy = 0.

MISCELLANEOUS EXERCISES

(This section is optional)

x y z 3) The plane - + - + - = 1 mccts the axes in the poinls A, B and C. Find the .

a b- c equations dclermining the circumcircle of the uiangle ABC (see Fig. 1).

In this section we have gathered some problems related to the contenls of this block. You may like to do them toget a better understanding of these contents. Our solutions to the questions follow the list of problems, in case you'd like to counter-check your answers. A

4) Prove that if every planar section of a surlace given by a quadratic equation is a circle, the surface must be a sphcrc.

X - 2 2-4

Fig. 1

Z

5 ) Find an equation of the set of points which are twice as far from the origin as from

1) Find the equations to the planes through the line - = Y-3 = -- , which are C

2 4 5 parallel to the coordinate axes.

x + l y- 3 2 + 2 2) Find thc equation of the plane passing through - = - = - and the point

1 -3 2

2- y z + 2 (0,7, -7). Also check i l x = -- = - lics in this plane.

3 2

(-1 3 1, 1).

6 ) If he sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 cuts x2+ y2 + z2 + 2u'x + 2v'y + 2w'z + d' = 0 in a great circlc, then show that 7 2 (uu' + vv' + ww') - (d + d') = 2r12, where r' is the radius ol the second sphere.

7) Find the equation of thc sphere inscribed in the tetrahedron whose faces are x = 0, y = 0, z = 0, x + y + z = 1 (see Fig. 2).

A 8) Show that the sum of thc squares of thc intercepts nladc by a given sphere on any three Fig. 2

mutually perpendicular lines through a fixed point is constant. (Hint: Take the fixed point Lo bc the origin.)

9) Find the cquations to the lines in which the plane 2x + y - z = 0 cuts the cone 4x2 - y2

+ 3z2 = 0.

10) If x = = z represents one out of a sct of three mutually perpendicular generiitors of 2

thYe cone 11 yz + 6zx - 14xy = 0, find the equations of the other two. (Hint: Take a plane through the given line, and apply the condition for the lines of inlcrsection o l this plane and the cone to be perpendicular.)

1 1 ) Find the equation of the right circular cone with vertex (1,1,3), axis parallel to the

X Y Z line - = - = - 1 2 2

and with one of i 8 generators having direction ratios 2, 1, -1.

12) Find themuation of the cone which passes through the common gcnerators of the . cones x2+ 2y2+ 32' = 0 and 5xy - yz + 5x2 = 0 and Lhe line with direction ratios l , O , 1. (Hint: The cone passing through the common gcnerators of h e cone C1 = 0 and C2 = 0 is C1+kC2 = Q, where k E R.)

The Sphere, Cone and Cylinder

fig. 3: The cylinder cnvelops the sphere.

Pig. 4: The plane Is tanfent to the cyllnder.

13) Find the equa~iop of the right circular cylinder that is generated by lines which are 2

parallel to 5 = 1 = 2 , and which are tangent to the sphere x +3+z2 = 3 (see Fig. 3). a b c

(Note: Such a cylinder is called the enveloping tylinder of the sphere.)

x y + l 2-1 14) The axis of a cylinder of radius 3 has equations 3 = 7 = - 1 . Find the equation

of the cylinder.

15) Provc that for a cylinder the tangent plane at any point is pmllel to its axis (see Fig. 4). --- - -

SOLUTIONS

1) The equation lo any plane through the given line is a(x-2)+ b(y-3)+c(z-4)=O,where2a+4b+5c=O . . Ifthis is parallel to the x-axis, we nlust have n(l)+b(O)+c(O)=O*a=O. Thus, the equation of such a plane is 5y - 42 + 1 = 0: Similarly, you can c h ~ k that the planes pafallel to h e y and z axes are 5x - 22 - 2 = 0 . . and 2x - y - 1 = 0, respectively.

2) Theplanewil lbea(x+l)+b(y-3)+c(z+2)=0, ......( 1) where -3a + 2b + c = 0. , . .. . .(2) Since (0,7, -7) lies on it, we have a+4b-5c =O. . .. . . .(3) Eliminating a, b and c from (2) and (3), we get

a b - c a b c -=-=-*-=-=- -10-4 1-15 -12-2 1 1 1' :. ( I ) becomes 1 (x + 1) + l(y - 3) + l(z + 2) = 0 3 x + y + z = O . -

x y- 2 2 + 2 The linc - = - = - will lie on this plane, if it is parallel to the plane and any 1 -3 2

point on it lies on the plane. Since 1 (1) + (-3) (1) + 2 (1) = 0, the line is parallel to the plane. Also, (0,2, -2) is a common point. Thus, !he line lies in the plane.

3) The circurncircle will bc the intersection of the given plane with any sphere passing , lfirough A, B and C; So, let us take the sphere through 0, A, B and C. The coordinates

of these points are (0,0,O), (a, 0, O), (0, b, 0) and (O,O, c). You can check that the equation is x ~ + ~ ~ + z ~ - ~ x -by-cz=0. Thus. the equations that give the circurncircle are

4) Let Ihe equation of the surrace bc ax2 + by2 + cz2 + 2hxy'+ 2gzx + 2fyz + 2ux + 2vy + 2wz + d = 0. It intersects z = 0 in the conic ax2+by2+2hxy+2ux+2vy+d=0. This will be a circle ilf a = b and h = 0. Similarly, on intersecting with x = 0 and y = 0 we will get a = b = c and f = 0 = g = h. Thus, Ihe equation of the surface reduces to ~ ( X ~ + ~ ~ + Z ~ ) + ~ U X + ~ V ~ + ~ W Z + ~ = O ;

which rcpresenls a sphere.

5 ) La (x, y, z) be any point in h e set Then

6) If the lwo spheres are S = 0 and S = 0, then (-ul,-v', -wf) lies on S - S, = 0, that is, o n 2 ( u - u ~ x + 2 ( v - v ~ y + 2 ( w - w ' ) z + d - d ' = 0 . :. 2(u - u') U' + 2(v - v') V' + 2(w - W ~ W ' = d - d'. A ~ ( u u ' + v v ' + w w ~ - ~ + ~ ' = ~ ( ~ ~ + v ' ~ + w ~ ) = ~ ~ + ~ ~ ' A 2 (uu' + vv' + ww) - (d + d') = 2r".

7) Let the equation be ~ ~ + ~ ~ + z ~ + 2 u x + 2 ~ ~ + 2 w z + d = 0 . Since the given plarlcs are tangent to il, the distance of (-u, -v, -w) from these planes

~ i ~ r = - , / ~ ~ + ~ ~ + ~ ~ - ~ . T h u s , w e s e e t h a t

3 + J5 Solving these equations, we gel r = -.

6 Thus, the equation of [he spherc is

8) Let us assume that the fixed point is (O,0, 0) and the lhree lines are thc axes. Thcn let thc spherc be given by x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0.

Its intercept on the x -axis, that is, y = 0 = z, is 2 d z .

Similarly, the other inlcrccpls ore 2-,/= and 2dW2_d.

Now, ( 2 4 ~ ) ~ + ( 2 ~ ~ 7 + (2 .~=)~= 4 (u2 + v2 + w2- 12d).

which is a constant, since the sphere is a given one.

x y z 9) Let a linc of intersection bc - = - = -. Then 21 + m - n= 0 and u12 - m2 + 3n2 = 0.

1 nt n Solving these equations, we get

m=-21,n=Oorm=-41,n=-21. . Thus, thc two lines are

10) 2x - y + k (y - 22) = 0, k E R, gives any plane through the given line. This will cul the given cone in perpendicular lines if 11 (k-1) ( -2k)+6(-2k)(2) -14(2)(k- l )=O~k=-2,7 /11 . Thus,theplanesare2x-3y+4z=Onnd l lx-2y-7z=0.

Now, 2x - 3y + 42 = 0 intcrsccts Ule cone in two perpendicular lines of which one is . the given one which lies in the plane. Therefore, he olhcr one has to bc h c normal to

X z the plane at (0,O.O). This is - = = -. So this will be another of the requircd set of 2 -3 4

rnutual!y perpendicular gchcrators.

Similarly, the third genenitor will be the normal to 1 lx - 2y - 72 = 0 at (0, 0, 0), that

11) If 9 is its semi-vertical angle, then

Miscellaneous Exercises

x -1 y - I z - - 3 Also, the axis is given by - = - = - 1 2 2 .

The Sphere, Cone and Cylinder

Thus, the equation of the r.c. cone is 4

((x- 1) +2(y-1) + 2 (z-3))'= 9 ((x- 1)2 + I)'+ ( z - 3 ) ' ) ~

aP2 + 1 0 ~ ' e 10z2+ 12xy + 24yzat 12x2-50x- l ay-962 + 221 =0.

12) Let the cone be (x2 + 2y2 c 3z2) + k (5xy - yz + 5x2) = 0, where k E R. Since the line 4 with direction ratios I ,@, 1 lies on it, (1,0, 1) must satisfy it. This gives us k = -- 5

Thus, the required cone is 5 (x2+ 2y2 c3z2) - 4 (5xy - yz c 5x2) = 0.

13) Let (a, P, y) be any point on the cylind,er. A gcnerator through this will be given by

x- a y - p - z - y -..---=--- a b c

This line intersects the sphcre in (ak + a, bk + P, ck i- y), wl~cre k is given by (ak + a)2 + (bk + P12 + (ck + y)2 = 2. This quadratic equation in k gives two values of k, which correspond to two points of intersection. Thus, the generator will be a tangent to the sphere if thcse points coincide, that is, ilf (aa+ bp +cy12=(a2+ b2+c2) (a2+p2 +*-?), Thus, the locus of (a, P, y), which is the equation of thc cnvcloping cylindcr, is (ax + by + cz12 = (a2 c b2 + c2) (x2 + y2 + z2 - r2).

15) We can always assume thc equation of Ule cylindcr to be x2 + $ = ?. Its axis is the z-axis, that is, x = 0, y = 0. As in the case of a cone, you can show that its tangent plane at a point (xl, y,, zl) is XXl + yy1=?. This is parallel to the z- axis. Hence, the result.

NOTES '