2 Chapter 9. Cones and Cylinders

In this chapter, we study only quadratic cones; i.e. cones having its equation of second degree in x,y and z. Chapter 9 9.2 Equation of a cone Cones and Cylinders To find the equation of a cone with vertex V (α, β, γ) and whose guiding curve is the conic 9.1 Cone ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, z = 0. (9.1) Definition 9.1 A surface generated by a straight line passing through a fixed point and intersecting a given curve is called as a cone. We have to find the locus of points on lines which pass through the vertex The fixed point is called the vertex of the cone and the given curve V (α, β, γ) and intersects the given guiding curve. Observe that the equa- is called the guiding curve. A line which generates the cone is called tions of any line passing through the vertex V (α, β, γ) and having direction a generator. ratios l,m,n is given by,

x α y β z γ − = − = − (9.2) V l m n Any point on the equation (9.2) has the coordinates, (α+lt,β+mt, γ+nt), where t R. If the line (9.2) intersects the guiding curve given by (9.1), ∈ then we have,

a(α + lt)2 + 2h(α + lt)(β + mt)+ b(β + mt)2 + 2g(α + lt) (9.3) +2f(β + mt)+ c = 0, (γ + nt) = 0 A B Eliminating t,l,m and n between (9.2) and (9.3), we get D C a(αz + γx)2 + 2h(αz + γx)(βz + γy)+ b(βz + γy)2 +2g(αz + γx)(z y) + 2f(βz + γy)(z y)+ c(z y)2 = 0. (9.4) Figure 9.1: Cone with a guiding curve − − − The surface in the Fig. 9.1 is a cone with vertex V, The line V A as a This is the required equation of the cone. generator. The lines VB,VC are also generators. In fact every line joining Remark 9.2 From the equation (9.4) it can be seen that the equation of a V to any point of the guiding curve is a generator of the cone. quadratic cone is of second degree in x, y and z; i.e., Remark 9.1 If the guiding curve is a plane curve of degree n, then the 2 2 2 equation of the cone is also of degree n and we call it a cone of order n. ax +by +cz +2fyz +2gzx+2hxy +2ux+2vy +2wz +d = 0 (9.5)

1 . ) + ′ . is a (9.8) (9.7) vy , tz , ′ ) = 0 + OP d ,ty = 0 , ′ ux d tx be any point )+ ( ′ , + ) a cone. Hence 2 ′ tion = 0 This is possible t tz ) ( . lies on the plane. ′ are , z wz = 0 = 0 ′ ′ . ) y w R ′ ′ d y are zero. For if not ′ ,y +2 ∈ ′ , z + hx ′ OP = 0 is given by, w x t hxy hx t ( vy ) + 2 d ) ′ ,y ∀ So that the equation of a ′ ′ P + 2 ) ) . + 2 x ty ′ + 2 ′ and OP +2 ( ( ′ wz x tz v ′ x is the vertex, the line say Let ′ ux = 0 ( )( and gzx . t ′ gz u, v + 2

Chapter 9. Cones and Cylinders 0) gz d ′ , +2 ty = ) + 2 ( + 2 0 ′ + 2 = 0 = 0 vy ′ , f + 2 ′ z ′ z tx z. ′ z d hxy (0 ( ′ z Therefore, fyz + 2 ′ u = + 2 wz O . ′ = fy +2 ′ 2 and 0 fy y ) y + 2 ′ ux + 2 w 2 ) + 2 ′ + 2 tz ′ gzx = + 2 x,y ( Since 2 = ′ cz ′ c vy +2( . ty 2 ′ x x v +2 cz lies on the surface (9.7) , it would mean that the + )( + ′ cz and we also have + 2 2 ) 2 + ′ = ′ ) tx + ′ fyz 2 ( by ′ , z u 2 h ′ ′ ux ty = 0 + ( by +2 b ,y by 2 ′ 2 + x + , w + ) + 2 ( ax cz 2 ′ ′ 2 2 ′ ) + ′ tx is a generator, these coordinates satisfy the equation (9.7 ax = 0 2 ( ax )( tx would represent a plane and the point ′ ( by , v a OP tz + ( 2 This equation can be written as, ∴ Observe that (9.8) is a quadratic equation for Thus the coordinates of any point on the line = 0 g = 0 ax As the point surface (9.7) is a plane.u This is impossible as it represents cone with vertex at the origin is of the form, + 2 which is a homogeneous in We now claim that each of the constants only if each of the coefficient is wz 4 Since i.e. if at least one of the constants is not zero, then the equa on the cone given by (9.7) We will show that generator of the cone. The equation of the line . . 3 is ) = ct d, (9.6) = 0 be, hxy d 2+ and + From this +2 . bt, . 2 2 so c wz 2) − gzx s, . 1+ − u, v, w = + 2 = 1 z. +2 z t at, 2 vy ) )] 1 = 0 = ( 1 2 and fyz 2 = 1 − − − + 2 (3 + say z y ( 2 z +2 t 1)] x,y 2 2(  Thus, ux be the direction ratios of a + b . = − c cz − 2 . y 2 + 2 xz + − 6 is non-zero, then the equation is 2 = 0 2( c − a, b, c 1 d = 0 − by z − ct lies on the guiding curve. Therefore + 3[1 hxy  ) + 2 , z yz 2) Let = 2 and ct )] 3 + 2 2+ . + 3 3 2 1 If each of the constants − ax − 2 − − = 1 2) b z 2+ z x − , z. 2  2 and gzx 1 z y y a 2( , bt, c u, v, w 2 (3 = and + 2 − + 5 + 3 = 1 + 3[1( − 1+ V 3 2 2 2 2 3 y ) 2[3 x a x,y − fyz at,  2 bt 3)] x 2 + 3 − + 2 2 (3+ 2 x x , The equation of a cone with vertex at the origin is a homo- 2 Find the equation of a cone with vertex at the point (3,1,2) cz 2( R we obtain + 3(1 + + ∈ − , The vertex is 2 t 2 ) 2) by Let the equation of a quadratic cone with vertex at the origin at − + z After simplification, the required equation of the cone is, 2 is a homogeneous second degree equation in , ax 9.3 Cone with vertex at theRecall origin that the general equation of second degree equation i we get, Example 9.1 and guiding curve is From (9.6) 2[3( Solution The coordinates of any point on the generator are non-homogeneous in zero, then the resulting equation, generator of the cone. Then the equations of generator are, For some Theorem 9.1 geneous second degree equation. Proof. 2(3 + If any one of the constants 9.3. Cone with vertex at the origin 0 . . . ) R then ∈ , ) = 0 t β α, β, γ ( − ) = 0 . ) . y isfy the equa- nt of a generator . . where )( )( , = 0 ) = 0 α ) = 0 ) l,m,n = 0 = 0 − ( mt mt ( φ x )( hlm hlm f l,m,n In the new coordinate ( hxy hxy lt . h ( ) + 2 + 2 + 2 h lt,mt,nt + 2 + 2 2 ( ) ) + 2 gnl gnl α nt ) + 2 α, β, γ gzx gzx ( ( 2 c lt

Chapter 9. Cones and Cylinders − ) + 2 + 2 . V γ )( x + + 2 + 2 2 )( − nt ) ( γ t. z ) = 0 fmn fmn g ( fyz fyz − mt c ( +2 z + 2 + 2 b + ( + 2 + 2 2 2 g 2 x,y,z 2 2 + ) ( cn cn 2 β φ cz cz ) + + lt − ) + 2 + + ( 2 2 γ 2 2 y a ( b − by by bm bm z + + + + ∴ + )( 2 2 2 2 2 ) β al α al The general equation of a quadratic cone with vertex at the ax ax ( − 2 ∴ − t y ( x ( f a ∴ This is a general equation of a quadratic cone with vertex at Let the origin be shifted to the point Thus, it follows that the direction ratios of a generator sat Also, it is easy to see that if the direction ratios + 2 system the above equation becomes, origin is, 6 equation of the cone. Let the equation of the cone be, Remark 9.4 The coordinates of any point on the line are These coordinates satisfy the equation. the equation of the cone is tion of the cone. of a cone with vertex at the origin satisfy the equation This equation holds of all values of ) 5 ′ ty The rep- (9.9) Thus . )( (9.10) ′ . ) ′ tx ( x,y,z Therefore Thus (9.9) h x,y,z . , tz satisfies the ′ . Thus, the ori- = 0 . ,ty ′ ) + 2 = 0 ′ ) be any other point ′ ′ tx is a generator of a l,m,n tx y ( ) y hxy ′ ′ ′ )( z n ′ , z hx hx ′ are + 2 tz = ( satisfies (9.9) ,y ) ′ g + 2 + 2 ) y ′ x ′ m gzx ′ OP ( x x say ′ ′ P ( , tz = ′ ) + 2 t + 2 ′ gz gz of (9.9) we have, l x tz = Let ,ty ′ . ′ + 2 )( fyz + 2 ′ z z ′ ′ tx lies on the surface given by (9.9) z ( z ty ′ ′ ( + 2 = Therefore the equation (9.9) represents a L.H.S. f ′ 2 fy fy y OP y cz . + 2 + 2 = + 2 2 OP. + lies on the surface given by (9.9) 2 ) ′ 2 ′ ′ ′ 2 x x is, cz tz cz by ( OP c + + + OP 2 2 2 + ′ ′ 2 (9.9) ) by by ′ Every second degree homogeneous equation in . . . + ty + If the line with equation, ( 2 2 b ′ ′ i.e. ax + ax ax 2 ( (0) . ) 2 2 Consider the homogeneous equation of second degree in ′ ∴ t t lies on the surface given by (9.9) tx Now we show that the line The coordinates of the origin satisfy the equation (9.10) The coordinates of any point on the line O ( a = = = 0 equation of the line is the surface generates by cone with vertex at the origin. Remark 9.3 on the surface given by (9.10) any point on the line Hence, the point with coordinates gin substituting these coordinates in Note that the converse of this theorem is also true. Theorem 9.2 resents a cone with vertex atProof. the origin. 9.3. Cone with vertex at the origin cone with vertex at the origin, then the direction ratios . A θ. and AP AP AP AV Thus ) Let = α. 2 α. θ ) γ. θ Then = γ 2 2 α.β, γ − ( α. − )] tan cos V γ z  As the direction β, z 2 ) − ) . γ + ( − γ z 3) ( 2 . V AP, − ) − n 9 z β z ( α, y be any point on the cone. n − )+ − ) + ( y β and the plane x 2 with a semi-vertical angle )+ ) − , + ( β β N y see F ig. are γ are constant. it follows that 2 ( ( Chapter 9. Cones and Cylinders x,y,z n − ) VN − − ( θ θ z α m y y P ( V P is V P θ − = tan m )+ + ( x β ( we have, α 2 VN m − ) and )+ y p − Figure 9.3: α 2 α x n = − and ( − l AV on the section of the cone by the plane [ x + α x l,m,n, ( l In the right angled triangle ( 2 −  P l V P x ) m 2 are n Since, + V A. 2 + θ. l VN 2 √ m tan = + be any point of the section of the cone by the plane θ 2 AV l be the axis of the cone and let ( P = cos VN The following example will make the above proof more clear. Therefore, the equation of the right circular cone is, The angle between Let AP whose axis is the line, ratios of the axis the section is a circle. Let 8 9.4.1 Equation of a rightTo find circular cone the equation of a right circular cone with a vertex ∴ The direction rations of the generator be the point of intersection of the axis is constant for all points is perpendicular to . 7 A − AP AP AP AV Thus Let with a = α. θ α. θ Then the semi α. tan point is called be a plane perpen- α V AP, is called as θ and the plane are constant. it follows that VN θ V B A tan and N AV on the section of the cone by the plane is a circle. In the right angled triangle P α see Fig. 9.2 of the cone. We show that the section Since, α V A. VN θ. Figure 9.2: Section of a cone with a plane of the cone. A right circular cone is a surface generated by a straight of the right circular cone, the fixed straight line is called tan Every section of a right circular cone by a plane perpendicu- of the cone and the constant angle be any point of the section of the cone by the plane be the semi - vertical angle of the cone and AV P θ = Let Let the vertex the axis AP vertical angle Remark 9.5 lar to its axis is a circle. fixed straight line passing through theas given point. The fixed as of the cone by the plane the section is a circle. dicular to the axis 9.4 The Right circular Cone Definition 9.2 line passing through a fixed point and making a constant angle ∴ be the point of intersection of the axis is constant for all points 9.4. The Right circular Cone is perpendicular to P 2 ) 1 = 0 n 2 c ) mz e generator 2 is given by, = 0 ) − )+ mz 2 = 0 . 1 1 n 1 c = 0 mz y − cn n z ( . n z . mz b − − of the generator are 1 ny )+ )+ ( y − = 0 n ( b = )+ 1 . n b mz c 1 1 mz But this point lies on the 1 y 1 mz is, z ( . d.r.s y z n − + )+ f − l n − )+ n P mz 0) − 1 − − , y y fy z ( mz 0 y n 1 − ny mz f ( = 1 n ) + 2 −

Chapter 9. Cones and Cylinders = = y + 2 1 mz − fn 1 n 1 )( lz ny y y − y 1 Hence the ) + 2 gx and , y 1 )( )( n m . − m − lz − n 1 lz z lz n n ) + 2 n 1 ,y lz y + 2 y lz 1 − 1 x − lz z 2 − ( n n − 1 = = lz − = g x − x − by ( x 1 1 ( 1 ( nx − y g m x x ( h +2 = + x h 1 nx l h l x ( 1 − = +2 − = ( x + 2 + 2 x x x gn l hxy + 2 l 2 x 2 − ) ) 2 are, 1 ) x +2 + 2 be any point on the cylinder. Then generator through n n lz ∴ lz lz 2 ) = 0 − − − ax ∴ z x 1 ( x nx a ( x,y,z ( ( a a The equation of the generator through P ∴ Let Therefore the coordinates of the point of intersection of th l,m,n. 10 is parallel to the line The point of intersection of the generator and the plane which when simplified, gives the equation of the cylinder as, guiding curve and the plane Hence, the locus of P is, . 9 be 4) The , θ 1 . 4 − Let , 2 − the given . (2 2 2 + 320 = 0 1 , z ts the conic, z V 4) 4) − + 4) we have, , z 2 − − + 1 120 , 4) z z − 1 − ,y − y 2 y + ( + ( z are 2 2 − + 2 4 Remark, 1)( + 4 1 x x + 16 − z − and are parallel to the line, − + 1) + 1) x z as the axis and semi-vertical = ( − y y are From 80 y 4 =  = 0 . 1 2 − + ( + ( 6 4 − V P − √ 2 2 4) + 2 , z + 1) + ( z 2 xy + 1 2) 2) x − = y y = = 0 z − − θ = 4 c = x x ( ( + ( + cos 2 2 zx + 1 2 2 p p 2) + 2( y 1 − 6 fy The straight lines which generate the cylinder − − x of the cylinder and the given curve is called as √ = + 1) x + 2 be any point on the cone with vertex yz 2 y = 4 ) 1( gx 1 6 − − 4 + ( 1+4+1 . x √ 2 2 + 2 z √ 6) x,y,z 2 2) ( A surface generated by a straight line which always re- √ a cylinder. = by ∴ Find the equation of the right circular cone with vertex at − P / . θ + 15 + x (4 the generators z 2 n ( the line 1 y Let  , cos − = hxy 4) 16 , cos y m + 12 1 + 2 2 Hence, the required equation of the right circular cone is, − 2 = x , l x direction ratios of the axis i.e. 15 Then the direction ratios of a generator is called as the semi-vertical angle. Hence, Example 9.2 (2 the guiding curve. 9.6 Equation of a cylinder To find the equation of the cylinder whose generator intersec 9.5 Cylinders Definition 9.3 mains parallel to thecurve is given called fixed line and which intersects to ax Solution. 9.5. Cylinders angle By and . Now, 3 lies on r. ) = 3 ; = 2 is the line P M α, β, γ 2 ( P M P M r ) A 2 γ 3) 2 ) = = 3) − γ − z z − 2 2 2 − ( 2 ) be a point on the cylinder z  2( n z γ ) 2 2) ) The point − n γ MA − − + ( + ( )+ be any point on the cylinder. r. 2 z + − − 2 β ) ) + ( 2

Chapter 9. Cones and Cylinders 2 z x,y,z + 1) β ( 2 ( − + ( m n + 1) y 2 y P − 2 + 6 AP ) ( , we get y + + 2 n y β y x,y,z 2 m )+ ( 2 2 l Let + 1 + ( − β P + ( √ 3 2 and having direction cosines proportional − 2 y 2) + 2( )+ p 2 − and radius is y m ) AMP 2) α 3) y − α and γ + ( ( , △ + − − 1 x + 2 2 l,m,n. − 2 n m − ) x l 3) x − x 1( projection of AP on the axis ( , x α z , l √ projection of AP on the axis 1 )+ (2 − = − α = = = = ( , x = = = ( ( β − (2 2 − m x A 2 ( Find the equation of the circular cylinder of radius y l MA MA MA AP MA MA  perpendicular to the axis of the cylinder. Then AP 4 perpendicular to the axis of the cylinder. Then Let = . . whose d.r.s. are ∴ 9 − 2 α ∴ L − l P M P M , − 2 x , 1 : Now, from the right angled Draw which is the required equation of the cylinder. Example 9.4 axis passing through L 12 9.7.1 Equation of a rightTo find circular the cylinder equation of the right circular cylinder whose axis the line Draw see F ig. Solution. to the distance formula, l 11 = 1 2 (9.11) y For some +2 . if its guid- ) circle pass- 2 If we take a t x , f the cylinder, s called as the lar to the plane +3 1 ) 3 = 0 . 9 = 0 t, z − − . + say = 9 Therefore the equation = 0 ( . 1 = 1 . t yz 2 = 1 1 yz ) 3 z 4 2 = 9 z 2 , z 1 3 ) t,y = ) 12 t z 2 − − = 1 ) 1 3 z − + − z z +2 = 1 1 y = xz 1 1 1 3 2 − t 4 − − y y x xz y = ( 1 z y − y 12 axis of the cylinder. 2 x right circular cylinder 2 + 2 = ∴ we get + 2( y − , 2 , + 2(3 2 1 2 + 2( x ) + 2(3 2 y t y ) 2 + 2 2 1 = 0 1 ) ) − 2 z t 1 z y 3 + 2 + 6 2 y z be any point on the cylinder. Then the gener- 2 1 2 are, 2 ) in (9.11) − + 3 y = x − 1 + 6 ( t 1 P − 1 1 x 2 z , z x and whose guiding curve is the ellipse 1 x x 1 (3 x 3 (3 + 18 z 3 2 ( ∴ − ,y 2 1 ∴ is, = x x i.e. x 9 ( P 1 y is parallel to the line, A cylinder is called a P Find the equation of a cylinder whose generators are paralle i.e. P = 2 x Let . = 0 z this point lies on guiding curve ator through also, t, Hence, locus of Solution. and Example 9.3 to the line of the generator through Coordinates of any point on this line are Substituting the value of 9.7 Right circular cylinder Definition 9.4 9.7. Right circular cylinder ing curve is a circlecontaining and the its circle. generators are The linesing normal perpendicu through to its the centre plane issection of of called the the as guiding cylinder the bythen a plane this perpendicular section to will the beradius axis a of o circle. the cylinder. The radius of this circle i e are rtex rtex axis (9.14) (9.13) − axis X as genera- . . − f the form Substituting f the form’ . 1 Co-ordinates . z Z = 0 − . 1) = 0 , 1) of the 2 = 0 = , = 0 xy − 2 c ) hxy 1 , h of the , k. y 2 1 3 + − + 3 (1 say . d.r.s − and − − ( ( = k zx g d.r.s = gzx and 4 3 x = 0 = − . and + − 1) = 0 = 0 and f 3 , , b h 3 1 1) 0 and − yz Chapter 9. Cones and Cylinders , , , k, h fyz ,we get 1 z 3 1 hxy hxy = 0 (1 = , , h i.e. in (9.13) we get the equation of the 0 a + + = 4 i.e. 4 g = (1 and h we get the general equation of the cone 2 = and are = 0 y , gzx gzx − = 0 1 and f + + k, g − = 0 f, g = kxy axis − h 3 hxy 1 x f, g − fyz fyz = + − in (9.12) + 2 f g Y + kzx = 0 axes satisfy the equation (9.12) gzx f ∴ c z of the + 4 + 2 Find the equation of the cone passing through the co-ordinat Find the equation of the cone which passes through the axes ∴ and and kyz d.r.s fyz − As the cone passes through the three co-ordinate axes, the ve As the cone passes through the three co-ordinate axes , the ve 2 ; = 0 x,y 0 From these conditions we get , 0 , b . , 1 Solving the equations for 1 , 0 = 0 , required cone as, Substituting the values of 0 axes and having the lines, Example 9.7 a 14 ratios of are Example 9.6 of co-ordinates and contains the points which passes through the three co-ordinate axes as tors. Solution. of the cone is at the origin. Hence the equation of the cone is o Solution. The cone passes through the points of the cone is at the origin.Hence the equation of the cone is o of these two points satisfy the equation (9.13) . - 13 , = 9 (9.12) 2 e axes. MA +9=0 − es. z 2 30 = 0 AP − y 6 hxy − + 2 x we have 48 gzx − Let the equation of the cone be, we get + 2 xy AMP, 4 P △ − fyz x,y,z. MA, in zx + 2 Figure 9.4: 2 2 and + 4 cz AP yz + A M 2 axes are the generators of the cone. The direction + 8 Axis by z 2 z + 2 and + 5 Find the general equation of the quadratic cone with the ver- ax 2 y x,y The vertex of the cone is the origin.Hence its equation is a ho + 5 2 x The cone given by (9.12) passes through the three co-ordinat 8 which is the required equation of the cylinder. 9.8 Illustrative Examples Example 9.5 tex at the origin and passing through the three coordinate ax mogenous equation of degree Solution. Therefore Substituting the value of 9.8. Illustrative Examples Now, from the right angled triangle + be one ) ′ 2) = 1 yz (9.17) (9.16) Hence 3 , z − 2 x ′ . ′ ′ − y ,y ′ , z ′ = 1 x xy 1)( ( e cone. Then with vertex at 3) + 4 = 0 n the new co- ,y ′ − − x ′ + 2 , since ′ , z and x z 2 Now the section of ) +4=0 z . z 2 = 1 = 0 homogeneous with the  + 2( ′ 2 Show that section of the + 2 z = 0 2 x y 2 + 6 ′ . x,y,z 2 2) + 6( 2  y ( 3) y − y z 3 = 16 2 − × = 2 − 11 ′ − − x − ′ 2 Let ′ y 4 2 z z − . 2 ,x Chapter 9. Cones and Cylinders y y ′ x x 3) + = 16 x 11( 4 − 2 − 2 + 2( Consider the equation of the guiding i.e. − Substituting in to the equation (9.16) 2 , y z = 16 2 + 2 2 x + 12 1 2 4 1) 2 + 2) − ′ z = 1 , 2 − z yz − − 1 ′ y + 3 ′ ′ x,y,z. z 2 − x We make one of the equations homogeneous y + 2 − ( is, y ( . , z = represents a cone with with vertex at the point 2 ′ ∴ − xy y is a hyperbola. 2 , z = 1 2 = 2 1 = 0 1 − z 1) 2 + 2 2 × ′ 3) + 12( 1 ,x − = 1 2 − x ′ z z − ′ 4 − y +4=0 ′ x 2 Find the equation of the cone whose vertex is at origin and z z = 16 = 16 y = + 2 Show that the equation 4( . 2 2 2 − 2)( z y z + 6 ,y 3) 3) 2 1 Since the vertex of the cone is at origin, the equation of the c = 2 − y + − − − x + ′ , , 4 2 − 2 x 2 2 y 2 11 y ′ x y − − x 4 3( − , , The equation (9.17) is a homogeneous equation in On simplification gives − 1 1 x = ∴ − − help of Example 9.10 the guiding curve is a circle the equation (9.17) represents a cone with vertex at( origin i ordinate system. Thus the equation (9.16) represents a cone cone by the plane we get respectively the old and newx co-ordinates of the points on th is a homogeneous equation in with the help of the other. We make curve Solution. Shift the origin to the point the cone by the plane 16 Example 9.9 12 Solution. Hence the equation of the cone is ( . 3 15 , 1 s the (9.15) − . , 2 = 0 . are xy z 3 3) = 0 . ) satisfy the equa- 8 h . = + 3 k. are the generators . 3 z 3 2) = 0 1) = 0 (8 3 z 1 say y zx − − ( − − = is the generator of the = k = 15 g (2)(3) 3 1 = z 3 + 4 2 y = y (1)( (3)( = 0 is a generator of the cone − we get the equation of the − 2 − − x h h , 3 yz = z 3 h xz 3 = = − 1) 3) k, h 1 . y = we get 1 x − − − 8 2 x . 20 = 1 h y − (8 4 g − xy = i.e. and f 1)(3) + (3)(1) + = in 2 x − and = g = of (9.15) we get, and of the line ( + 4 h + 4(2)( g 1 = 0 2 2 x f 2 z − = 0 f, g z 3 . of the line and + h k, g kxy + 3 1) + = 2 . d.r.s 2 2)(3) + L.H.S. 2 15 − y 2 = 0 − y f, g − − − 1) ( d.r.s + g + 15 1)( f − xz 2 = of the lines satisfy the equation = − x Thus, the line ( − f + 3 1 x . + ( kzx 0 f ∴ 2 xy 6 20 2 d.r.s ∴ − − Show that the line + 4 and f 2 then we can say that the given line is a generator of the cone z ∴ , kyz The equation of the cone is, + 15 2 − y which is a homogenous equation. Hence the vertex of the cone i + 2 Example 9.8 required cone as Substituting the values of of the cone, thus Solving these two equations for origin. If we show that Solution. Given that the lines given cone. Which is equal to Substituting these values in 9.8. Illustrative Examples x tion (9.15) given by the equation (9.15) , . . ∴ 2 + = 6 and 2 , 3+ 3 , − 1) 2 (9.21) (9.22) 2 , − MA = 9 , 1 V P 5) − 4) 2 2 2 f the right − x −  − +294 = 0 = 9 5) 6 2 z z is the radius z is a generator Thus, 2 14 √ − = ( . 2 on the axis. + 3 6 + ( are z − 126 2 , y √ 2 V P 4 z MA 3 P M − 5) AP + ( + 3) y − − = AP − V A , 2 7 x + 6 − z 2 − 2 2 of 3) y 280 z y + proportional to 3 − − AP + 2 y = 3( V A − x + ( 2 y 2 d.r.s + ] is the angle between 2 x 42 1) 2 d.c.s 2 be the semi vertical angle. The x θ + ( 2) − −

Chapter 9. Cones and Cylinders 5)  θ 2 we get be a point on the cylinder. Draw + 3) + 1( − − xy ) 2) − + ( y x z Then
( 2 − 1) + 1(2) 2 . 1 and has p 5 +12 x − + ( 3) ( 3 AMP, p x,y,z 3) pbe the projection of 2 2 − ( , zx − √ 2) + 1( △ p By Distance formula, 2 3) as d.r.s of the axis are 2 P z , . Therefore the 24 , z .. − + 1 = − 3) (1 − MA 2 x 5) 6 + 1 + ( − y and + 3 , z 1(1) + 1( 2 2 = 2 2 1( yz √ respectively. Let 3 2 + 1 3 Let 3) ,y + ( 2) +6 2 − 2 , . 2 2 , √ + 1 2 2 1 2 . 2)+6( − +36 3) 2 , P M (2 − − 2) 2 √ − 3) 1 and y y 14 z (1 V x 1 − √ +( 3( Find the equation of the right circular cylinder of radius − = A 2 − be a point on the right circular cone.Then + ( − − x 2 z θ ) , = z are 2 1) +13 and √ 1 2 2[( − θ Let 1) x y + ( cos 3) 7 + 6 i.e. , 2( − 2 d.r.s y cos 2 x,y,z 3 x 3 ( 2) perpendicular to the axis of the cylinder. Then = +40 − i.e. ( ∴ − , P 2 Simplification of the equation gives the required equation o From (9.21) and (9.22) we get Now from the right angled − − x 5 ∴ is the angle between the axis of the cone and (1 x , y 2 MA ( P M of the cylinder so Solution. which is the required equation of the rightExample circular 9.13 cone. Let 2 θ whose axis passes through circular cylinder as 45 the axis and its 18 A , 17 Let one . = 0 1 (9.18) (9.19) (9.20) 3 = 0 ,  ows, 1 − z , get z 1 + y + 4 4 and whose axis y + = 0 = 1 x 5) + 2 2 z , xy 3 x  + 2 − 3 3 = 0 + 4 y 11 = 0 , 4 − of the axis are − − 2 (2 + − zx z z  x z 2 12 + z d.r.s 2 + 4 − + 6 + y y . y y + 4  yz i.e. 2 The equation of the cone is obtained + z + 2 x 18 + 4 x x + x 2 − 2 11 = 0 y 4  2 − z − + 2 + z x,y,z. 8 = 0 2 2 z whose vertex is at x 29 z z − 2 + 4 − + z 3) +  + 6 ,  2 2 2 x 2 y y y z y 2 + 2 − 21 + + , y + − + 4 2 y 2 (1 − 2 4 x x z x 2 + 2 + Find the equation of the cone with its vertex at the origin Find the equation of the right circular cone which passes x x + x + 2 4 2 2 12 2 y We are given that axis of the right circular cone makes equal z  Since the vertex of the cone is at origin, the equation of the c y + + 2 2 +2 x y Subtracting the equation (9.18) from the equation (9.19) we using (9.20) we make the equation (9.17) homogeneous as foll + 2 homogeneous with the help of the equation is a homogeneous equation in Solution. makes equal angles with co-ordinate axes. Solution. angles with co-ordinate axes .Therefore the by making the equation x and whose guiding curve is given by which is a hyperbola. Example 9.11 through the point which is the required equation of the cone. Example 9.12 9.8. Illustrative Examples Simplification yields, . , + 2; 2) , 2 , Test 2 = 0 . z 2 . , z 2 2 , 1 + . guiding n +5=0 (1 the lines = 2 vertex at = 0 z os ich passes y 2 containing d direction 1 +5 y − + 2 − xy 3 = 0 3 y 2 and one of its s 2 whose axis s 2 whose axis m = 4 x − 2 − z 3 x z − 2 − x zx 2 − + l = 4; (d) 3 + y +6 z = p. 2 z . . 6 − yz ) y , xy = = 2 5 6 + 3 x 3 8 − , 3 y y − − 3 1 − nz 2 , 2 , = = . 2 z 7 + zx − 6 2

Chapter 9. Cones and Cylinders x 2 − x 4 + 9 = 0; 5 − x 2 ( and the semi vertical angle of the z 2 − my − y 3 + (c) 1 2 = yz 2 √ + . z z 8 z 3 1 + 2 1 1 , lx √ +3 2 = − + − + 1 y ; 2 3 x 1 , 2 2 y z . y − 1 √ y y and has d.r.s. = 2 = 1) + 1 Show that the axis of the right circular cone , + = 2 3) x 2 2 − 2 2 . , 2 x . x x − 2 12 , , by (b) , z (2 (1 4 = 5 + , z 2 3 = and y ax axis parallel to the line + 4 − , and 2) y 3 1) − = 6; , , , x 3 1 − 3 , , x , lies along the line 2 (2 passes through through the curve whether the following lines are generators(a) of the cone. curve is the circle ratios of whose generators satisfy the equation through them has d.c.s. cone is —-. Also obtain the equation of the cone. 2 generators having d.r.s. represents a cone whose vertex is drawn from the origin(2 to cut the circle through the points the curve 15. Find the equation of the right circular cylinder of radiu 19. Determine the equation of the right circular cone having 13. The equation of a cone is 11. Find the equation of the right circular cylinder of radiu 16. Obtain the equation of the right circular cylinder whose 14. Find the equation of a cone with vertex at the origin and wh 17. Lines are drawn through the origin having direction rati 12. Find the equation of a cone whose vertex is at the origin an 18. Show that the equation 20. Find the equation of the right circular cone generated by 21. Find the equation of the cone with vertex at the origin and 20 19 find as a akes ough s and , 2 5) taining allel to a circle , , 2 +9=0 3 − z − through the and guiding and passing , , 2 2 and having 1 5 (2 2) 3) n − − , , z P y 1 1 Find the equation , , = 1 (1 + 8 4 − is a circle. ( − m x y and radius 5.Also show 2 plane. . = 0 − − = 12) 0) x 3 , , l zx 0 − 2 6 , x − ZOX , . − (13 3 in (1 2 − 2 yz z a 4 . . . and whose guiding curve is the ellipse o = = − z 2 30 = 5 2 = 0 = 4 2 1 − y z z z y y + = 2 + with centre . 2 = 2 x y = 4; = 1; 1 = 1; y 3 2 = 3 − 2 2 2 − x y y z z = 12 + 2 + − + z = 2 2 2 2 x = 1; x x x x 6 3 4 2 y + 2 2 vertex; axis PQ which makesthe equal semi angles vertical with angle coordinate is axe of the surface so formed. points on the circle curve is as axis the line in the plane represents a cone with vertex at the curve that the section of any plane parallel to through equal angles with thethe coordinate line axes. drawn from If the the origin cone with passes direction thr rations the equation of the cone. x the line 8. Find the equation of the right circular cone having 1. Find the equation of a cone whose vertex is at 7. Find the equation of the right circular cylinder of radius 9. Show that 2. Find the equation of a cone with vertex at the origin and con 3. Find the equation of a cone whose vertex is at 4. The axis of a right circular cone with vertex at the origin m 5. Find the equation of the cylinder whose generators are par 6. Lines are drawn parallel to the line 10. Find the equation of a cone with vertex the origin and base

9.9. Exercise 9.9 Exercise . . . + 21 y as its + rdinate . 19 = 0 2 . x z d -2,-2,1. oordinate − . − nerated by + 75 = 0 + 26 = 0 z z z = . . = 1; 5 14 1 y + 69 = 0 + 22 + 30 z − = 0 z y + 278 = 0 y + 24 = 0 = y = 196 + y 24 z 2 xy y 3 16 . x . − 18 + 38 7) − − − − + 150 y x + 86 − − . = 0 x x zx . . 3 x 34 y 46 y xy and 4 = 0 4 − xy = 0 − 24 − are not generators. 6 3 z = 0 = 0 + 8 + 22 − + 2 ) − x − yz d z − xy x . ( lzx = 32 xy xy xy x xy . zx 2 8 4 2 zx − + 4 1 y a = 0 (18) 24 − . + (3 + 5 − − + 6 + 9 and + 12 . 2 − . 2 ) − xy zx = ) m xy a zx zx zx zx zx zx yz x ( 5 = 0 8 8 2 = 0 myz 2 yz x 3 = + 8 5 = 0 2 − + 5 − − − + 9 + 12 2 + 6 xz − . ) xy − . − yz z zx + 10 yz y 13 yz yz yz yz yz (25) 6 yz 2 + ny 8 4 . 2 = 0 . = 0 + 10 − + 6 12 z nz − 2 + 4 zx − − + 8 + 9 + 4 2 2 + 4( 2 2 2 z − − = 0 2 2 2 2 2 z z z z 2 4 = 0 − 2 + 5 yz 4 z z z z z − 2 about the y-axis. mz 3 2 z 2 z 7) ( 2 2 − are generators; xy z − z + 5 + 6 + 3 + + y ) ) − + + 5 + 5 + 3 + 4 − − 2 2 2 2 2 2 2 − c − + ) bpy 2 2 2 2 2 2 2 y y y y y ( z y 2 2 y 4 y 2 y y y y 4 y ly zx + + 29 y 5 y 8 + + + − − + 6 + 4 − +1=0 2 2 − and 2 y − + − + + 5 + 8 + 6 + 4 − 2 2 2 2 2 2 x z x ) 2 2 2 2 2 2 2 2 x x x x x x axes and has the lines revolving the line whose equations are generators. 3 axes and has two generators having direction ratios 1,2,2 an b apx x x yz x x x x x x mx 22. Obtain the equation of the right circular cone which is ge 24. Obtain the equation of the cone which passes through the c 23. Find the equation of the cone which passes through the coo (22) (20) (16) 26 (15) 2 (14) (24) (21) 8 (13) ( (23) (17) (12) 3 (11) 9(2 (10) 6 (8) 5 (6) ( (7) 5 (3) 12 (2) 25( (5) 3 (4) 4 9.10. Answers 9.10 Answers (1) 12