Analytical Geometry of Two and Three Dimensions 9788173815416, 8173815410

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ANALYTICAL GEOMETRY of

TWO AND THREE DIMENSIONS

"This page is Intentionally Left Blank"

For the students of Mathematics Honours of all Indian universities

ANALYTICAL GEOMETRY

of TWO AND THREE DIMENSIONS [with Objective Questions]

Dr A N Das MSc (Gold Medallist), PhD Reader in Mathematics, Alipurduar College Ex-Lecturer in Mathematics, St Joseph's College, Darjeeling

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ANALYTICAL GEOMETRY OF Tl/VO AND THREE DIMENSIONS. AN Das

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Contents Preface

xiii

ANALYTICAL GEOMETRY OF TWO DIMENSIONS Chapter 1. Transformation of Coordinates and Invariants 1.1. Transformation of coordinates 1.2. Change of origin 1.3. Rotation of Axes 1.4. Transformation due to both translation and rotation of axes 1.5. Invariants under orthogonal transformation Examples Exercise

3-11 3 3 4

4 5 7

10

Chapter 2. Pair of Straight Lines 2.1. Equation of a pair of straight lines 2.2. A homogeneous equation of nth degree represents n straight lines through origin 2.3. Bisectors of the angles between the pair of straight lines ax 2 + 2hxy + by 2 = 0 2.4. Condition that the general second degree equation may represent a pair of straight lines 2.5. Point of intersection of the lines given by ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 2.6. To find the equation of the pair of straight lines joining origin to the point of intersection of lx + my + n = 0 and ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 Examples Exercise

13-40

Chapter 3. Circle 3.1. Definition 3.2. Equation of a circle 3.3. Position of a point with respect to a circle 3.4. Equation of tangent and normal at any point P(h, k) on a circle 3.5. Length of the tan~ent drawn from (h, k) to the circl~_x + y 2 + 2gx + 2fy + c = 0 3.6. Equation of chord of contact of the point P(h, k) with respect to the circle x2 + y2 = a2

41-61 41 41 42 42

[v]

13 13 14

15 17

18 19 29

43 44

3.7. Equation of the common chord of two circles 3.8. Equation of pair of tangents from a given point P(h, k) 3.9. Parametric equation of a circle 3.10. Orthogonal circles 3.11. Radical axis of two circles

44 45 46 46

3.12. Properties of radical axis 3.13. Coaxial circles and its various forms

46 47

Examples

48 48

Exercise

57

Chapter 4. Parabola 63-82 4.1. Definition of conic 63 4.2. Parabola 63 4.3. Some important results associated with the standard form of parabola 64 4.4. General equation of parabola 64 2 4.5. Position of a point P(h, k) with respect to the parabola y = 4ax 64 4.6. Equation of tangent and normal at any point P(h, k) on a parabola 65 4.7. Equation of chord of contact of the point P(h, k) with respect to the parabola y 2 =4ax 66 4.8. Equation of pair of tangents from a given point P(h, k) 67 4.9. Parametric equation of a parabola 67 4.10. Conormal points 67 Examples 68 79 Exercise Chapter 5. Ellipse 83-100 5.1. Definition 83 5.2. Some important results associated with the standard form of ellipse 84 5.3. General equation of ellipse 84 5.4. Position of a point P(h, k) with respect to the ellipse x2 /a 2 + y 2 /b 2 =1 84 5.5. Equation of tangent and normal at any point P(h, k) on an ellipse 85 5.6. Equation of chord of contact of the point P(h, k) with respect to the ellipse x2 / a 2 + y 2 /b 2 = 1 86 5.7. Equation of pair of tangents from a given point P(h, k) 87 5.8. Parametric equation of an ellipse 88 5.9. Auxiliary circle and eccentric angle 88 5.10. Conormal points 89 Examples 91 Exercise 98 [vi]

101-111

Chapter 6. Hyperbola 6.1. Definition

101

6.2. Some important results associated with the standard form of hyperbola

102

6.3. General equation of hyperbola 6.4. Position of a point P(h, k) with respect to the hyperbola x2 /a 2 -y2 /b 2

102

=1

102

6.5. Equation of tangent and normal at any point P(h, k) on a hyperbola

103

6.6. Equation of chord of contact of the point P(h, k) with respect to the hyperbola x 2 I a 2 - y 2 /b 2 = 1 6.7. Equation of pair of tangents from a given point P(h, k)

104 105

6.8. Parametric equation of a hyperbola

106 106

6.9. Rectangular hyperbola 6.10. Chord, tangent, normal of the rectangular hyperbola xy

= c2

106

6.11. Conjugate hyperbola 6.12. Auxiliary circle

107

6.13. Conormal points

107

Examples

108

Exercise

110

107

113:132

Chapter 7. Pole and Polars 7.1. Definition

113

7.2. Pole of the line Ix + my + n = 0 with respect to a curve

114

7.3. Properties of pole and polar

115

7.4. Special properties of pole and polar for circle

116

7.5. Condition for conjugate lines

117

7.6. Self-conjugate or self-polar triangle

118 118

Examples Exercise

127 133-166

Chapter 8. Diameters 8.2. Diameters, conjugate diameters and principal diameters

133 134

8.3. Diameters of conic sections

135

8.4. Properties of diameters of a conic

137

8.1. Equation of a chord of a conic in terms of its middle point

8.5. Properties of conjugate diameters of ellipse

138

8.6. Equi-conjugate diameters

140

8.7. Properties of conjugate diameters of hyperbola

140 143

Examples Exercise

156 [vii]

167-196 167

Chapter 9. Polar Coordinate System 9.1. Introduction 9.2. Distance between two points (r 1, 81) and (r 2, 92)

168

9.3. Equation of the directrix 9.4. Equation of chord, tangent and normal

172

9.5. Chord of contact Examples

173 174 175

Exercis~

l';JU

Chapter 10. General S~cond Degree Equation 10.l. Introduction 10.2. Conics having origin as the centre 10.3. Centre of a conic 10.4. Reduction of a central conic to standard form 10.5. Reduction of a non-central conic to standard form 10.6. Contact of two conics Examples Exercise

197-214 197 199 199 200 201 202 203 212

ANALYTICAL GEOMETRY OF THREE DIMENSIONS Chapter 1. Coordinates 1.1. Introduction 1.2. Distance between two points 1.3. Coordinates of a point which ~ivides the line joining two points in a given ratio 1.4. Projection of a segment 1.5. Direction cosines 1.6. Direction ratios 1.7. Projection of the straight line joining P(x 1, y 1, z 1) and Q(x2, y2, z 2) on another line 1.8. Angle between two straight lines E~amples

Exercise Chapter 2. The Plane 2.1. Introduction 2.2. Equation of a plane through three given points 2.3. Normal form of a plane 2.4. Angle between two planes 2.5. Equation of bisectors of the angle between two planes [viii]

217-230

217 218 219 221 221 222 223 223 224 229 231-243 231 232 232 233 233

2.6. 2.7. 2.8. 2.9.

Plane through the line of intersection of two planes Condition that the three planes may have common line of intersection Pair of planes To find the necessary condition that the equation ax 2 + by2 + cz 2 + 2fyz + 2gxz + 2hxY. = 0 may represent a pair of planes Examples Exercise

233 234 234

235 236 241

245-276 Chapter 3. The Straight Line 245 3.1. Introduction 245 3.2. Symmetrical form 246 3.3. Distance of a point from a line 3.4. Expression of line of intersection of two planes in symmetrical form 246 3.5. lntersection of a straight line and a plane 247 3.6. Necessary and sufficient conditions of coplanarity of two straight lines 247 249 3.7. Skew lines and the shortest distance between them 251 3.8. Equations of two skew lines 3.9. A straight line intersecting other given lines 252 Examples 253 Exercise 272 Chapter 4. Transformation of Axes 4.1. Transformation of axes 4.2. Invariants 4.3. Area of a triangle and volume of a tetrahedron 4.4. Tetrahedron formed by four given planes Examples Exercise

277-287 277 279 280 281 282 285

Chapter 5. The Sphere 289-312 5.1. Definition and equation of a sphere 289 289 5.2. Equation of a sphere when the end points of a diameter are given 290 5.3. Plane section of a sphere 5.4. Spheres through a given circle 291 S.S. Intersection of a sphere and line 291 5.6. Length of the.chord and tangency of a line'. 291 5.7. Tangent plane 292 5.8. Condition of tangency of a plane 292 293 5.9. Plane of contact of the tangent planes to a sphere 5.10. Polar plane of a point with respect to a sphere 294 5.11. Pole of a plane 294 [ix]

5.12. 5.13. 5.14. 5.15.

Properties of pole and polar planes Orthogonal intersection of two spheres Radical plane Properties of radical planes Examples Exercise

295 295 296 296 296 308

313-331 Chapter 6. Cone 313 6.1. Definition 6.2. Equation of cone when the vertex and the base is given 313 6.3. Equation of a cone with vertex as origin 314 6.4. Condition for the general second degree equation represents a cone 315 6.5. General equation of the cone passing through the coordinate axes 315 6.6. Intersection of a cone with a plane 316 6.7. Right Circular Cone 317 6.8. Condition that the general second degree homogeneous equation represents a right circular cone 318 Examples 319 328 Exercise

Chapter 7. Cylinder 7.1. Definition 7.2. Equation of cylinder when the axis and the base is given 7.3. Right circular cylinder 7.4. Equation of right circular cylinder Examples Exercise

333-337 333 333 334 334 334 336

Chapter 8. Surface of Revolution 8.1. Definition 8.2. Equation of the surface of revolution Examples Exercise

339-341

Chapter 9. Conicoids 9.1. Central conicoids 9.2. Shapes of central conicoids 9.3. Ellipsoid and its properties 9.4. Hyperboloid of one sheet and its properties 9.5. Hyperboloid of two sheets and its properties 9.6. Shapes of non-central conicoids 9.7. Elliptic paraboloid and its properties 9.8. Hyperbolic paraboloid and its properties

343-350 343 343 344

[x]

339 339 340 341

345 345 346 346 347

9.9. Paraboloids are limiting case of central conicoids Examples Exercise

348 349 350

Chapter 10. Tangent, Normal, Pole and Polar 351-379 10.1. Equation of a tangent to the surface 351 10.2. Tangent plane and condition of tangency 352 10.3. Condition of tangency of the plane Ix + my+ nz == 0 to the cone ax 2 + by2 + cz 2 + 2hxy + 2gxz + 2fyz == 0 353 10.4. Equation of the plane of contact of the tangent planes to a conicoid 353 10.5. The pole and polar plane 354 354 10.6. Director sphere 355 10.7. Reciprocal cone and its equation 356 10.8. Enveloping cone and its equation 10.9. Enveloping cylinder and its equation 357 10.10. Equation of a normal to a conicoid 357 10.11. Normals from a given point 358 10.12. Cubic curve through the feet of the normal 358 359 10.13. Cone through the six normals to a conicoid 10.14. Cone through the five normals to a paraboloid 360 361 Examples 372 Exercise Chapter 11. Diameter 11.1. Diameter 11.2. Conjugate diameters of a central conicoid Examples Exercise

381-389 381 383 383 386

Chapter 12. Generating Lines 391-406 12.1. Ruled surface and generating lines 391 391 12.2. Condition for a straight line to be a generator of a conicoid 392 12.3. Generating lines of hyperboloid of one sheet and their properties 12.4. Generating lines of hyperbolic paraboloid and their properties 394 12.5. Locus of the point of intersection of perpendicular generators 396 Examples 397 Exercise 403 Chapter 13. The General Second Degree Equation 13.1. The general equation 13.2. Principal planes and principal directions 13.3. [>rincipal directions and their properties 13.4. Centre and reduction of central conicoid 13.5. Classification of quadrics

407-421 407 407 408 410 411

13.6. Canonical form of the equation of qua 0. Find also A, B. 6. What does the equation x 2 - 2xy + 4y 2 + 7x - 18y + 32 = 0 become when the origin is moved to the poin•. (2, 3) and the axes are rotated through an angle 45°? 7. If ax +by and ex +dy are changed to a'x' +b'y' and e'x' +d'y' respectively for rotation of axes, show that ad - be = a' d' - b' e'. 8. If ax 2 + 2hxy + by 2 is changed to a'x' 2 + 2hx'y' + b'y' 2 respectively for rotation of axes, show that (a - b) 2 + 4h 2 =(a' - b')2 + 4h' 2 . 9. Show·that there exists only one point whose coordinates do not alter due to transformation of axes. 10. If under translation, the expressron ax 2 + 2hxy + by 2 + 2gx + 2fy + e transforms to 2 a'x' + 2h'x'y' + b'y' 2 + 2g'x' + 2f'y' + c', then show that (a)a=a', (b) b = b', (c) h = h', (d) !i = !i'.

11

CH. 1: TRANSFORMATION OF COORDINATES AND INVARIANTS

11. If under rotation, the expression ax 2 + 1.hxy + by 2 + 2gx + 2/y a'x' 2 + 2h'x'y' + b'y' 2 + 2g'x' + 2/'y' + e', then show that (a) a+ b =a'+ b', 2 (b) ab - h 2 = a'b' - h' , (c) 1 2 + g2 = !'2 + g'2, (d) be+ ab+ ae - / 2 - g 2 - h 2 = b'e'

+ e transforms to

2 2 g' - h' . 12. lfby rotation ofrectangular axes about the origin, ax 2+2hxy+by 2 and ex 2+2gxy+dy 2 2 be changed to a' x' 2 + 2h'x' y' + b' y' 2 and e' x' 2 + 2g'x' y' + d' y' respectively, then show that ad + be - 2hg = a'd' + b' e' - 2h' s' and ae + bd + 2hg = a' e' + b' e' + 2h' g'. 13. Transform the equation 2x 2 - xy + y 2 + 2x - 3y + 5 = 0 to new axes of x and y given by the straight lines 4x + 3y + 1 = 0 and 3x - 4y + 2 = 0 respectively.

+ a'b' + a'e' - f' 2 -

[Hints: Solving 4x + 3y + 1=0and3x - 4y + 2 = 0, we get x = -2/5 and y = 1/5. So, shifting origin to the point (-2/5, 1/5), we get the transformed equation as

2(x' - 2/5) 2 or,

-

(x' - 2/5)(y' + 1/5)

+ (y' + 1/5)2 + 2(x' - 2/5) - 3(y' + 1/5) + 5 = 2 2 50x' - 25x'y' + 25y' + 5x' - 55y' + 101 = 0.

0

Now, to take the line 4x + 3y + 1 = 0 as x-axis we have to rotate the axes through an angle tan- 1 ( -4 /3). In this case the equations of transformation are I 3X +4Y x=--5

and

I

y =

3Y-4X 5

Using these results in the above equation, we get 46X 2

+ 31XY + 29Y 2 + 47X -

29Y

+ 101

= O.]

14. If (x, y) and (x', y') be the coordinates of the same point referred to two sets of rectangular axes with same origin and if ~x + vy becomes u' x' + v' y', where u and v are independent of x, y, then show that u 2 + v 2 = u' 2 + v' 2.

"This page is Intentionally Left Blank"

Chapter 2

Pair o~ Straight Lines

2.1

Equation of a Pair of Straight Lines

The general equation of a straight line is given by Ax + By + C == 0, where at least one of A and B is non-zero. Consider two similar equations of straight lines

ax +by +c == 0

a' x

(I)

+ b' y + c' == 0.

(2)

Next consider the equation

(ax+ by+ c)(a'x

+ b'y + c') =

0.

(3)

Obviously, the points on the lines (I) and (2) will satisfy the Eqn. (3) which is second degree equation in x, y of the form

Ax

2

+ 2H.(1· + By2 + 2Gx + 2Fy + C =

0.

(4)

Hence, equation of the form (4) may represent a pair of straight lines, where it is assumed that at least one of A, 8, H is non-zero.

2.2

A Homogeneous Equation of nth Degree Represents n Straight Lines through Origin

Let the homogeneous equation of 11th degree be

aoyn

+ a1yn--lx +a2yn-2x2 +- .. +anxn == 0

(5)

which can be written as

(6) where z == y / x. If the above algebraic equation has n roots (real or imaginary and all of them may not be distinct) m 1, m2, ... , mn then Eqn. EM can be written as

ao(z - m 1)(z - m2) .. . (z - mn) = 0 or,

ao(y - m1x)(y -- m2x) .. .(y -- mnx) == 0.

13

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

14

Hence, Eqn. (5) represents fl straight lines given by

y - m 1x = 0; y - m2x

= O;

y-m11X

=0

all of which clearly pass through origin. For fl = 2, let us consider the equation

ax 2

+ 2hxy + by2 = 0.

(7)

If y - m 1x = 0 and y - m2x = 0 be the individual equations of the line represented by Eqn. (7), then comparing both sides of

ax 2 + 2hxy

+ by 2 = b(y -

we obtain

m1 + m:.

2h

= -b;

m 1x)(y - m2x)

m1m2

a

= b.

(8)

The straight lines will be real, imaginary or coincident according as m 1, m2 are real, imaginary or equal. Considering Eqn. (7) as quadratic equation in y we get

y

=

-h

± ../h 2 -ab b

x.

Hence, the straight lines will be real, imaginary or coincident according as h 2 - ab >,

+ 8ABxy +. (B 2 -

Now the length of the perpendicular from.O on AB is p = The area of the triangle

= (1 /2) pAB = pAD =

,, " = 0. 3"A-).r

(I)

C

J A2 + B2

p · p tan 30°

c2 ~

of a'x 2

11. If one of the lines given by ax 2 + 2hxy + by 2 = 0 coincides with one of the lines + 2h'xy + b'y 2 '-" 0 and the other lines are perpendicular, then show that,

2h'ab 2ha'b' ~- - = - - = v -aa'bb'. b- a b' - a' Solution: Ii y = m 1x and y ax 2 + 2hxy + by 2 = 0, then

m2x be the individual equations of the lines given by

2h

and

m1+fn,,=·--

-

b

( 1)

y

1 y=--x m2

x I

If we take y = m 1x and y = - - x as the individual equations of the lines given by m2

a'x 2 + 2h'xy

+ b'y 2 =

0

then I

2h'

m2

b'

m1--=----

(2)

and

mi m2

a'

= b'

(3)

CH.

2:

25

PAIR OF STRAIGHT LINES

From Eqn. (2) and Eqn. (3) we get -m2 _aa' _ I -

(4)

bb' a.a'bb'

-m,2 = bii}2 m1 =

(5)

,,/-aa'bb' bb'

(6)

Using Eqn. (6) in Eqn. (I) we get

mz =

ab' J-aa'bb'

(7)

.

Now, using Eqn. (6) and Eqn. (7) in Eqn. (2) we get

J -aa'bb' (~ _ ~) = _ 2h'. b' b a b'

(8)

Again, using Eqn. (6) and Eqn. (7) in Eqn. (I) we get

(_!_ _ _!_) =

,,/-aa'bb' b b'

_ 2h.

.a'

(9)

b

From Eqn, (8) and (9) we get the result

2h'ab = ~ = J-aa'bb'. b-a b' -a'

12. If the lines ax 2 + 2hxy + by 2 = 0 be the sides of a parallelogram and the line Ix + my = I be one of the diagonals, show that the equation of other diagonal is y(bl - hm) = x(am - hi). ~

Solution: Let OABC be the parallelogram formed by the lines ax 2 + 2hxy + by 2 = 0 and the lines parallel to them. Let

and

l1x+m1y=O

(1)

l2x +m2Y = 0

(2)

be the individual equations of the lines given by ax 2 + lhxy

+ by 2 =

0, then we have

Obviously, the diagonal given by the equation

Ix+ my= I

.

is not passing through the origin. So the otlier diagonal will pass through the origin. If A (x1; YI) be the point of inters~ction .~f Eqns. (I) and (3), then

x, = lm1 - ml1 ; m1

-/1 YI=----·-.

lm1 - ml1

(3)

26

ANALYTICAL GEOMETRY OF

Two

AND THREE DIMENSIONS

B

x

. Similarly, tlic point of intersection C(x2, y2) of Eqns. (2) and (3) is given by

If D be the middle poir1t of AC, then coordinate of U is

[~Cm2 ~ ml2 + /m1 ~ m/J' ~Cm2--~ ml2 + lm1 ~ mlJ l 1

1 2

1

2

1

On simplification we get the coordinate of D as ( b/ 2

-

bl - hm am - hi ) 2hlm + am 2 ' b(2 - 2hlm + am 2

·

Hence, equation of the other diagonal is bl - hm

x-0 y-0

----,.-------.,.. - 0

b/ 2 - 2hlm + am 2 am - hl -2- - - - - - 0 bl - 2hlm + am 2

= x (am

which gives the equation of other diagonal as y(bl - hm)

- hi).

~

13. Find the product of the perpendiculars from (p, q) to the lines represented by 2 ax + 2hxy + by 2 = 0. Solution: If y = m 1x and y = m2x be the individual equations of the lines given by

ax 2 + 2hxy

+ by 2 = 0

then

2h b

m1+111-,=--

and

(I)

a

m1m2-= ,;·

(2)

If PI, P2 are the perpendiculars drawn from ( p. q) on y = m 1x and y = m:.x, then (q - pm 1 )(q -- prrf'2) PIP" = - - - - - . = = = ·

- Frnr)(1 + m~)

q 2 - (m 1

/1 +

+ m2)pq + m 1m2p 2

+ m~ + m~m~ ap 2 t 2hpq + bq 2 Jca - b) 2 + 4h 2 mr

CH.

2:

27

PAIR OF STRAIGHT LINES

~ 14. Find the length of the line joining the feet of perpendiculars from (p, q) on the lines ax 2 + 2hxy + by 2 = 0.

Solution: Let OA and OB be the lines represented by ax 2 + 2hxy + by 2 = 0 and P(a, /3) be the moving point. PA, PB are the perpendiculars drawn from P on OA and OB; then 0, A, P, B lie on a circle having OP as diameter. Let l.AOB = e and l.OBA =¢.Then we have tane =

2./h 2 - ab . a+b

(I)

y

x

()

From the triangle .1AOB we get

AB sin()

OA sin¢·

(Z)

OP sin 90°

OA sin¢

(3)

Again, from triangle !1.AOP we gel

From Eqns. (2) and (3)

~ 15. If the straight line ax +by + c = 0 bisects an angle between a pair of straight lines of which one is Ix +my+ n °-=-: 0, show that the other line of the pair is

(Ix+ my+ n)(a 2

+ b 2 ) -- 2(al + bm)(ax +by+ c) = 0.

Solution: Let the equation of AB be Ix +my+ n = 0

(I)

ax+ by+ c = 0.

(2)

and equation of AC be Then equation of the other line, i.e. AD can be written as

Ix +my+ n

+ ),(ax+ by + c) = 0.

Now, slopes of the lines (I), (2) and (3) are m1=

m

I -t- a,.i mJ-= - - - -

m

+ bA. ·

(3)

28

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Now as angle between Eqns. ( 1) and (2) is same as angle between Eqns. (2) and (3), we have

a

l +aA. a m +bA. m b --~---= . al · a(l + aA.) b

1+---b(m

1+-

+ bA.)

bm

On simplification which yields

+ bm) + b2 .

A. = _ 2(al a2

(4)

Using Eqn. (4) in Eqn. (3) we get the required result. ~

16. Find the equations of the diagonals of the parallelogram formed by the lines

= 0,

(I)

ax+ by +d = 0,

(2)

+ b' y + c' =

(3)

ax+ by+ c

a' x

0

and a'x+b'y+d'=O. Show that the parallelogram will be a rhombus and area of the parallelogram i·s

(4)

(a' 2 + b' 2 )(c - d) 2 ;

if (a 2 + b 2 )(c' - d') 2 =

(c -- d)(c' - d')

.

ab' - a'b

Solution: Equation of the straight line through the intersection of Eqns. ( 1) and (4) is ax+ by+ c

+ A.(a'x + b'y + d') = 0

and equation of the straight line through the intersection of Eqns. (2) and (3) is ax+ by+ d

+ µ(a'x + b'y + c

1 )

= 0.

If the above lines are identical, then that line will be a diagonal; so A. d + µc'. Hence equation of the diagonal is (c' - d')(ax +by+ c)

+ (c -

d)(a'x

=µ

and c

+ Ad' =

+ b'y + d') = 0.

Proceeding in the same way we get the equation of the other diagonal is (c' - d')(ax +by -t- d) - (c - d)(a'x

+ b' y + c') =

0.

The parallelogram will be rhombus if the distances between the pairs of parallel lines are same. Now distance between the I st pair of parallel lines is p=

c-d

.Ja 2

t

b2

and that between the 2nd pair of parallel lines is •

pI

c' -d' Ja12

+ b'2

CH.

2:

PAIR OF STRAIGHT

29

LINES

So the required condition is

c' -d'

c-d ,Ja2+b2

i.e.,

(a

2

+b

2

Ja'2+b'2

2

)(c' - d') = (c/

2

+ b' 2 )(c -

d) 2 .

The required condition can also be obtained using the condition that the diagonals of a rhombus are perpendicular to each other. If()

be the angle between the straight lines ax + by + c tan

e=

= 0 and a' x

+ b' y + c'

= 0, then

ab' - a'b ---aa' + bb'

Hence the area of the parallelogram c- d c' ~ d' [ I+ (aa' + bb') ] l/Z -Ja2 + b2 Ja'2 + b'2 ab' - a'b 2

= pp'cosec() =

(c - d)(c' - d') ab' - a'b

EXERCISE I. Prove that the equation of the bisectors of the angle between ax 2 + 2hxy + by 2 = 0 is (a - b)(x 2 - y 2)

+ 4hxy

= 0.

2. Prove that two pairs of straight lines ax 2 + 2hxy + by 2 = 0 and a 2x 2 + 2h(a + b)xy + 2 b y 2 = 0 have the same bisectors. 3. Prove that two pairs of straight lines ax 2 + 2hx y + by 2 = 0 and a'x 2 + 2h' xy + b' y 2 = 0 have the same bisectors if h(a' - b') = h'(a - b). . 4. If two pairs of straight lines x 2 - 2 px y - y 2 = 0 and x 2 - 2qx y - y 2 = 0 be such that each bisects the angle between the other pairs then prove that pq = -1. 5. Show that lines joining origin to the point of intersection of ax 2 +2hxy +by 2 +2gx = 0 2 and a'x + 2h'xy + b'y 2 + 2g'x = 0 will be at right angle if g'(a + b) = g(a' + b'). 6. Show that lines joining origin to the point of intersection of the parabola y 2 = 4ax and the straight line y = mx + c are (a) at right angles if c + 4am = 0, (b) coincident if a c=-.

m 7. Find the angle between the lines joining origin to the point of intersection of~+~= I a

8. 9. 10. 11.

b

and the circle x 2 + y2 =ax+ by. Show that the two straight lines x 2 (tan 2 & +cos?-()) - 2xy tan()+ y 2 sin 2 () = 0 make with the x-axis angles such that the difference of their tangents is 2. Show that the equation of the lines through the origin, each of which makes an angle a with the line v = x is x 2 - 2xv sec 2a + v2 = 0. Show that fo~r lines given by.(y - mx)i = c~(I + m 2 ) and (y - nx)2. = c2(1 + n2) form a rhombus. Show that the pair of lint>S (a - b)(x 2 - y 2 ) + 4hxy = 0 and the pair of lines 2 2 h(x - y ) - (a - b)xy = 0 are such that each pair bisects the angle between the other pair.

30

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

12. Show that the area of the triangle formed by the straight lines ax 2 + 2hxy + by 2 = 0 . Jh 2 - ab . and lx +my = 1 1s 0 am- - 2hlm +bl-0 13. Show that the triangle formed by the straight lines ax 2 +2hxy+by 2 = 0 and lx+my = 1 is (a) right-angled if (a +b)(a/ 2 + 2hlm + bm 2 ), (b) isosceles if h(l 2 -m 2 ) = (a -b)lm. 14. If the equation ax 2 + 2hxy + by 2 + 2gx + 2.f y + c = 0 represents two intersecting straight lines, then show that the square of the distance of their point of intersection from origin is c(a+b)-f 2 -g 2 ab -- h 2 15. Prove that the straight lines ax 2 + 2hxy + by 2 + l(x 2 + y 2 ) = 0 have the same pair of bisector. Interpret the case when l =-(a+ b).

[Hints: Equation of the bisectors of the given lines is given by 0

or,

0

x- - y-

xy

(a+ l) - (b + l) 0 0 x- - yxy a - b h

h (I)

which is independent of l and hence for all values of l the given equation has the same pair of bisectors. When l = -(a +b ), the given equation becomes bx 2 -2hx y +ay 2 = 0 which has the same bisector given by Eqn. (I). I 16. A triangle has the lines ax 2 + 2hxy + by 2 = 0 for two of its sides and the point (p, q) for its orthocentre. Prove that the equation of the third side is (a + b)(xp + yq) = bp 2 - 2hpq + aq 2 .

[Hints: Taking the equation of third line as~+~ = 1 and using the results of Example 8 a fJ we get pa

p -

-

p

= qfJ = (a

aa. 2 q

fj = ;-

= l

b

a+b 2h

+ b )a.[3 2 2ha.{J + b[J 2 (say)

.ccc}

aq - 2hqp + = ------2

a

a

-+ p2 a.2 a{J

bp 2

and

pA

=

and

(a+ b)p

(a+ b)a. 2 [J q = aa.2 - 2ha[J + b[J2 (a+ b)qp 2 .

--~-----~­

aq2 - 2hqp

fJ

= aq2 -

+ bp2 2hqp + bp2

(a+ b)q

. . x y Puttmg these values m - + - = 1 we get the required result.] a. fJ

17. If the lines ax 2 + 2hxy + by 2 = 0 form an equilateral triangle with the line x cos a+ y sin a. = p, show that

1 - 2 cos 2a

h =--2 sin 2a.

b I+ 2cos2a. ·

CH.

2:

31

PAIR OF STRAIGHT LINES

[Hints: Putting A= cosa; B =sin a; C =-pin the Eqn. (I) of Example 10 we get -~ a )~ x(cos-~ a - 3 sm-

==>

(I -

2 cos 2a)x 2

-

· + 8 cos a smaxy + c·ry sm- a -

+(I + 2 cos 2a)y2

4xy sin 2a

3 cos-~ a )~ .V- =

o

= 0.

Now comparing this result with ax 2 + 2hxy + hy 2 = 0 we obtain given result which was to prove.] 18. If the equation ax 2 + 2hxy + by 2 + 2gx + 2f y +.c = 0 represents a pair of straight lines, then show that the equation to the third pair of straight lines passing through the ~ · ~ 4fgxy points where these meet the axes is ax- -· 2hxy +by-+ 2gx + 2fy + c + - - - = 0.

c

[Hints: Given that the equation ax 2 + 2hxy

+ by2 +

2gx + 2fy + c = 0

(!)

represents a pair of straight lines. Equation of coordinate axes is xy

= 0.

(2)

Now equation of the third pair of straight lines passing through the point of intersection of Eqns. (I) and (2) can be written as

+ 2hxy + by 2 + 2gx + 2fy + c + 2hy = 0 ax 2 + 2(h + A.)xy + hy 2 + 2gx + 2fy + c = 0. ax

or,

2

(3) ·

Since Eqn. (3) represents a pair of straight lines, we have abc

+ 2fg(h

+A.) - af 2

-

bg 2 - c(h

+ A.) 2 = 0

which on simplification gives A.

=

2(fg - ch)_

(4)

c

Using Eqn. (4) in Eqn. (3) we get the result.] 19. If the equation ax 2 + 2hx y + hy 2 + 2gx + 2fy + c = 0 represents two straight lines equidistant from origin then show that f 4 - g 4 = c(bf 2 - ag 2 ).

[Hints: If l1x + m1y the lines given by

+ c1

= 0 and l2x

+ tn2Y + c2

ax 2 + 2hxy t by2

= 0 be the individual equations of

+ 2gx + 2fy + c

=:cc

0

then we have l1l2 =a; m1m2

= b;

c1c2

= c;

l2m1 +.f1m2 = 2h; l2c1 +l1c2 = 2g; c2m1+·c1m2=2f.

(I)

Since the lines are equidistant from origin, CJ

-·---

jt? +mi jti + m~ ciui

+ m~) =du? +mi)

2 2 c.2 2 _ CI 12 ·- 21I -

2

2

Czm1 ·-

Using Eqn. (I) in Eqn. (2) we get the required result.]

2

2

C1m2

(2)

32

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

20. Show that the pair of straight lines ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 form a rhombus with the straight lines ax 2 + 2hxy + by 2 = 0 if h(g 2 - f 2 ) = gf (a - b). [Hints: Equation of the diagonal AC can be written as ax

2

+ 2hxy + by 2 + 2gx + 2fy + c + A.(ax 2 + 2hxy+ by 2 ) == 0.

Since equation of AC is a linear equation in x, y, there would be no second degree term in the above equation which leads that A. = - I. Hence equation of AC is (I) 2gx + 2/y + c = 0. The point of intersection of the lines given by the equation ax 2 + 2hxy + by 2 + 2gx + . o f.mtersection . o f the 1mes . . . (hf - bg , hg - af) an d the pomt given 2 y + c = 0 1s

!

ab- h 2 ab- h 2 · by the equation ax 2 + 2hxy + by 2 = 0 is (O, 0). So the slope of the diagonal OB is hg-af) . g . . ( hf- bg and that of AC 1s - -f . As the diagonals of a rhombus are perpendicular to

each other, we have _ _!(hg - af) =-I f hf -bg . •

h(g 2 - f 2 ) = gf(a - b).]

==:}

21. Show that the pair of straight lines (ab -- h 2 )(ax 2 + 2hxy + by 2 + 2gx + 2fy) + af 2 + bg 2 - 2fgh = 0 represents a pair of straight lines and form a rhombus with the straight Jines ax 2 + 2hxy + by 2 = 0, if h(g 2 - f 2 ) = gf(a - b). [Hints: We know that ax

2

+ 2hxy + by2 + 2gx + 2fy + c =

0

(I)

represents a pair of straight lines if

+ bg 2 + ch 2 - 2fgh - abc = 0 af + bg 2 - ~f gh - c(ab - h 2 ) = 0 af 2 + bg 2 - 2fgh = c(ab - h 2 ) af 2 + bg 2 - 2fgh af 2

i.e., if i.e., if

2

(2)

i.e., if C = - - - - -ab- h 2

Using Eqn. (2) in Eqn. (I) we get the first result and for the second part see problem no. Ex. 20.} 22. Prove that if one of the lines ax 2 + 2hxy + by 2 = 0 be perpendicular to one of the lines of a'x 2 + 2h'xy + b'y 2 = 0, then (aa' - bb') 2

+ 4(ah' + hb')(a'h + h'b) =

0. ·

23. Prove that if one of the bisectors of the angle between ax 2 + 2hx y with one of the lines of Ax 2 + 2Hxy + By 2 = 0, then h 2 (A

+ 8) 2 =

{2hH -- B(a - b)}{2hH

+ A(a -

+ by 2 = 0 coincides b)}.

CH.

33

2: PAIR OF STRAIGHT LINES

24. Prove that if one of the lines ax 2 + 2hxy a'x 2 + 2h'xy + b'y 2 = 0, then (ab' - ba') 2

+ by 2 =

= 4(bh' -

0 coincides with one of the lines of

hb')(a'h...:. h'a).

25. A parallelogram is formed by the pair of straight lines ax 2 + 2hxy + by 2 = 0 and the straight lines through (p, q) parallel to those. Show that the equation of the diagonal, which does not pass through the origin, is (2x - p)(ap

+ hq) + (2y -

q)(bq +hp)= 0.

[Hints: The equation of the pair of str!light lines through the point (p, q) and parallel to the lines (l) • ax 2 + 2hxy + by 2 = 0 is

a(x - p) 2 + 2h(x - p)(y - q)

+ b(y -

q) 2

= 0.

(2)

As the diagonal, which does not pass through the origin, passes through the point of intersections of Eqns. (I) anq (2), we can write its equation as a(x - p) 2 + 2h(x - p)(y - q)

+ b(y -

q) 2 + A.(ax 2 + 2hxy

+ by2) =

0.

(3)

Since equation of a straight line is a linear equation in x, y, value of A. should be such that there would be no second degree term in x, y in Eqn. (3) which leads that A. = -1. So from Eqn. (3) we get 2hyp - 2hxq + 2hpq + bq 2 - 2bqy = 0 2apx + 2hxq - ap 2 - hpq + 2hyp + 2bqy - bq 2 - hpq = 0 -2apx

=> =>

+ ap2 -

(2x - p)(ap

+ hq) + (2y -

q)(bq +hp)= O.]

26. Show that the equation of the line joining feet of the perpendiculars from the point (d, 0) on the straight lines ax 2 + 2hxy + by 2 = 0 is (a - b)x + 2hy + bd = 0.

[Hints: Let y =mix and y = m2x be the individual equations of the lines given by ax 2 + 2hxy + by 2 = 0.

Then we have m1 +m2 =

2h

-b

and

a

m1m2

= ;;·

(l)

Equation of the perpendicular drawn from (d, 0) to the line y =mix is

Solving the equations

y= mx and x + my= d we get 1

1

A: Similarly,

F-3

(1:mr.1:m~r)·

f B 1s: . ( - -d. , , dm2 . the~oord mateo -) I +mi I+ m 22

(2) •

(3)

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

34

Therefore, equation of AB is d

d

.x-~ =

dm1

dm1

+ mi)x -

d

~~~~~-

m1

=

+ m2

2h

~~~-

(1 +mi )y - d m 1

(a - b)(I

dm2

I +mi - I + m~

y - I +mi (I

d

}+~-~

m1m2-I

+ mT)x -

+ 2h(1 + mi)y -

b-a

d(a - b)

(4)

2dhm1 = 0.

Since y = m1x is one of the lines given by ax 2 + 2hxy

+ by 2 =

0, we have

a+2hm1 +bmi=O

(5)

b(I +mi)= b- a - 2hm1.

Rewriting Eqn. (4) as (l + mi)[(a -b)x + 2hy] = d[a -b + 2hmi] and using Eqn. (5) we get

(I+ mr)[(a =:}

(a - b)x

b)x

+ 2hy]

+ 2hy + bd =

= -db(I

+ mf)

0.]

27. Find the area of the triangle formed by the x-axis and the bisectors of the lines ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0. [Hints: The equation of the bisectors of the pair of straight lines ax 2 + 2hxy + by 2 + 2gx

+ 2fy + c = 0 is given by

(x - a) 2 - (y - {J) 2

a-b

=

(x - a)(y - fJ)

h

where the point of intersection (a, fJ) of the lines ax 2 +2hxy +by 2 +2gx +2fy +c = 0 is given by a=

j 12

-be;

(l)

h 2 -ab

If the bisectors meet x-axis at (x1, 0) and (x2, 0), then x1 - a and x2 - a are roots of (x - a) 2 -

a-b

fJ 2

(x - a)(...:.p)

=---h

i.e. of the equation h(x - a) 2 +(a - b)(x - a)fJ - h{J 2

= 0.

So, we have (x1 - a)+ (x2 - a)=

-fJ(a - b) h

and

(x1 - a)(x2 - a)=

-p 2 .

(2)

CH.

2:

35

PAIR OF STRAIGHT LINES

Now, (x1 - x2) = (x1 -- a) - (x2 - a)= /3J

(a•~ b)

2

+ 4.

So, area of the triangle = (x1 - x2)/3/2=132 .J[(a - b)2 + 4h2] 2h 2 .J[(a - b) + 4h 2] g 2 - ac = 2h h2 - ab·

28. Prove that the equation ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a pair of parallel straight lines if~= ~ = }·Find the distance between them.

[Hints: Let ax 2 + 2hxy

+ by 2 + 2gx + 2fy + c = (Jax+ Jby + ci)(.Jax + Jby + c2).

Comparing both sides we get 2.J(ab) = 2h Ja(c1 Jb(c1

+ c2) = + c2) =

c1c2

(I)

=c

(2) (3) (4)

=}

(5)

2g 2f

Froin Eqn. (I) we get ab= h 2

Dividing Eqn. (2) by Eqn. (3) we get /a_ab=fg Vb -

=> af 2 = bg 2 => abf 2 = b 2" g~

=>

h2 !2 = b2g2

=>

hf =bg h g -=b

[by (5)]

(6)

f

By Eqns. (5) and (6) we get

h g = h b f Now, difference between two parallel straight lines = difference of their distances mea. . CJ Cz sured tirom ongm = r::-71:: ,...., r::-71:: ..;a+b va+b a

- =-

g 2 - ac

=2 - - a(a

+ b)

{by Eqns. (2) and (4) }.]

36

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

29. A point moves so that the distance between the feet of perpendiculars frcm it on the lines ax 2 + 2hxy + by 2 = 0 is a constant distance 2d. Show that its locus is

(x 2 + y 2 )(h 2 - ab) = d 2 [(a - b) 2 + 4h 2]. [Hints: Let OA and OB 1}e the lines represented by ax 2 + 2hxy + by 2 = 0 and P(a, ft) be the moving point. PA, PB are the perpendiculars draw from P on OA and OB; then 0, A, P, B lie on a circle having OP as diameter. Let LAOB = and LOBA = ,.PTQ =

=-

~TP x TQ

a

(

../h2

+ k2 -

) .

a2

x sin20

= TP x TQ x sinfJ x cos8 = (h 2

+ k2 -

a(h2

a

a 2) x - - - = x Jh2

+ k2

"lh 2

+ k2

.../h2

-

a2

+ k2

+ k2 - a2)J/2 h2 + k2

~ 8. If Si = 0 and S2 = 0 represent two circles, then find the equation of the curve which passes through the points of intersection of Si = 0 and S2 = 0. Solution: Let S1 = x 2 + y 2 + 2gx + 2fy + c and S2 = x 2 + y 2 + 2g'x + 2f'y + c'. Then

Si

+ kS2 = x 2 + y 2 + 2gx + 2fy + c + k(x 2 + y2 + 2g' x + 2f' y + c') =(I + k)x 2 +(I + k)y2 + 2(g + kg')x + 2(/ + kf')_v + (c +kc').

CH.

3:

53

CIRCLE

So, S1 + k S2 = 0,

x

i.e.,

2+ v- + 2 (gt kg') --- x + 2 (j -+ -kj') - y + ·(C -+kc') - x = 0. 'l

.

l+k

l+k

l+k

As the equation S1 + kS2 = 0 is satisfied by those points which satisfy both S1 = 0 and S2 = 0, this is the equation of the curve which passes through the points of intersection of S1 = 0 and S2 = 0, and represents a circle if k -=fa -1. When k = -1, then S1 - S2 = 0 gives 2(g - g')x + 2(g - J')y + c - c' = 0, which represents the common chord of the circles S1 = 0 and S2 = 0. ~ 9. If S = 0 and P = 0 represent a circle and a straight line respectively, then find the equation of the circle described the chord P = 0 of the circle S = 0 as diameter.

= x 2 + y 2 + 2gx + 2f y + c and P = lx +my+ n, then

Solution: Let S

S

+ kP

2

+ y2 + 2gx + 2fy + c + 2k(lx +my+ n) = x 2 :+- y 2 + 2(g + kl)x + 2(f + km)y + c + 2kn.

= x

As the equation S + k P = 0 is satisfied by those points which satisfy both S = 0 and P the equation of the circle passes through the points of in~rsection S = 0 and P = 0 is

x 2 + y2

+ 2(g

t kl)x

= 0,

+ 2(f + km)y + c + 2kn = 0

whose centre is the point { - (g +kl), -(f +km)} and this point must lie on the diameter

Ix +my+ n = 0 i.e.,

l(g +kl)+ m(f +km)= n k(f 2 + m 2 ) = n - lg - mf k = n - lg -mf 12 + m2

or, or, So, the equation of the circle is ,,

x-

+ y 2 + 2gx + 2fy + c + 2

n - lg - mf + m 2 (Ix+ my+ n) 12

= 0.

~ 10.A chord of circle x 2 + y 2 - 2ax = 0 is drawn to pass through the origin. Show that the locus of the centres of the circles described on this chord as diameter is a circle passing through the centre of the given circle.

Solution: Let us take y = mx as chord of the given circle, where m is a variable. Then by Problem No. 9, the equation of the circle on this chord as diameter is given by x 2 + y2 or, whose centre is (a,

k

/J)

--2 =

x

2

+ y2 -

2ax (2a

+ k(y - mx) = 0 + km)x + ky = 0

= (a +km /2, --k /2). If this point lies on the line y

m(2a +km) 2

=>

k(I

,,

= mx, then

2am

+ m-) = -2am => k = ----. 1 + m2

54

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

y

x 2+/-ax = 0

)' =

mx

x

x 2 +/-2ax = 0 am 2

Now, a= a +km/2 =a- - 2

l+m

Eliminating m we get

a2 or,

a

2

+ p2 =

+jJ

2

-

a

am

l+m

l+m

= - -2 and/1 = -k/2 = - -2 .

(-a-)2 + (_!!!!!__)2 + + I

m2

I

m2

=

~2 = aa I

+m

aa '·' 0.

Hence, locus of the centre, i.e. of the point (a, /J) is x 2 + y 2 - ax passing through the point (a, 0), i.e. the centre of the given circle.

=

0, which is a circle

[Proved] ~

11. Prove that the equation of the circle on the common chord of the circles

and x 2 +(y-k) 2 =k 2

(x-h) 2 +y2=h 2 as diameter is (x 2

+ y 2 )(h 2 + k2 ) = 2hk(kx + hy).

Solution: Equation of the common chord is

or,

(x - h) 2

+ y2 -

xh - yk

= 0.

h 2 - [x 2 + (y -- k) 2 - k 2 ] = 0

y

x

55

CH. 3: CIRCLE

Now, equation of the circle, which passes through the points of intersection of the circle (x - h) 2 + y 2 = h 2 and the line xh - yk = 0 is

+ y2 - h 2 + ).(xh - yk) = 0 x 2 + y2 + (). - 2)xh - ).yk = 0 (x - h) 2

or, whose centre is (

(I - ~ )h, ).2k). If this point lies on the line xh -

=

yk

0, then we have

or, or, Hence, the equation of the ci_rcle is

x-,

h-

+ k-

2k 2

, ,, x-+y--,, h-

or,

2

)

2 xh -

2h , ,, yk "-'- 0 .. h- + k-

2h 2

xh2

+k 2 2 2 (x + y )(h + k 2 ) =

or, ~

2

+ y-, + ( ,2h , -

,,vk=O h 2 + k-· 2hk(kx t hy). [Proved]

12. If two plane curves

+ by 2 + 2gx + 2/y + c = 0 Ax 2 + 2H xy + By 2 + 2Gx + 2Fy + C =

ax 2 + 2hxy and

0

intersect in four concyclic points, prove that H(a-b)=h(A-B). Solution: Equation of the curve which passes through the point of intersection of the given curves can be written as

+ 2hxy + by 2 + 2gx + 2f y + c +k(Ax 2 + 2Hxy + By 2 + 2Gx + 2Fy + C) (a+ kA)x 2 + 2(h + kH)xy + (b + kB)y 2 + 2(g + kG)x + 2(/ + k F) y + c + kC = 0.

ax 2 or,

= 0

If the above equation represents a circle, then we have Coeff. of x 2 i.e.,

= Coeff. of y2

a+kA =h+kB

Eliminating k we get H(a - b)

and

and

Coeff. of xy

=0

h+kH =0.

= h(A - B).

[Proved]

~ 13. Prove that two circles each of which passes through two fixed points (0, k) and (0, -k) and touch the line y = mx + b will cut orthogonally, if b 2 = k 2 (2 + m2 ).

56

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Solution: If one of the circles be x2 +

y2 +

2gx + 2f y + c

=0

then we have From these equations, we get f = 0 and c = -k 2 . So, the equation of the above circle becomes x 2 + y2

+ 2gx -

k2 = 0

J

whose centre is (- g, 0), radius = g2 + k2. Now, if the above circle touches the line y = mx b - mg= j(g 2 + k2)(1

or, or,

( 1)

+ b, then we have

+ m2)

b2 - 2mg + m2g2 = g2(1 + m2) + k2(1 + m2) g 2 + 2mg + k 2(1 + m 2) - b 2 = 0.

If g1, gz be its two roots, then

Now, two circles of the form (I) will cut orthogonally if

+ 0 = -k 2 k 2 (1 + m 2) - b 2 = b 2 = k2(2 + m2 ).

k2

• 2.g1g2

or, or, ~

-k 2

[Proved]

14. If S = 0, S' = 0 be two circles of radii r and R, then prove that the circles

S S' - ± - = 0 will intersect orthogonally. r R Solution: Without loss of generality, we assume the equation of the circles as S = (x :_ a ) 2

and

s' = (x + a )

2

+ y2 + y2 -

r2 = 0

R2

= O.

. . "d uaI equations . . Now, "h t e md1v1 of the circles -S ± -S' = O are r R

(i)RS+rS'=O, i.e.,

R[(x - a) 2 + ,.,

r)x~

y2 - r 2 ] + r)y~

r[(x + a) 2

or,

(R +

or,

-rR(R+r)=O 2 x + y 2 - 2af(R - r)/(R

and (ii) RS - rS' = 0, i.e.

+ (R +

,.,

+ y2 -

R2 ] = 0

- 2a(R - r)x + (R + r)a

+ r)]x + a 2 -

rR = 0

2

CH.

3:

57

CIRCLE

,. ,

2

,. ,

R[.(x-a)-+y -r-]-r[(x+a) or, or,

2

2

,. ,

,. ,

+y~-W]=O

,. ,

,. ,

(R - r)x + (R - r).V- -- 2a(R + r)x + (R - r)a~ x 2 + y2- 2a[(R + r)/(R - r)]x + a 2 + rR = 0.

+ r R(R -

r) = 0

Now, we see that the condition of orthogonality, i.e.

2g1g2

+ 2f1f2 =CJ+ C2

S S' is satisfied for the circles (i) and (ii). Hence the circles - ±- = 0 will intersect orthogonally. r R [Proved] ~ 15. Find the circumcentre of the triangle formed by the lines x + y = 0, x -y = 0 and lx +my= I. If land m vary such that 12 + m 2 = I, show that the locus of the circumcentre of the triangle is (x 2 - y2)2 = x2 + y2.

Solution: Solving the equations x + y = 0, x - y time, we get the vertices of the triangle are A(O, O);

B ( -I- -I - )

= 0 and lx +my =

and

L+m'l+m

I, taking two at a

-I- ) . C ( - I- , l-m l-m

If S: (h, k) be the circumcentre then we have SA 2 = SB2 = SC2 and hence

h 2 + k 2 = ( h - -I1+ m From h-,. , .

. + k2 =

)2 + (

k - -Il+m

( h - -I-

l+m

h 2 + k 2 = ( h - - -I 1 -m

)2 = (h - 1-m - - )2 + ( + - - )2 l-m J

k

)2 + (k ._ -I- ) 2, we get h + k l+m

)2 + (k + -I- ) 2, we get h l-m

J

= -Il+m

and from

I k =--.Solving these we get l-m

I -m ) . , l 2 -m 2 l 2 -m 2 Again, given that 12 + m 2 = I, using this, we get

. . ( the c1rcumcentre 1s

h2

+

k2 =

2 + m (/2 _ m2)2

12

=

I

2

2 2

(12 _ m2)2 = (h - k ) ·

Generalizing, we get the required locus is (x 2 - y 2 )2 = x2

+ y2.

EXERCISE I. Show that the lengths of the intercepts on the axes by the circle x 2 + y 2+2gx +2fy+c = 0, are 2Jg 2 - c and 2J f 2 - c. 2. Show that if a circle x 2 + y2 = a 2 cuts off from the straight line y = mx + c, a chord of length 2b, then c 2 = (I + m 2 )(a 2 -- b2).

[Hints: Putting y

= mx + c in x 2 + y 2 = a 2 , we get (I

+ m 2 )x 2 + 2mxc + c'J -

a 2 = 0.

58

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

If the line y "': mx

+r

cuts the circle x 2 + y 2 = a 2 in (x 1, YI) and (x2, y2), then

.,

.,

2mc x2 +xi= - - - 5

c- ·--a-

t.,x 1 == -I+ --.m2 .

and

I +m-

Now, given that (x2 - xi)

2 + (y2 - yi) 2 = 4b 2

or,

(x2 - x1) 2[ 1+m 2]=4b 2

or,

[(x2

or,

41m-c- - (c- -- a-)(1+m-)J=4b (I+ m )

+ x1 ) 2 ")

"I

"I

")

.,

..,

2

.,

2 .,

.,

4b 2

+ m 2] =

4x1x2Jl l

or,

m-c- - c- - m c- = -(I

or,

c2 = (1

+ m2)(a2 -

+m

2

.,

2

2

)(a- - b )

h2).l

3. Prove that the equation of the circle described on the chord x cos a + y sin a = p of the circle x 2 + y 2 = a., as diameter is x 2 + y 2 -- a 2 - 2p(x cos a + y sin a - p) = 0. [Hints: Equation of the circle described on the chord x cos a + y 2 = a 2 as diameter can be written as

+ y sin a =

p of the circle

x2

x 2 + y2

-

a 2 + 2k(x cos a

+ y sin a

-- p) ~.: 0

whose centre (-k cos a, -k sin c() lies on the line x cos a + y sin a = p, etc.] 4. Show that two circles x 2 + y 2 + 2gx + 2fy = 0 and x 2 + y 2 + 2g' x + 2f' y tou

a 2h f(t1

=>

a-h- - 2 - 4h) a a

Generalizing, we get

, ,(k

+ tz) 2 2

,

(given)

4t1t2] = 4A

2

. , = 4k

,

.x 2 [y--4.xa]=4k.

~ 11. Find the area of the triangle formed by the tangents drawn from R(h, k) to the parabola y 2 = 4a.x and their chord of contact.

Solution: Let the tangents are dnrwn from the point R(h, k) to the parabola y 2 = 4ax and touch it at P(atr, 2at1) and Q(ati, 2at2), then the tangents at P, Q are yt1 = x and

· ytz

h = at1t2

and

=x

')

+ atl

(I)

+at?.

(2)

Solving Eqns. (I) and (2) we get k = a(t1

+ tz).

(3)

74

ANALYTICAL GEOMETRY OF

Two

AND THREE DIMENSIONS

Now, length of PQ

=

J(atr - ati) 2 +

(2at 1 - 2at2)2

t2)J(t1 + t2) 2 + 4

= a(t1 -

= aJ(r;-+-i2)'.~--=Mit;J(fi"+t2) 2 = a

+4

j

~-~-:-: Vr,;;·~ ;2 ~~a a1ri· ---····-- - · - -

==

v lk 2 --- 4ah)(k2 + 4a2)

----·-----------~---

a

Now, equation of PQ (being chord of contact of the point (h, k) w.r.t. the parabola) is yk = 2a(x

+ h).

So, k 2 - 4ah

RS=

_ ~4a 2

and the area of the triangle I

t\PQR =

l PQ

x RS

(k 2 ~ 4ah) 3 12

= - --a-2

~ 12. Find the locus of the middle points of the portion !f the tangents intercepted between the curve and directrix.

Solution: Equation of the tangent at P(at 2 , 2at) to the parabola y 2 o= 4ax is yt

It meets the directrix x

=

=

-a at Q: ( - a, a

then we have h=

~-(t 2 -

1)

x +at 2 .

(t - ~)). ~ (3t - ~).

and

U (h, k) be the middk: point of PQ,

k=

Eliminating t from the above results we get k 2 (2h +a)= a(3h

+ a) 2 .

y2(2x +a)= a(3x

+ a) 2 .

Generalizing, we get

~ 13. Show that from a point three normals can be drawn to a parabola y 2 = 4ax and feet of the normals lie on a rectangular ityperbola. Further, show that sum of the ordinates of the feet is zero.

Solution: We know that the equation of the nor.nal to the parabola y 2 = 4ax at (am 2 , -2am) IS

y=mx-am 3 -2am.

75

CH. 4: PARABOLA

If it passes through (h, k), then

k

am 3 - 2am

= mh -

am 3 +m(2a-h)+k=0.

or,

The above cubic equation in m implies that from a point (h, k) three normals can be drawn lo the parabola y 2 = 4ax. Now, if the normals cut the parabola at (amr, -2am 1), (am~, ~2am2) and (am~, -2am3), then we have

0 Sum of the roots= m1 + m2 + mJ = - = 0. a

Now, sum of the ordinates of the feet= -2a(m 1 + m2 + mJ) = 0. Again, consider the equation of normal at {a, fJ), which is fJ y - fJ = - -(x - a). 2a If it passes through (h, k), then

or,

k - fJ = _!!__(h - a) 2a 2ak - 2afJ = -(Jh + (Ja

or,

(Ja

+ (2a

-- h)fJ - 2ak

= 0.

Hence, locus of the foot of the normal is xy

+ (2a

- h)y

+ 2ak = 0

which represents a rectangular hyperbola. ~ 14. Find the locus of the point through which three normals can be drawn to the parabola y 2 = 4ax such that two of the normals coincide. Solution: Let (h, k) be the point through which three normals are drawn to the parabola y 2 =

4ax and ifthe normals cutthe parabola al (amr, -2am 1), (am~, -2am2) and (am~, -2am3), then we have

m1+m2+m3=0, 2a -h m1m2 + mzm3 + m1m3 = - - a k and m1m2m3 = --. a

Now, putting m 1 = mz in Eqns. (I )-(3) we get m3 = -2m1, 3 i. -2mj

and or,

= ---a

2a - h 2 2 m1-m3=--

a

2a - h -3m 21 = - - . a

(I)

(2) (3)

76

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Now, eliminating m 1 we get k2

(2a - h) 3 4a2 27a3 2 27ak + 4(2a - h) 3 = 0.

or,

Hence, the locus of the point (h, k) is

27ay2

+ 4(2a -

x) 3 = 0.

~ 15. Three normals are drawn from a point to the parabola y 2 = 4ax meet the axis in points whose distances from the venex are in A.P. Show that the locus of the point is

27ay2

+ 2(2a -

x) 3 = 0.

Solution: Equation of the normal to the parabola y 2 = 4ax at (am 2 , -2am) is

y = mx - am 3 - 2am. It meets the axis at (am 2 + 2a, 0). Let (h, k) be the point through which three normals are drawn to the parabola y 2 = 4ax and if the normals cut the parabola at (amr, -2ami), (am~, -2am2) and (am~, -2am3), then we have

m1+m2+m3=0, 2a -h m1m2 + m2m3 + m1m3 == - - a

Given that

2(am~ 2

+ 2a) = 2

ami

+ 2a +am~+ 2a

2

2mz =m1 +m3 2 2 2 2 2a-h 3m 2 =m 1 +m 2 +m 3 = 0 - 2 - a ., 2a -h m].=-2--3a

or, or, Again

or, or,

(2)

(3)

and

or,

(I)

. 2m 22 =m 21 -tm 23 =m 22 -2m1m3 2 2k m2=-am2 3 2k m.,=-. ~

a

Eliminating m2, we get 4k 2

or,

8(2a -h) 3 a2 27a 3 2 27ak + 2(2a - h) 3 = 0.

=

77

CH. 4: PARABOLA

Generalizing, we get 27ay~

+ 2(2a -

x) 3 = 0.

[Proved]

~ 16. Find the locus of the point through which three normals can be drawn to the parabola y 2 = 4ax such that two of the normals make complementary angles with the axis of the parabola.

e,

Solution: Let one of the normals makes an angle then another will make an angle 90° with the axis. So, mt m2 = tan 0 x cot 0 = I. Now using the following results m1 +m2+m3 =0 2a -h m1m2 + m2m3 + m1m3 = - - a

e

(I)

(2) (3)

we obtain

~

k m3. = --· a'

2a -h

1-m] = - - - . . a

Eliminating m3 we get k2 a2

2a - h

)--=--

a

k2 h -a --.,. = - -

or,

a

a~

Generalizing, we get

y2 =

a(x - a).

17. If the normals at P, Q, R on the parabola y 2 = 4ax meet in the point (h, k), 2(h - 2a) ) , 0 and orthocentre of the then show that the centroid of the triangle /l..PQR is ( 3 ~

~riangle /l..PQR is (h - 6a, -~).

Solution: If (h, k) be the point where the three normals to the parabola y 2 = 4ax meet and if the feet of the normals are P: (amT, -2am1), Q: (am~, -2am2) and R: (am~, -2am3), then we have m1 +m2 +m3 = 0 2a -h m1m2 + m2m3 + m1m3 = - - a

k

/J)

is given by

a=~ LmT = ~ [ (Lm1) =

2 -

2Lm1m2]

~(0 _ 2 2a:h) 2a -h 3

=...-2---

(2) (3)

a

Now, centroid of the triangle (a,

(I)

78

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

and fl

=-

2a" L., m 1

3

. . (2(h - 2a) ) = 0. So, the centroid of the tnangle l':!.PQR 1s ,0 3

.

Now, equation of the line through P and perpendicular to QR is

.

and equation of the line through Q and perpendicular to PR is

Solving the above equations we get

= x(m2 - m1) -a(m1 -m2)(m1m2 2a - h -4a - a - - = h - 6a

4a(m1 -- m2)

or,

x

and

(y

=

a

+ 2am1)(m1 + m3) -

+ 2am2)(m2 + m3) = 2(m2 - ml)(m2 + m3)(m1 + m3) (m 1 - m2)Y = 2a[m2(m2 + m3) - m 1(m1 + m3)] a , , + 2(mz - mJ)(m2+m3)(m1 + m3) (mi - mz)y = 2alm~ - mf + m3(m2 - m1}] a , 2 + 2==< NP

·~

2

k 2 >=< y2 = b 2 (1- h 2 /a 2 ) h 2/a 2 +k 2/b 2 -1>= b) is x 2

+ y 2 = a 2.

89

CH. 5: ELLIPSE

Let P be any point on the ellipse and PN be a perpendicular to the major axis. If NP is ' extended to meet the auxiliary circle at Q, then clearly abscissae of both P and Q are same. As the points (x, P N) and (x, Q N) lies on the ellipse and the auxiliary circle respectively, we have x 2 · PN 2 a2

+ b2 =

1

2

xi+ QN = a2

and which give

= ~QN

PN

a

this shows that any ordinate on the ellipse is

~ times the ordinate of the corresponding point

a

on the auxiliary circle. If the line joining center of the ellipse and the point Q on the auxiliary circle makes an angle

1) e being the eccentricity, then the equation of hyperbola becomes x2

y2

a2

b2 -

----1 where b 2 = a 2(e 2 - l), which is the standard/basic/normal/canonical form of hyperbola.

Another Definition Hyperbola is the locus of a point which moves in such a way that difference of its distances from two fixed points, called foci, is constant 2a, length .of the transverse axis.

K'

S'

A'

y

K

A(a,O)

(O,O)C

101

·x

102

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

6.2

Some Important Results Associated with the Standard Form of Hyperbola

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

Centre: C(O, 0), Foci: S(ae, 0), S'(-ae, 0) Transverse axis of the hyperbola: x-axis, Conjugate axis of the hyperbola: y-axis, End points of transverse axis (±a, 0), End points of conjugate axis (0, ±b), Eccentricity: e = J 1 + b2 / a 2 , Equation of the directrices: x = ±a/e, 2b 2 (ix) Latus rectum= - , a

2

~ ) and L' ( -ae, ± ~ ), 2

(x) End points of lat us rectums: L ( ae, ±

(xi) Focal distance of any point (x, y): a± ex.

6.3

General Equation of Hyperbola

If we take the line ax+ by+ c = 0 as directrix and the point (h, k) as focus, then for a point P (x, y) in the hyperbola

SP = and

j (x -

h )2

+ (y -

k )2

ax+ by +c PM=-===,.Ja2+b2

Using SP = ePM, we get or, or,

[(x - h) 2 + (y - k) 2][a 2 + b 2] = e 2(ax +by+ c) 2 [a 2 + b 2][x 2 + y 2 - 2xh - 2yk + h 2 + k 2] = ei[a 2x 2 + b 2y2 + 2~bxy + 2axc + 2bcy Ax 2 + 2Hxy + By 2 + 2Gx + 2Fy + C = 0

+ c2 ]

where A= a 2 (e 2 - l)-b 2 , B = b 2(e 2 - 1) - a 2, H = abe 2 , G = (a 2 + b 2)h + e 2ac, F = (a 2 + b 2)k + e 2bc and C = e2c2 - (a2 + b2)(h2 + kz).

6.4

Position of a Point P(h, k) with Respect to the Hyperbola x2/a2 - y2/b2 = 1

Let Q(h, k) be any point on extended NP, where P(h, y) is any point on the hyperbola x2 y2 2a - 2b = 1 as shown in the following figure.

103

CH. 6: HYPERBOLA

y

Q(h,k)

V(O, 0)

x N(h,0)

Now the point Q is outside, on·or inside the hyperbola according as NQ >==< NP 2 k2 >=< y2 = -b2(1 - h2 /a2)

or,

h 2 /a 2 -k 2 /b 2 -1>= b) is x 2 + y 2 = a 2.

Conormal Points 2

2

- y = 1 at the point (x1, y1) is a2 b2 x -x1 y-y1 - y1/b 2 • x1/a 2

Equation of normal to the hyperbola x

If it passes through (h, k ), then h -xi

k-y1

x1/a

Y1/b

- - 2 = - - - 2 =).(say).

Simplifying, we get

a 2h xi = -a2-+-A.

and

YI

-b 2k = -b 2 +)..

As the point (x1, YI) lies on the hyperbola, we have

a2h2 b2k2 -----1 (a2 + J.)2 (-b2 + J.)2 from which we get four values of ). and which indicates that four normals can be drawn to the ellipse from the point (h, k) and feet of the normals are given by 2

2

-kb ) ha ( a 2 + A.; , -b 2 + A.; Now, the results a 2

+ ). =

a2h

-

and b 2

Xj

-

).

,

i = I, 2, 3, 4.

b 2k = give YI

a 2 h b2 k a2+b2=-+x1 YI which shows that the feet of the normals lie on the curve • 2

or, which is .a rectangular hyperbola.

2

a2+b2 =ah+~ x y xy(a 2 + b 2) = a 2hy + b 2kx

108

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Another Method

Equation of normal to the hyperbola :: xaC£Js8

~: =

1 at the point (a sec 8, b tan 8) is given by

+ ybcot8 =

a2

+ b2 .

If it passes through (h, k), then we have

ha cos() + kb cot 8 = a 2 + b 2. Introducing the variable t = tan(8 /2) in the above equation we get

or,

I - t2 1 - t2 a h - -2 +bk--= a 2 +b 2 1+ t 2t 2 2aht(I - t ) + bk(l - t 4 ) = 2(a 2 + b 2)t(I + t 2)

from which we get four values of t, i.e. of 8 and which indicates that four normals can be drawn to the hyperbola from the point (h, k). Simplifying the above equation in t, we get bkt 4 + 2(ah + a 2 + b 2)t 3 + 2(-ah + a 2 + b 2)t - bk·~ 0. If the feet of the normals are given by (a cos 8;, b sin 8; ), where t;

by theory of equation

= tan (

i),

then we have

·

L

-2(ah + a 2 + b 2) t1=------bk

Now, by basic trigonometry, we have 81 ()2 83 ()4) :Lt1 - :Lt1t2t3 t a( n - + - + - + - = 2 2 2 2 l-:Lt1t2+t1t2t3t4

=OO

i.e. (81 + 82 + 83 + 84) is an odd multiple of 7r. So, four normals can be drawn from a point to the hyperbola and the sum of the eccentric angles of their feet is an odd multiple of n. Examples ~

1. In a rectangular hyperbola, prove that SA x SA' = a 2, where S is one of the foci

and A, A' are the vertices of the hyperbola. Solution: Let the equation of the rectangular hyperbola be x 2 - y 2 = a 2. Now, So,

SA= ea - a= J2a - a

and

SA'= ea+ a= J2a +a. SA x SA'= (J2a - a)(J2a +a)= 2a 2 - a 2 = a 2. [Proved]

.109

CH. 6: HYPERBOLA

~ 2. In a rectangular hyperbola, prov~ that SP x S'P = CP2, where Pis any point on the hyperbola, S, S' are the foci and C is the centre of the hyperbola. Solution: Let the equation of the rectangular hyperbola be x 2 - y 2 = a 2. If P(h, k) be any point of the hyperbola, then SP=eh-a=.j2h-a

S'P=eh+a=.j2h+a.

and

SP x S'P = (.j2h - a)(.j2h +a)= 2h 2 - a 2 = 2h2 - (h2 - k2) = h2 + k2 = c p2,

So,

[Proved]

~ 3. Find the equation of the chord of a hyperbola joining two points whose eccentric angles are 8 and¢. .

Solution: Equation of the line joining P(a sec8, b tan8) and Q(a sec, b tan) is b(tan ¢ - tan

y - btanO x-asecO

bcos

y - btan8 x-asecO

or,

a

(t+

13. If A and B he two points on the ellipse --., a~

. v')

+ b·~,

= 1 such thm three times the

eccelllric angle of the one is equal to the supplement of that of the other. then find the locus of the pole of AB with respect to the ellipse.

125

CH. 7: POLE AND POLARS

Solution: Equation of the line joining A: (a cos(}, b sin 0) and B: (a cos¢, b sin¢) is ay sin (

Now, given that¢+ 38

f!__~

+ (2 sin

2

2

+ [-4 cos W

+ y' cos 8)

- 3 sin W]x' y'

2

8 + 4 cos8 sin 8 - cos 8)y' = 6. 2

To removex'y' we need that

i.e.,

-4cosW- 3sin28 = 0 cosW sinW I =--= 3 -4 5

Hence, the above equation becomes 2

2

[(I+ 3/5) + 8/5 - (I - 3/5)/2Jx' +[(I - 3/5) - 8/5 - (I+ 3/5)/2]y' = 6 or, or,

3x' 2 x'2

-

2y' 2 = 6

y'2

2-3=1. 4. Reduce the following equation to its canonical form: 7x 2

-

2xy + 7y 2 - 16x + I6y - 8 = 0.

Solution: Here a= 7,b = 7,c = -8,h = -1,g = -8 and f = 8. So, abc + 2fgh - af 2 - bg 2 - ch 2 = -256 =F 0, hence the given equation is not ofpair ·· · .· ... of straight lines. Also, ab - h 2 = 48 f. 0, so the given equation is of central conic. .· : , .,-: ;.:.~·. Let (a, /J) be the centre of the conic. Shifting origin to the centre of the conic, i.e. putting·'. '. ~· (bx - ayf = -2ab(ax+ by - a;) and the lines bx - ay = 0 and ax+ by - -ab 2

. . these l"mes = 0 are perpendicular. Taking

as X and Y axes we see that y

3. 4. 5. 6. 7. 8. 9.

=

bx -ay -;====:o Ja2 + b2

X=

and

ax+ by - ab/2 Ja2

+ b2

.

2ab So, the above equation becomes Y 2 = - - - - X , etc.J . . Ja2+b2 . Find the equation of axis and latus rectum of the parabola whose equation is given by 9x 2 + 24xy + 16y 2 + 8x - 156y - 95 = 0. Show that the curve 3x 2 - 2xy + 3y 2 - 4x -- 4y - 12 = 0 has a single centre. Show that the curve x 2 + 4x y + 4y 2 - 20x + I Oy - 50 = 0 has no centre. Show that the curve x 2 + 4xy + 4y 2 + 4x + 8y + 4 = 0 has infinitely many centres. Find the values of a and f for which the curve ax 2 -20xy +25y 2 -14x +2fy- 15 = 0 (i) has a single centre; (ii) has infinitely many centres. Reduce the equation x 2 + 2xy + y2 + 2x +2y +k = 0 to its canonical form and determine the type of the conic represented by it for different values of k. Show that the conic passing through the point of intersection of two rectangular hyperbola is also a rectangular hyperbola.

[Hints: If ax 2 +2hxy+by 2 +2gx +2fy+c = 0 and a'x 2 +2h'xy+b'y 2 +2g'x+2f'y+ c' = 0 represent two rectangular hyperbola. then we have a + b = 0 and a'+ b' -= 0. Equation of the conic passing through the puint of intersection of the above conics can be written as ax 2 + 2hxy + by 2 + 2gx

+ 2fy + c + k(a'x 2

t 2h'xy

+ b'y2

+2g'x + 2f'y + c') =

or,

0 2 (a+ ka')x + 2(h + kh')xy + (b + kb')y2 + 2(g + + 2(/ + kf')y + c +kc'

+ kg')x

=0

in which we see that Coeff. of x 2 +Coeff. of y 2 = (a+ka')+(b+kb') = a+b+k(a' +b') = 0.] I 0. Prove that in general two parabolas can be drawn through the point of intersection of two conics ax 2 +2hxy+by 2 +2gx+2fy+c = Oanda'x 2 +2h'xy+b'y 2 +2g'x+2f'y+c' = 0 and their axes are at right angles if h'(a - b) = h(a' - b').

[Hints: Equation of the conic passes through the point of intersection of the above conics can be written as ax 2 + 2hxy + by2 + 2gx + 2fy + c +2g'x or,

+ k(a'x 2 + 2h'xy + b'y2

+ 2f'y + c') = 0

(a + ka')x 2 + 2(h + kh')xy

+ (b +

+ 2(g + kg')x + 2(/ + kf')y + c

kb')y 2 +kc'= 0.

214

ANALYTICAL GEOMETRY OF

Two AND THREE DIMENSIONS

This will be a parabola if

+ ka')(b +kb') =

(a

(h

+ kh') 2

2 (a'b' - h' )k 2 +(ab'+ ba' - 2hh')k +ab - h 2 = 0

i.e. if

(I)

which is a second degree equation on k (if both the conics are not parabolas) giving two values of k. Hence, two parabolas can pass through the points of intersection of the given comes. If k1 and k2 be two roots of the above equation ink, then axes of the parabolas can be taken as Ja

+ k1a'x + Jb + k1b'y + k1 = 0

and

Ja

+ k2a'x + Jb + k2b'y + k2 = 0.

These will be at right angle if

')

+ kz)aa + k1k2a 12 = b 2 + (k1 + kz)bb + k1k2b t2 2 2 h' )a 2 -- (ab' + ba' - 2hh')aa' + (ab -- h 2 )a'

or,

a-+ (k1

or,

(a' b' -

I

I

2 2 = (a'b' - h' )b 2 - (ab'+ ba' - 2hh')bb' +(ab - h 2 )b'

or,

(ah' - a'h) 2 = (bh' - b'h) 2

or,

h'(a - b)

.

= h(a' -

b').]

.

11. Fmd the locus of the focus of the parabola

(x + - a

y b

1)

2

. van. ables = -4xy , a and b be mg ab

such that ab = c 2 .

[Hints: Proceed as Example 13.] 1 I 12. Show that + x+y-a x-y+a

. (a2' 2a) . JS

+

I = 0 represents a parabola whose focus x+y+a

ANALYTICAL GEOMETRY OF

THREE DIMENSIONS

"This page is Intentionally Left Blank"

Chapter 1

Coordinates

1.1

Introduction

In rectangular coordinate system, the position ofa point in space is defined by referring it to three fixed coRcurrent, non-coplanar straight lines which are perpendicular to each other. The common point is called origin and the lines are called coordinate axes. The planes through lines taken in pairs are called coordinate planes which divide the space in eight parts known as octants. If we take XOX', YOY' and ZOZ' meeting at 0(0, 0, 0) as the coordinate axes, then the position of any point Pis determined by the triad of numbers (x, y, z) as shown in the figure.

y

Z' P(x.y, z)

M

z

Y'

The points which satisfy the equations (I) y = 0, z = 0 lie on x-axis (2) x = 0, z = 0 lie on y-axis (3) x = 0, y = 0 lie on z-axis.' 217

2is

ANALYTICAL GEOMETRY OF TWCJ AND THREE DIMENSIONS

Similarly, the points which satisfy the equations (4) x = 0 lie on yz-plane (5) y = 0 lie on xz-plane (6) z = 0 lie on xy-plane. The whole space is divided by the coordinate planes in eight oclants, namely, OXYZ, OX' YZ, OXY' Z, OXYZ', OX' Y' Z, OX' YZ', OXY' Z' and OX' Y' Z'. The signs of the coordinates of a point determii•e the octant in which it lies, i.e. the sign depends on octant. The following table shows the signs for the octants: Octant~

-

OXYZ

ox'rz

+ + +

-

OXY'Z OXYZ' OX'Y'Z OX'YZ' OXY'Z' OX'Y'Z'

Coordinates..;,

x y

z

1.2

+

+ -

+ +

+

+

-

+

-

+

-

-

-

-

-

-

-

+

Distance between Two Points

Let the coordinates of two points P and Q be {x, y, z) and (x', y', z'). Draw PL parallel to OX, LM parallel to OY and MQ parallel to OZ respectively.

z Q(x',/.z')

M

x

0

y

Then from the above figure we have PM 2

=

PL2

+ LM2

=(Difference of the di~·'.anccs of P, Q from YOZ-plane) 2 +(Difference of the distances of P, Q from XOZ-plane) 2 = (x' - x)2 + (y' - y)2 and

PQ 2 = PM 2

+ QM 2

2

c-=

PM +(Difference of the distances of P, Q from YOX-plane) 2

=

(x' - x) 2 + (y' ·- y) 2 + (z' - z) 2 .

219

CH. l: COORDINATES

So, PQ =Jex' - x) 2 + (y' - y) 2 + (z' - z) 2. If P coincides with 0, then distance of the point Q(x', y', z') from origin is given by OQ

=

Jx'z

+ y'z + z'z.

z Q(x',y',z')

M

x

()

y'

y l

From the above figure, distance of the point Q from X-axis is QM, where

Distance of the point Q(x', y', z') from X-axis is Y and Z axes are

1.3

Jy'

2

+ z' 2 . Similarly, distance of Q fro~

J x' 2 + z' 2 and Jx' 2 + y' 2 respectively.

Coordinate of a Point Which Divides the Line Joining Two Points in a Given Ratio

Let P(xi, Yi, z1) and Q(x2, yz, z2) be two points and R(x, y, z) be a point on PQ such that PR: RQ =A.:µ.

·Q

z

R

µ

x

220

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Let the planes through P, Q, R and parallel to the YOZ-plane meet the OX-axis at P', Q', R'. Then from the above figure

P'R' PR x -xi RQ R'Q' X2 - X A. X - XJ µ x2 -x Axz + flXI X= A.+µ -

=}

=}

Similarly, AZ2

+ flZJ

z=----

and

A.+µ

. R 1s . t he mt'ddl e po mt . o f PQ, th en Its . coord'mates are (XJ If ,._, = µ, 1.e.

z

1

;

z

2

).

Again, if we put A.

=

k 11

=

+ x2 , YI + Y2 , 2

2

k, i.e. if R divides PQ in the ratio k : I, then its

coordinates are

(

XI+ kx2 YI+ kY2 ZI + kz2) k+I ' k+I ' k+I .

If A (xi, YI, z 1), B(x2, yz, z2), C(x3, y3, z3) be the vertices of the triangle ABC, then the · o f CB ·ts, 1.e. · o f D 1s · given . b y (. XJ· + x2 , ---·-, Y3 + Y2 Z1 ·+-z2) . 1t . 1s . mt'ddl e pomt - . F rom statics,

2

2

2

well known that the CG of the triangle is a point on the median which divides it in the ratio 2: l. So, if (x, y, z) be the CG of the triangle ABC, then

x

2 x2 + x3 + 1 . x I = - -2- - - -

2 +I

Similarly,

YI+ Y2 3

+ YJ

y=-----

B

D

and

x1+x2+x1 = -----

3

ZJ + Z2 + Z3 z=-----

3

~

c

C

E

A

Let A(x1, YI, zi), B(x2, y2, z2), C(x3, y3, z3.) and D(x4, y4, z4) be the vertices of the tetrahedron ABCD. Using the above result, we get the CG of the triangle ABC as F: x1 + x2 + x3 YI + Y2 + Y3 z1 + z2 + , . Now, from statics, we know that the CG of ( - - -3 - - , 3 3 . the tetrahedron ABCD is a point on the line joining any vertex to the CG of the opposite face and divides it in the ratio 3 : 1. So, if (x, y, z) be the CG of the tetrahedron ABCD, then

Z3)

3 x1 + x2 + x3 + 1 . x 4 3 XJ + X2 + x = -------·- = --3+1 4

.

X3

+ \'4

CH.

221

I: COORDINATES

Similarly,

y=

1.4

YI

+ Y2 + Y3 + Y4 4

and

z=

ZI

+ Z2 + Z3 + Z4 4

Projection of a Segment

If A' be the foot of the perpendicular drawn from A to a plane, then A' is called the projection of A on the plane. Similarly, if B' be the projection of the point B on the same plane, then we say A' B' is the projection of the line segment AB on the plane.

B

lf B be the angle between AB and A'B', then the length of A'B' is ABcosO. If A, /), C, .. ., M, N be any number of points in a plane and A', B', C', .. ., M', N' be their projections on a line PQ and these points being on the same straight line, we have A' B'

+ B c' + c' D + ... + M' N' =A' N' 1

1

where A' B' is the projection of AB on PQ. Similarly, B' C' is the projection of BC, C' D' is so of CD and so on. Hence sum of the projections of AB, BC, CD, ... on PQ, i.e. projection of AN on PQ is A'N'. It is to be noted further that projection of BA is B 1A 1 , i.e. projections of AB and BA differ in sign only.

1.5

Direction Cosines

In three-dimensional coordinate geometry, the angle between two non-intersecting lines AB and CD is measured as the angle between CE and CD where CE is parallel to AB, i.e. if a straight line AB is parallel to OP, where 0 is the origin, and OP makes a, f3, y with the coordinate axes, then AB is assumed to make the same angles with coordinate axes. Cosines of the angles which a straight line makes with the positive direction of axes are called direction cosines of the straight line. If a, /3, y be the angles which the line OP makes with x, y, z axes, then its direction cosines are cos a, cos f3, cos y and those of PO arc -- cos a, - cos f3, - cosy as it makes TC - a, TC - f3, y with OX, OY and OZ respectively. The x-axis makes 0°, 90°, 90° with x, y, z axes, so its direction cosines arc (1. 0, 0). Similarly, direction cosines of y and z axes are (0, 1, 0) and (0, 0. I) respe

2

[cos 2 a+ cos 2 fJ

+ cos2 y]

= 169

r = 13.

The direction cosines of the line segment are 3/r, 4/r, 12/r, i.e. 3/13, 4/13, 12/13. ~ 3. A line makes angles 60° and 45° with y-axis and z-axis. Find the direction cosines of the line.

0

If the line makes an angle a with x-axis, then we have,

cos 2 a + 1/4+ 1/2 "'=I

=> => =>

cos 2 a= 1/4

cosa = ±(1 /2) a= 60° or 120°.

So, the direction cosines of the line are ±cos 60°, cos 60°, cos 45°, i.e. ±( l /2), l /2, I/ J2. ~ 4. A line in the XOY-plane makes an angle 30° with the x-axis. Find the dircction cosines of the line.

CH.

I:

225

COORDINATES

0 Since, the line is on the XOY-plane, it makes 90° with z-axis. Again ••as it makes 30° with x-axis, it will make 60° with y-axis. Hence, Lhe direction cosines of the line is cos 30°, cos 60°, cos 90°, i.e. J3 /2, I /2, 0. ~ S. Find the projection of line segment joining the points (2, --1, 4) and (0, I, 5) on the straight line joining (3, J.., 5) and (5, 4; 3).

0 Direction ratios of the line segment joining (3, 3, 5) and (5, 4, 3) are -2, -1, 2. So, the direction cosines of that line are -2/3, - I /3, 2/3. So, the projection of the line segment joining the points (2, -1, 4) and (0, I, 5) on the line whose direction cosines are -2/3, - I /3, 2/3 is -2(2 - 0)/3 - 1(-1 - 1)/3 + 2(4 - 5)/3 = -4/3. ~ 6. If 11, m1, n1: 12, m2, n2 are two mutually perpendicular straight lines, show that the direction cosines of the line which is perpendioular to both of them are

m1n2 - m2n1, n1l2 - /1n2, m2l1 - m1l2.

0 Let/, m, n be the direction cosines of the line which is perpendicular to both the given lines, then

+ mm 1 + nn 1 = 0 l/2 + mm2 + nn2 = 0. /11

Solving we get

m n J12 + m2 + n2 ----- = = = = cosec90° =I. m1n2 - m2nt n1/2 - n2lt m2lt - m1l2 J~)m 1 n 2 - m 2ni)2 Hence, the direction cosines of the line which is perpendicular to both the given lines are m1n2-m2n1,n1/i-l1n2,m2lt -m1'2. ~ 7. Show that the straight lines whose direction cosines are given by al + bm +en = 0 and ul 2 + vm 2 + wn 2 = 0 are perpendicular or parallel according as . 2

a (w

D

a2

+ v) + b 2 (u + w) + c 2 (u + v) = 0

Eliminating l from al

+ + bm

en

= 0 and ul 2

or.

+

u

b2

c2

+ -v + -w

= 0.

vm 2 + wn 2 = 0 we get

u(bm + cn) 2 ---.,--+vm 2 +wn 2 =0 a~

or,

(b

2

u

+a v)(:f + 2

2ubc:

+

(c 2u

+ a 2 w) =

O

. m . -m . If -m1 an d -m2 be .its two roots, then we have . h.1s second degree equation wh1c

n

m1m2

=

n1n2 m1m2

c2u

n1

nz

c 2u + a 2 w b 2u + a 2v n1n2

1112

+ a2w = b2u + a2v = c2v + b2w

[By symmetry]

(I)

226

ANALYTICAL GEOMETRY OF Two AND THRBE DIMENSIONS

Now, the lines will be perpendicular if

+

/1/2 m1m2 + n1ni= 0 ' 2v)+(c-u+a ' 2 w)+(b-u+a-v) ' ' (b-w+c 2

a (w

+ v) + b

2

+ w) + c

(u

2

(u

=0

+ v) = 0.

Now, the lines will be parallel if m1

n1

m2

n2 m2

-=m1

-=-

:::::}

i.e., the roots of the Eqn. ( 1) are equal, i.e. if

=>

+ c 2u)(b 2 u + a 2v) a 4 wv + a 2c 2uv + b 2a 2 wu + b 2c 2u 2 a 4 wv + a 2 c2 uv + b 2 a 2 wu = 0

:::::}

-+-+=0. u v w

4u 2b 2c 2 = 4(a 2 w :::::}

~

2 u 2b c 2 =

b2

a2

c2

8. Show that the angle between the straight lines whose direction cosines are given

by I

+m +n =

0

Eliminating n from I+ m

0 and f mn

7r I = 0 is - if -

+ gnl +him

g(

+ n = 0 and

~)

fmn

. h.1s a second degree equation . .m wh1c ·

f- g - h)

I

g

0.

(~I ) + f

=c·.

0

(I ) -

m

. If -/1 and -/2 be .its two roots, then we have m1

11

m2

(I)

g

mJ1n2

and

I

+ - + -h =

+ gnl +him= 0, we obtain

2

+ (/

f

3

/2

-+-=m1 m2

f +g - h g

.

From Eqn. (1) we see that

From Eqns. (I) and (2) we get 11 m1 :::::}

m2

(/1m2 - /2mJ) 2

g

= k2 [(f + g -

h) 2 - 4/g].

(2)

227

CH. I: COORDINATES

Similarly, (l1n2 - /2ni) 2 = k 2[(f + h, - g) 2 - 4fh] (n1m2 - nzm1) 2 = k 2[(h + g - f) 2 - 4hg].

and 2

Now,tan

(TC) 3

2

L,(11m2 - /2mi) 2 =

=

(L,l1l2)2

k 2 L,[(f + g - h) - 4fg] . k2(J+g+h)2 gives

+ g + h) 2 = ~)U +. g - h) 2 - 4fgJ = [(f + g - h) 2 - 4fgJ + [(! + h - g) 2 -- 4fhJ + [(h + g - f) 2 - 4hg] 3rt 2 + g 2 + h 2 + 2(fg + gh + fh)] = 3lf 2 + g 2 + h2 J -- 6[fg + fh + gh] 12[/g + fh + ghJ = 0 Jg+ fh + gh = 0 I I I - + - + -- = 0. 3(f

~ ~

f

g

h

~ 9. If 11, m 1, n 1; 12, m2, nz are the direction cosines of two concurrent lines; show that the direction cosines of the bisectors of the angle between them are proportional to (11±12, m1 ± m2, n1 ± nz).

0 Let 0 be the origin and A, B be two points on the given lines such that OA =OB= r, then the coordinates of A and Bare 01r, m1r, n1r) and (l2r, m2r, nzr) respectively. We take a point A' on the extension of OA such that 0 is the middle point of AA'. Then coordinates of A' will be (-fir, -m1r, -n1r).

Now, if P, Q be the midpoints of AB, and A'B then OP and OQ are the internal and external bisectors of the angle AOB. Hence, coordinates of P, i.e. direction ratios of OP are

r(/1 (

+ /z) 2

r(m1

'

+ m2) 2

r(n1

'

+ nz)) 2

.

Similarly, direction ratios of OQ are

(

r(/z-/I) r(m2-mi) r(n2-n1)) 2 ' 2 ' 2 .

.

Hence, the direction cosines of the bisectors of the angle between OA and OB are proportional ~

228

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

~

10. If l1, m 1, n 1; 12, m2, n2 are the direction col·im:s of two concurrent lines; show

l1 +Li· m1 +m2 n1 + n2 ) . that the line with direction cosines ( . , , ---.(}-- bisects the angle 2 cos(O /2) 2 cos(B /2) 2 cos( /2) (), between the given lines. In the Ex. 9 it is shown that the direction cosines of the internal bisector of the angle between the given lines are proportional to (/1 + 12, m1 + m2, n1 + n2). Now,

0

JTt1 + /2) 2 + (m 1 + m2) 2 +(n1 + n2) 2 If+ mf + nf +

v'f+ I + 2 cos(),

l~ + m~ + n~ + 2(/1/2 + m1m2

+ n1n2)

where 0 is the angle between the given lines

= 2 cos(() /2). So, the direction cosines of the internal bisector are

l1 +l2 mi +mi n1 +n2 2 cos(O /2)' 2 cos(O /2)' 2 cos(O /2) · ~ 11. The direction cosines of a moving line in two adjacent positions are l, m, n and I + JI, m + Jm, n + Jn. Show that the small angle JO between the positions is given by

(J0)2 = (Jl)2

0

+ (Jm)2 + (Jn)2.

We have, cos JO= 1(1 +JI)+ m(m + Jm) + n(n +Jn) 12 + m 2 + n 2 = I and (1 + Jl) 2 + (m + Jm) 2 + (n + Jn) 2 = I.

Now,

I -- cos JO= -(lJl

J:) (2J0)1

::::}

2 sin 2 (

::::}

2

+ mJm + nJn)

=-(/JI+ mJm

= -(!Jl

(Jt/) 2 ~-= -2(/Jl

+ nJn)

+ mJm+ nJn)

+ mJm + nJn).

(I)

Again,

+ Jl) 2 + (m + Jm) 2 + (n + Jn) 2 = t 2 + m 2 + n 2 (Jl) 2 + (Jm) 2 + (Jn) 2 = -2(lJl + mJm + nJn). (1

(2)

From Eqns. (I) and (2) we get the result. ~ 12. If lk. mk. nk; k = I, 2, 3 be the direction cosines of these mutually perpendicular lines then show that · /1 m1 n1 12 1112 nz = ±I. /3 m3 n3 1

I

229

CH. 1: COORDINATES

D

Given that /1/2

+ m1m2 + n1n2 =

/1/3

+ m1m3 + rqn3

= 0

/3/2

+ m3f!l2 + n3n2

= 0

121

2 1

+m +n

2 _ 1-

0

I

Ii + m~ + n~ =I 2 /3

+ m32 + n32 =

1 ·

Solving the first two equations for l 1, m 1, n 1 we get /1

m1

-----= mzn3 - m3n2 nz/3 = =

n1

= l2m3 nJl2

/3m2

+ m1m1 + n1n1 ---------------------!, (m2n3 -- m3n2) + m 1(n2/3 - n3/2) + n 1(/2m3 - /3m2) /1/1

1

where f). is the determinant whose value is to be determined. From the above result we obtain /)./1 = m2n3 - mJn2, /).m1 = n2/3 - n3/2,

/).n1 = l2m3 - /3m2.

Squaring and adding we get f).

:::::}

2

(/f + mT + nT) =

~:)m2n3

- m3n2)

2

= sin 2 90° = I

/).=±1.

EXERCISE I. 2. 3. 4. 5.

6. 7. 8. 9. 10.

Find the direction cosines of the line joining the points (2, I, 5) and ( -1, -2, 2). Can the numbers I/ .j2, I/ .j2, I/ .j2 be direction cosines of a line? Find the direction cosines of line which is equally inclined to the coordinate axes. A line in the plane YOZ makes 45° with y-axis. Find the direction cosines of the line. Find the projection of the line joining the points (2, 3, 4) and (4, 1, 5) on the line joining (5, 8, 0) and (-1, 2, 6). . Prove that the angle between the diagonals of a cube is cos- 1(1/3). Prove that the angle between ttie lines whose direction cosines satisfy l + m + n = 0 and / 2 + m 2 - n 2 = 0 is 7r /3. Find the angle between the lines whose direction cosin'!s satisfy 2/ + 2m - n = 0 and mn + Im + ln = 0. Prove that the angle between the lines whose direction cosines satisfy l + m + n = 0 and mn - 2lm - 2/n = 0 is cos- 1(1/2). Show that the straight lines whose direction cosines are given by a 2 / + b 2m + c 2n = 0 and mn +Im +In = 0 are parallel if a + b + c = 0.

230

ANALY:rICAL GEOMETRY OF Two AND THREE DIMENSIONS

11. Show that the straight lines whose direction cosines are given by al + bm + en = 0 and f mn + gnl +him = 0 are perpendiculal' or parallel according as

f h - + ~- + -a b c

=0

. or

M

+

fog+ ../Ch = 0.

12. If l1, m 1, n1; l2, m2, nz and [3, m3, n3 be direction cosin~s of three mutually perpendicular straight lines, then find the direction cosines of a line whose direction ratios are l1 + /2 + [J, m 1 + m2 + m3, n 1 + n2 + n3 and show that the line is equally inclined to the given lines. 13. If the edges of a rectangular parallelopiped be a, b, c, show that the angles between the ±a2±b2±c2] . four diagonals are given by cos- 1 [ ~ . a-+ b 2 + c2 14: A line makes angles a., fJ, y , J with four diagonals of a cube; show that ~

1

~

~

cos- a. +cos- p +cos- y +cos- J = 4/3.

Chapter 2

The Plane

2.1

Introduction

A plane is a surface such that any point on the line segment joining any two points on it is also a point on it, i.e. the line joining any two points on it lies who.Hy on the surface. We shall now show that the surface represented by the linear eq-uation in x, y, z, ax +by

+ cz + d =- 0

(I)

where a, b, c, dare constants and not all of a, b, care zero, is a plane. Let (x1, }'1, z 1) and (x2, v~, z2) be two points on the surface (I), then we have

+ by1 + cz1 + d = 0 ax2 + byz + cz2 + d = 0. ax1

and

(2) (3)

Multiplying Eqn. (3) by k( f --1) and adding it to Eqn. (2) we get

+ kx2) + b(y1 + ky2) + c(z1 + kz2) + d(I + k) = 0. . (XI +kx2 , YI +ky2 , ZI +kx2) which . he . that the points

a(x1

.

. on the lme k+l k+I · segment joining (x1, y1, z1) and (x2, y2, z2) satisfy the Eqn. (1), i.e. the line segmentjoining (x1, YI, zi) and (x2, y2, z2) lie wholly on the surface (1 ). So, Eqn. (1) represents a plane. Subtracting Eqn. (2) from Eqn. (I) we get From this we sec

k+I

a(x -x1) +b(y-y1) +c(z-zi)

=0

which represents the equation of the plane through (x 1, YI, z1). F. om this we get the equation of the plane through origin as ax + by + cz = 0. Further we note that the Eqn. (I) involves four unknown constants but dividing the equation by d we see that the equation of a plane consists of three unknown constants. To determine three constants we need three equations which can be constructed using three conditions given in the problem like (i) The plane passes through three given non-collinear points, {ii) The plane passes through two given points and perpendicular to a given plane, (iii) The plane passes through a given point and perpendicular to two given planes.

231

232

ANALYTICAL GEOMETRY OF Two AND THREE DiMENSIONS

2.2 .Equation of a Plane through Three Given Points Let the equation of the plane be

ax+ by+ cz

+d =

0.

(4)

If it passes through the points (x1, YJ.ZJ), (x2, }'2, z2) and (x3, y3, z3), then we have

+ by1 + cz1 + d = 0, ax2 + hY2 + cz2 + d = 0 ax3 + by3 + cz3 + d = 0. ax1

and

(5) (6)

(7)

Eliminating a, b, c, d from Eqns. (4H7), we get

x y X1

I

i

YI ZI

x2 Y2

Z2

Y3

ZJ

X3

=0

which is a first degree equation in x, y, z and represents the plane through the points (x1, YI, z1 ), (x2, y2, z2) and (x3, YJ, Z3). If a plane intercepts the lengths a, b, c from the coordinate axes respectively, then it passes through the points (a, 0, 0), (O, b, 0) and (O, 0, c) and its equation is

x

y z I a 0 0 I =0 ObOI 0 0 c I or,

~+~+.:=I. a

b

c

This is the intercepting fonn of a plane.

2.3 Normal Form of a Plane Let ON be the perpendicular drawn from origin 0 to the plane ABC whose direction cosines are/, m, n. If P(x, y, z) be any point on the plane, then ON is perpendicular to PN, i.e. ON is the projection of OP on ON.

z

x y

=

=

=

If ON p, then p projection of OP on ON l(x - 0) + m(y - 0) Ix +my + nz = p, which is the equation of the plane in nonnal form.

+ n(z -

0), i.e.

CH.

2:

233

THE PLANE

Comparing this equation with the general equation of the plane, i.e. Ax+ By+Cz+ D = 0 we see that A, B, C are proportional to I, m, n, i.e. proportional to the direction cosines of the normal to the plane. Hence, in the general equation of the plane A, B, Care the direction ratios of the normal to the plane.

2.4. Angle between Two Planes As the angle between two planes is equal to the angle between the normal to the planes and the coefficients A, B, C of x, y, z in the general equation of a plane are the direction ratios of the norm·a1 to the plane, the anglt! (} between two planes

a1x a2x

+ b1y + c1z + d1 = 0 + b2Y + c2z + d2 ~ 0

is given by

So, if the planes are perpendicular, then

a1a2

+ b1b2 + c1c2 =

0

and if the planes are parallel, then

2.5

Equation of Bisectors of the Angle between Two Planes

Let the equations of the planes be

a1x+b1y+c1z+d1 =0 a2x

+ b2y + c2z + d2 = 0

where the constant terms are of same sign. If (x, y, z) be any point on the bisectors of the angles between the given planes, then the lengths of the perpendiculars from it to the bisectors are same, i.e.

a1x

+ b1y + c1z + d1

Jaf+bf+c~

_

-

± .a2x +b2Y + c2z + d2 Jai+bi+ci.

The positive sign in the r.h.s. will give the equation of the bisector of the angle which contains the origin and negative sign will give the e'luation of the bisector of the other angle.

2.6 Plane through the Line of Intersection of Two Planes Let the equation of the planes be

+ C]Z +di= 0 + b2y + c2z + d2 = 0.

a1x +b1Y a2x

(8)

234

ANALYTICAL GEOMETRY OF TWO AND THREE DIMENSIONS

As any point, whose coordinates satisfy the above planes also satisfy the equation

a1x

+ hJY + c1z + dJ + k(a2x + b2y + c2z +dz)= 0

(9)

for any value of k, the Eqn. (9) represents the plane which passes through the line of intersection of the planes given by Eqn. (8). In fact, Eqn. (9) represents a system of planes which passes through the line of intersection of the planes given by Eqn. (8). The constant k is determined using some prescribed condition, viz. "The plane passes through a given point" or "The plane is perpendicular/parallel to a given plane," etc. ·

2.7 Condition That the Three Planes May Have Common Line of Intersection Let the equations of the plane be

+ bJ y + crz + dr a2x + bzy + c2z + d2 a3x + b3y + c3z + dJ aJX

= 0 = 0 = 0.

Any plane through first two planes is

a1x

+ hJY + c1z + dJ + k(azx + b2y + c2z +dz)= 0.

If it is identical to the third plane, then we have aJ+ka2

b1+kbz

=

c1+kc2

dr

+ kd2

- - - -= ), (say). d3

C3

Then

a1

+ ka2 -

bi

+kb~

c1

+ kc2 -

A.c3 = 0

dr

+ kd2 -

Ad3 = O.

Aa3 = 0 -A.b3 =0

Eliminating k, A from the above set of equations taking three at a time we gyt

a1 a2 a3 br b2 b3 = 0,

hr bz bJ

er c2 c3

dJ dz dJ

CJ

c2 c3

=0

a1 az a3

and

d.1 dz dJ

= 0,

hJ b2 bJ CJ Cz CJ = 0. di dz dJ

2.8 Pair of Planes Consider the following equations of planes:

+d~ 0 a'x + b'y + c'z + d' = 0. ax+ by+ cz

(10)

·

(11)

2:

CH.

235

THE PLANE

Next consider the equation (ax+ by+ cz

+ d)(a'x + b'y + c'z + d') = 0.

(12)

Obviously, the points on the plane.; (I 0) and/or (I I) will satisfy the Eqn. (3) which is second degree equation in x, y, z of the form Ax 2 + By 2

+ Cz 2 + 2Fyz + 2Gxz + 2Hxy + 2Ux + 2Vy + 2Wz + D =

0.

(13)

Hence, equation of the form (4) may represent a pair of planes, where it is assumed that at least one of A, B, C, F, G, H is non-zero. Comparing Eqns. (12) and (13) we get A= aa', B

= bb',

C

= c!::+

+ 6y + l 5z = 6 + 2y + 5z = 2.

or,

3x

or,

x

12. Show that the straight lines

and

y-a ·z-a-d x-a+d =--= a - r5 a a+ r5 y-b x-b+c z-b-c =--= fJ - y fJ + y fJ

are coplanar and they lie on the plane x D

+z -

2y

= 0.

The given lines will be coplanar if the determinant

b-c-a+db-a b+c-a-d a - r5 a a +'5

fJ

{J-y

fJ + y

vanishes. Now, performing C1 + C3 we get 2(b - a) b - a b+c-a-d

2a

a

a+ r5

2/J

fJ

fJ + y

So, the condition of coplanarity is satisfied.

=0.

CH.

3:

259

THE STRAIGHT LINE

Equation of the plane through the first line is

+ B(y - a)+ C(z J) + Ba + C(a + J) = 0.

A(x - a+ d)

where

A(a -

a - d) = 0

(1)

(2)

If the plane ( 1) contains the second line we have A(,8-y)

+ Bfi + C(,8 + y) =

(3)

0.

Eliminating A, B, C from (I), (2) and (3) we get

z-a-d

x-a+d y-a

Performing C1

+ C3 -

a -

c5

a

a+ c5

P-

y

/3.

[i+y

=0.

2C2 we obtain

.x

+z-

2y y -a

0 0 or,

x+z-

z-a-d

a.

a+ c5

p

p+y

2y = 0

which rs the equation of the plane containing given lines. ·~

13. Show that the straight lines

x

px qy rz =-=l m n

z

y m

x pl

- = - = -;

n

y

= qm

=

z rn

are coplanar if p = q or q = r or r = p.

D Let A, B, C be the dir.:ction ratios of the normal to the olane containing the given lines, then we have Al

+

Bm

p

Al+ and

+ Cn

Bm

= O,

r

q

+en= 0

Alp+ Bmq

+ Cnr =

Eliminating A, B, C from the above equations we get

l/ p

m/q n/r m

n

Ip

mq

nr

=0.

Taking common l, m, n from C1, C2, C3 respectively we get

l/p l/q l/r =0. p

q

r

0.

260

ANALYTICAL GEOMETRY OF Two AND THREE DIMENSIONS

Performing C3 - C2 and C2 - C1 we get I/p I/q - I/p 1/r -1/q p

1

q--p

r-q

q-p

I

=

r-q

1

(q - p)(r -

q)(_!_ - -

i.e.,

rq pq (q - p)(r - q)(p - r) = 0

i.e.,

p = q or q

=r

or,

~

0

1/q _ 1/ p i/r _ 1/q

or,.

=0

0

or r

0

) = O

= p.

14. Prove that the square of the distance of the point (xo, yo, zo) from the line u 1 CZ + d = 0 and V '= a' X + b' y + C Z + d' = 0 is given by

=

ax + by +

(a'uo - avo) 2 + (b'uo - bvo) 2 + (c'uo - cvo) 2 (be' - b'c) 2 +(ca' - c'a) 2 +(ab' - a'b)2 where uo and vo are the values of u and v at (xo, yo, zo).

D Any plane through the line of intersection of u = 0 and v = 0 can be taken as u +kv = 0. If (xo, yo, zo) lies on it, then we have uo + kvo = 0. So the equation of the plane through the given line and the point (xo, yo, zo) is vuo - uvo = 0, i.e. (avo - a'uo)x + (bvo - b'uo)y + (cvo - c'uo)z +constant term= 0.

UVo~O

If()

be the angle between this plane and the plane u = 0, then

.J[ L,! (bvo -

.

b' uo)c - (eve - c' uo)b J2 ]

sm () = --:=:;:::=:=::;;=~---:~:;------.::-::--

J a2 + b 2 + c 2.J{ L,