I I ~ A COMPREHENSIVE TEXT BOOK OF APPLIED MATHEMATICS Rakesh Gupta (M.Sc. B.Ed.) ABHISHEK PUBLICATIONS CHANDIGARH (INDIA) ( All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from the publishers/copyright owner. ISBN Copyright First Edition Publislied by : 978-81-8247-225-9 : Author : 2009 Abhishek Publications SCO 57-59, Sector 17-C, Chandiharh. Phone: 0172-5003768 Telefax: 0172-2707562 e-mail: abhpub@ yahoo.com, www.abhishekpublications.com Printed at : Shiva Offset. Naveen Shahadra, Delhi PREFACE ______________________________________ __ This book" A Compresensive Text Book of Applied Mathematics-II" has been written for the polytechnic diploma level students preparing for a carrier in Engineering. It covers the latest syllabi of various state boards of technical education in North Region. The subject has been developed in a systematic, logical and concise form to meet the requirements of all types of students. The book has been written in a style which makes it extremely useful for self study. The examples are so arranged that the easier problem come first and the difficult ones later. At the end of chapter a large number of multiple choice questions have been provided. This book covers the complete syllabus but some typical examples in some of the topics have been included to maintain the flow of ideas. The author lay no claim to the original research in preparing this book. Available sources on the subject has been used frequently. The subject matter has been arranged in a proper sequence and presented in a simplified manner. I feel that no work is perfect and there is always a scope for further improvement Errors might have been crept in inspite of utmost care. Corrections and suggestions for further improvement will be thankfully acknowledged and will be implemented in the subsequent editions. In last but not least, my thanks are due to my publisher" Abhishek Publications, Chandigarh", for bringing out this book in such a short period of time. Suggestions for further improvement of text are again welcome. Rakesh Gupta (M.5c. B.Ed.) Lecturer, S.C.N. Polytechnic, Chourmastpur, Ambala "This page is Intentionally Left Blank" SYLLABUS ____________________________________ __ APPLIED MATHEMATICS-II RATIONALE Applied Mathematics forms the backbone of engineering discipline. Basic elements of permutations and combinations trigonometry, vector, complex number and statistics have been included in the curriculum as foundation course and to provide bases for continuing education to the students. DET AILED CONTENTS 1. Co-ordinate Geometry (25 hrs) 1.1 Area of triangle, centroid and incentre of triangle (given the vertices of a triangle), simple problems on locus. 1.2 Equation of straight lines in various forms (without proof)with their from to another angle between two lines and perpendicular distance formula(without proof) 1.3 Circle: General equation and its characteristic's given: The centre and radius Three points on it The Co-ordinates of the end's of the diameter 1.4 Conics (parabola, ellipse and hyperbola), standard equation of conics (without proof) , given the equation of conics to calculate foci, directrix, eccentricity, laws rectum, vertices and axis related to different conics Differential Calculus. 2. Differential Calculus (30 hrs) 2.1 Concept of function Four standard limits Lt x n _ an L t sin x Lt a x_I Lt 1 --------- -- - - ---- (1 + x) x x a 'x->O x 'x->O x 2.2 Concepts of differentiation and its physical interpretation Differential by first principle of xn, (ax + b)n, Sin x ,Cos x, tan x, cosec x and x x cot x, e ,a log x Differentiation of sum, product and quotient of different functions Logarithmic differentiation, Successive differentiation excluding nth order Application of derivatives for (a) rate measure, (b) errors, (c) real root by Newton's method, (d) equation of tangent and normal, (e) finding the maxima and minima of a function (simple engineering problems) 3. Integral Calculus (20 hrs) 3.1 Integration as inverse operation of differentiation 3.2 Simple integration by substitution by parts and by partial fractions 3.3 Evaluation of definite integrals(simple problems) by using the general properties of definite integrals 3.4 Application of integration for Simple problem on evaluation of area under a curve where limits are prescribed for circle, ellipse, parabola and straight line calculation of volume of a solid formed by revolution of an area about axis (simple problems) where limits are prescribed for sphere and cylinder to calculate average and root mean square of a function area by Trapezoidal Rule and Simpson's Rule 4. Differential Equations Solution of first order and first degree differential equation by variable separation and their simple numerical problem. CONTENTS __________________________________ __ Co-ordinate Geometry 1. The Point 2. The Straight Line 3 The Circle 4. The Conic Section Differential Calculus 1. Functions and Limits 2. Derivation of Functions 3. Tangents and Normals 4. Rate of Change of Quantities 5. Maxima and Minima Integral Calculus 1. Indefinite Integral 2. Definite Integral 3. Application of Integration and Numerical Integration Defferential Equations 3 44 84 97 137 162 195 209 224 241 321 376 398 "This page is Intentionally Left Blank" Section I Co-ordinate Geometry "This page is Intentionally Left Blank" Chapter 1 THE POINT Co-ordinate geometry is the branch of mathematics, which deals geometry algebraically. Let X'OX and YaY' be two mutually perpendicular straight lines intersecting at a point '0' is called as origin. The horizontal line X'OX is called X - axis and vertical line YaY' is called Y-axis. These two axes divide the plane into four parts which are called quadrants as shown in the following fig. Y lInd Quadrant X' lllrd Quadrant 1st Quadrant O(Origin) TVth Quadrant Y' Y-axis x X-axis The X- axis and Y-axis taken together are called as rectangular axes or co -ordinate axes Rectangular Co - ordinates of a point Let P be any point in the plane of rectangular axes. Through P draw PL perpendicular on X - axis and PM perpendicular on Y-axis as shown in fig. Y M y X' o L X Y' /4/ The horizontal distance (PM = OL) of point P from Y- axis denoted by 'x' is called abscissa of point P. The vertical distance (PL = OM) of point P from X-axis denoted by 'y' is called ordinate of point P The abscissa and ordinate taken together (x,y) is called co-ordinates (or rectangular co-ordinates) of point P. Sign conventions for rectangular co-ordinates 1. All horizontal distances (i.e. x-eo-ordinates) to the right of Y-axis are taken as positive and left of Y-axis are taken as negative. 2. All vertical distances (i.e. y-co-ordinates) above the X-axis are taken as positive and below the X- axis are taken as negative y lInd Ist ~ y+ve I x +ve I y +ve X' 0 X IIIrd IVth ~ y -ve I x +ve I y -ve Y' Note: (i) Co-ordinates of origin are taken as (0,0). (ii) For any point on X-axis, ordinates (i.e. y-coordinate) is zero. (iii) For any point on Y-axis, abscissa (i.e. x-coordinate) is zero. POLAR CO- ORDINATES OF A POINT P(r,O) O ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ X Let 'a' be the origin and OX be the fixed straight line. Take any point 'P' join OP. The distance of point P from 'a', denoted by 'r' is called radius vector. The anglf XOP denoted by '8' is called vectorial angle. The ordered pair (r,8) is called the polar co-ordinates of point P. Note: The vectorical angle '8' is taken as positive in anticlock wise direction and as negative in clockwise direction. /5/ RELATION BETWEEN RECTANGULAR AND POLAR CO - ORDINATE Let (x, y) be the rectangular co-ordinates and (r, 8) be the poldf co-ordinates of the point 'I" y p(x,y) X' Draw PL.l OX, then OL = x, PL = Y Join OP, then OP = rand LXOP = 8 Now in the right angled triangle OLP, i.e. OL x -- = cos e or - = cos 8 OP r x = r cos e PL -= sin 8 OP y . 8 or - = SIn r Y' ----(i) i. e. y = r sin e ---(ii) 'Squaring and adding (i) and (ii) we get x2 + y2 = r2 cos8 + r2 sin2 e = r2 (sin2 8 +cos2+ e) = r2 (1) (','sin2 e + cos2 8 = 1) or x2 + y2 = r2 or r = -+-y-2 (Rejecting -ve sign because r is positive) Hence r = + y2 Dividing (ii) by (i), we get y rsin8 -=-- = tan8 x rcos8 or 8 = x Note: The value of '8' depends upon the quadrant in which the point lies i.e. signs (+ve or -ve) of the values of 'x' and 'y' which will be clear from the following examples. /6/ X' Y' Here x= J3 y=l r = J( J3) 2 -(-1)-; c= 2 and 8 = tan-1 = tan-1( 53) 8 = 30 x X' Here and Casell y e Y' x = -J?, Y =-1 r = [(-J3)2 = 2 8 = tan-1 (-1 ) = tan-1 -=J3 8 = 30 or 210 x (As the point P lies in the third quadrant therefore angle '8' will be (180 + 30) i.e. 210 not 30) Hence 8 = 2 I 0 Method to change Cartesian (rectangular) form to Polar form Put x = r cos8 , y = r sin8 and simplify Method to change polar form to Cartesian form x Put cos8 = -r sin8 = y and r r = + y2 and then simplify to remove fractional powers. fJI DISTANcr; FORMULA To find the distance between two points whose co-ordinates arc given. Let P(Xl'Yl) and Q(x2, Y2) be two given points. From P and Q, draw PL and QM perpendiculars on x - axis. From P draw PR perpendicular to QM. Let'd' be the distance between the points P & Q. In right angle PQ! = PR2 + RQ2 - - -(i) y I.?-'" Q (X"Y2) (Using Pythagorous theorem) Now PQ= d PR = LM = OM-OL = X2-X1 RQ = QM-RM = Y -y

p r R 2 1 From (i) , d2 = (X2-X1)2 + (Y2-Yl)2 or d= Rejective -ve sign as d is the distance o d= Hence, distance between two points = of abscissa)2 + (Difference of ordinate)2 Some Important points to Remember ) L \1 x (i) When three points A, B, C are given and we have to prove that A, B, C dre collinear (i.e. A, B, C lie on the same straight line) then we will show that (ii) (a) AB +BC = AC or AC +BC = AB or AB +AC = BC When three points are given and we have to prove that an isosceles triangle show that any two sides are equal (b) an equilateral triangle show that all three sides are equal (c) right angled triangle show that sum of the squares of any two sides is equal to square of the third side. (iii) When four points are given and we have to prove that (a) a square show that all four sides are equal and diagonals are equal. (b) a rectangle show that pair of opposite sides are equal and diagonals are also equal. (c) a parallelogram /8/ show that pair of opposite sides are equal but diagonals are not equal (d) a rhombus show that all four sides are equal but diagonals are unequal. Example 1 Plot the following points and find the quadrants in which they lie. (i) P (3, 7) (ii) Q (3, -7) (iii) R (-3, -7) (iv) 5 (-3, 7) Solution S(-3,7) 8 R(3,7) $------- r - - - - - - ~ 6 5 4 3 2 1 -7 -6 -5 -4 - ~ -2 -1 1 2 -1 -2 -3 -:1 -5 -6 ------- -7------ 4 5 6 7 R(-3,-7) -8 Q(3,-7) (i) P(3, 7) lies in the 1st quadrant. (ii) Q(3, -7) lies the IVth quadrant. (iii) R(-3, -7) lies in the IIIrd quadrant. (iv) 5(-3, 7) lies in the Ilnd quadrant. Example 2 (i) (iii) Find the polar co-ordinates of the following:-(1, 1) (ii) (1,-1) (-1,-1) (iv) (-1,1) Solution (i) Given Cartesian coordinate is (1,1) Here x = 1, Y = 1 X' y P(l,l) x 1t or 8 =: 4 Because x and y both arc positive, therefore, poinllies in the 1st quadrant. I-Ience polar coordinate is ( (ii) Given cartesian coordinate is (1, -1) Here x = 1, Y = -1 r = y2 J(1)2- .J2-Y x /9/ x and 8 == = tan-1( = tan-1(-I) P(l,-l) 8 -= tan-1 (-1) Y' Because x is positive and y is negative, therefore point lies in the ivth quadrant 71t Y 8=4 Hence polar co-ordinate is ( .J2, 7: ) (iii) Given Cartesian coordinate is (-1,-1) X' Here x = -1, Y = -1 r = cc + Ji P(-l,-l) and = tan-l (1) Y' 5 -IT 4 Because x and y both are negative, therefore, point lies in the I1Ird quadrant 8 = 51t 4 Hence polar coordinate is (iv) Given Cartesian coordinate is (-1, 1) Here x = -1, Y = 1 r = +-y --= J( _1)2 =--li x /10/ and =tan-1(-1) Because x is negative and y is positive, therefore point lies in the lInd quadrant e = 31t 4 Hence polar coordinate is P(-l,l) X' Example 3. . Change the following into Cartesian co ordinates:-(i) (ii) Solution (i) Given polar co-ordinate is ( 2,- ) Here 1t r = 2 and e =--4 x = r cose = 2 cos ( - = 2 cos 1 =2xTz=-J'i and y = r sine = 2sinC- = -2 sin Hence Cartesion co-ordinate is (-J2,--J2) (ii) Given polar co-ordinate is ( 4, 231t ) Here 21t r = 4 and e = ---3 x = r cose = 4 cos ( .?31t) = 4 cos ( 1t - ;) Y X Y' 1t = -4 cos-3 1 == -4 x - =-2 2 [.: cos (1t-8) = -cos8] and y = r sin8 ,= 4 sin ( 231t) = 4 sin ( 1t - ;) 1t = 4 sin-3 ./3 = 4x-2 = 2./3 [.: sin (1t-8) == sin8] Hence Cartesian co-ordinate is (-2,2./3) Example 4 Solution or or Example 5 Solution or Transform the cartesian equation X2+y2 = 2ax into polar equation. The given equation is X2+y2 = 2ax ---- (i) Putting x = r cos 8 and y = r sin 8 in equation(i) we get, r2 cos28 + r2 sin28 = 2ar cos8 + cos28) = 2ar cos8 ( .: sin28 + cos28 = 1) r = 2a cosO is the required equation in polar form. Transform the polar equation r2 = a2 cos28 into cartensian form. The given equation is r2 = a2 cos28 r2 = a2 (cos28 - sin28) Putting cos8 = sin8 = y and r = + y2 , we get r r or or or Example 6 Solution r2(x2+y2) = a2(x2_y2) (X2+y2)(X2+y2) = a2(x2_y2) (X2+y2)2 = a2(x2_y2) is the required equation in cartesian form. Find the distance between the points (5,7) and (-3,1) Let A = (5,7) and B(-3,1) be the given points. /11/ /12/ Then AB = = +(1_7)2 = .J64 + 36 = .J100 = 10 Example 7 If the points (x,y) be equidistant from the points [(a+b),(b-a)] and [(a-b), (b+a)] show that bx = ay. Solution or or or 01' or or or Example 8 Solution A [(a+b),(b-a)] P(x.y) B [(a-b),(b+a)] Let P(x,y) be the point which is equidistant from the points A((a+b), (b-a)) and B((a-b),(b+a)). PA=PB PA2 = PB2 . [x-(a+b)F +[y-(b-a)F = [x-(a-b)F +[y-(b+a)]2 x2+(a+bf -2x(a+b) +y2 +(b-a)2 -2y(b-a) = xZ+(a-b)2 -2x(a-b) +y2 +(a+b)2 - 2y(a+b) -2x(a+b)-2y(b-a) = -2x(a-b)-2y(b+a) ax+bx+by-ay = ax-bx+ay+by bx+bx = ay+ay 2bx = 2ay or bx = ay is the required result. Show that the points (-2,3), (1,2), (7,0) are collinear. Let A( -2,3), B(l,2), C(7,O) be the given points AB = .J(1 + 2)2 + (2 -- 3)2 = /3-z t- (_1)2 = .J9--:;-i = .Jio AC = J(7t- il 3)2 -.CO Jsi +-9 =0 -.190 3./i6 BC = (7 - 1):2 + (0 -- 2):2 -'" b 2. + ( - 2):2 =.J 36 + 4 =- .J 40 = 2 JiO Here AB+BC = .Jio f 2-.1i-o = 3J10-= AC /13/ ~ A B ,10 I ~ ~ I ~ 2, 10 ~ I I ~ 3-10 ~ I Given points lie on the same line and hence they are collinear. Example 9 If P(at2, 2at) and Q( t ~ , - ~ a ) be the two given points and S(a,O) is another 1 1 point. Prove that SP + SQ is constant for all values of t. Solution The given points are (a -2a) P(at2, 2at), t2 '-t- and S(a,O) 111 1 - + - = + ---;======== .. SP SQ ~ ( a - at2)2 + (0 - 2at)2 /14j - --- ---- 1- ----------- 1 [1 t2 1 - a (1+t2) (t2+1) 1 1 1 = -- which is constant for all values of t. a lll'nn' ST; 1 5Q is independent of 't' and so is constant for all values of t. l:x90, mC'clsuJ'l'd dntidockwise) with the +ve direction of x-axis, then its slope is (a) 0 (b) -ve (c) +ve (J)-l 18. If a straight line subtends an angle 8 (measured anticlockwise) with thl' +ve direction of x -axis, then (a) 0< 8 < 900 (b) 900 < 8< 1800 (c) 0 < 8 < 1800 (d)OD 8 < 1800 /18/ 19. If a straight line subtends an angle 8 = 60 (measured anticlockwise) with the +ve direction of x -axis, then, its slope is (a) ../3 1 (b) ../3 (c) 1 (d) J3 2 20. If a straight line subtends an angle of 135 (measured anticlockwise) with the +ve direction of x-axis, then its slope is (a) J3 (b) 1 (c) -1 21. If a straight line is parallel to x-axis, then its slope is ' 1 (d) ---J3 (a) 00 (b) - 00 (c) 1 (d)O 22. If a straight line is parallel to y -axis then its slope is ( a) 00 (b) - 00 (c) 1 ( d) 0 23. Slope of line 3x -6y + 7 = 0 is 1 1 (a) -2 (b) 2 24. Slope of line ax + by + c = 0 is -a (b) --b 25. Slope of line y = c (constant) is 1 (a) /3- (b) -;;3 26. Slope of line x = c (constant) is 1 (a) ../3 (b) )3 27 Slope of line as shown in figure is (a) 1 (b) -1 (c) -2 b (c) a (c) 00 (c) (X) (d) 2 -b (d) ----a (d)O (d)O (c) 00 (d) None of these 28. Slope of line as shown in figure is 1 (a) J3 (b) F3 (c) -J3 1 (d) - F3 X' Y X ' + - - - ~ ~ - - ~ X Y' Y x 29. Slope of line y=O (x-axis) is 1 (a) ~ 3 (b) J3 30. Slope of line x=O (y-axis) is '1 (a) .J3 (b) J3 (c) 00 (c) 00 31. Slope of line passing through the points (5,3) and (2,-1) is 4 4 3 (a) - (b) -- (c) -3 3 4 (d)O (d)O 32. Equation of line passing through origin and having slope 1/3 is (a) Y = 3x (b) 3y = -x (c) 3y = x (d)y =-3x fl9f 33. Equation of line passing through origin and making an angle of 30 with the +ve direction of x-axis (measured anticlockwise ) is 34. (a) y =.J3x (b) .J3y = -x (c) .J3y = x (d) y = -.J3x y = mx+ c is __ from of straight line. (a) intercept form (c) one-point form (b) slope intercept form (d) Normal form 35. Y = mx represent the equation of a straight line (a) passing through origin (b) not passing through origin (c) cutting y-axis not at origin (d) none of these 36. (Y-Yl) = m(x-x1) is form of straight line (a) intercept form (b) slope intercept form (c) one-point form or slope point form (d) normal form 37. ~ + ~ = 1 is __ form of straight line (a) intercept form (b) slope intercept form (d) normal form (c) slope point form 38 ) yz - Yl (y - YI = ---(x - xl) is form of straight line Xz -Xl --(a) intercept form (b) slope-point form (one-point form) (c) two point form (d) normal or perpendicular form 39. x cosa + y sina = p is __ form of straight line (a) intercept form (b) slope point form (c) two point form (d) normal or perpendicular form X-Xl Y-Yt . . . --- -. - = r IS __ form of straIght lme. cosa sma 40. /80/ (a) intercept form (c) slope point fOfm (b) symmetric or distance form (d) normal form 41. Equation of straight line having x-intercept 2 and y-intercept 3 is (a) 2x+3y=6 (b) 2x-3y=6 (c) 3x+2y=7 (d)3x-2y=6 42. Equation of straight line having equal intercepts on axis is (a) x+y=a (b) x+y=-a (c) x-y=a (d) both (a) and (b) 43. Equation of straight line having equal but opposite in signs intercepts on axes is (a) x+y=a (b) x-y=a (c) x-y=-a (d) both (b) and (c) . 44. Intercepts of axes cut by straight line 3x-4y=12 are (a) a=4, b=3 (b) a=4,b=-3 (c) a=-4,b=3 45. Intercepts of axes cut by straight line 4y-5x=6 are -6 3 6 -3 6 3 (a) a = 5' b = 2" (b) a = :5 ' b = 2 (c) a = 5' b = "2 46. The straight line ~ + ~ = 1 has its no portion in (a) 1st quad. (b) IInd quad. (c) IIIrd quad. 47. The straight line 3x-2y=6 has its no portion in (a) 1st quad. (b) IInd quad. (c) IIIrd quad. 48. For the straight line 5x-2y=10; values of m and care 5 5 5 (a) 5'2 (b) 2,5 (c) 2,-5 (d) a=-4, b=3 a = -6 b = -3 (d) S ' 2 (d)IVth quad. (d)IVth quad. S (d) -S'2 2 49. Equation of straight line passing through the point (-2,4) and having slope "3 is (a) 2x+3y+12 = 0 (b) 2x-3y-12 = 0 (c) 2x+3y-12 = 0 (d)2x-3y+12 = 0 SO. Equation of straight line passing through the point (1,3) and making an angle 4S0 (measured anticlockwise) with the +ve direction of x-axis is (a) x-y+2 = 0 (b) x+y+2 = 0 (c) x-y+2 = 0 (d)x+y-2 = 0 51. Equation of straight line passing through the points (2,S) and (-1,3) is (a) 2x+3y+ll = 0 (b) 2x-3y+ll = 0 (c) 2x-3y-ll = 0 (d) 2x+3y-ll = 0 S2. The perpendicular drawn form the origin on a straight line makes an angle 60 (measured anticlockwise) with the +ve direction of x-axis and its length is 1/2 units. The equation of straight line '[' is (a) x-J3y=l (b) x+J3y=l (c) x+J3y=-l (d) x-J3y=-l /81/ 53. The equation of straight line PQ as shown in figure is y (a) J3x - y = 1 (b) J3x + y = 1 p (c) J3x+y=-l (d) J3x-y=-l 54. For the equation of straight line x+y+ 4..fi = 0 values of p and a (symbols have their usual meanings) are. (a) 4,135 (b) 4,45 (c) 4,225 X' (d) 4,-45 x 55. For the equation of straight line 3x-y-5 = 0, values of p and a (symbols have their usual meaning) are (b) p= J%, a = tan-l ( ~ 1 ) and in lInd quadrant (d) p=#, a=tan-1( - ~ ) and in IIIrd quadrant and in IVth quadrant 56. A point A at a distance J3 units from the point p(4,5) lying on the line which makes an angle of 30 (measured anticloekwise) with the +ve direction of x-axis is [11 J3+10) (a) 2' 2 [11 -J3+10) [-11 J3+10) [11 J3-10) (b) 2' 2 (c) 2' 2 (d) 2' 2 57. The co-ordinates of a point p on line '1' which is parallel to line I m' as shown in figure are (a) (..fi -3, ..fi -5) (b) (..fi +3, ..fi +5) (c) (..fi+3, ..fi-5) (d) (..fi-3, ..fi-5) y 58. The perpendicular distance of point (-3,4) from the straight line 3x+y-5 = 0 is 1 'm' (a) J8 (b) .flO (c) .flO (d) None of these 59. The distance between the parallel lines 3x+y-10 = 0 and 3x+y-5 = 0 is (a) .J5 (b) JI (c) # (d) ..fi 60. The foot of perpendicular drawn from the point (3,5) on the line x+y=6 is (a) (2,-4) (b) (-2,4) (c) (-2,-4) (d)(2,4) /82/ 61. Two straight lines are parallel if (ml,m2 are their slopes) 1 (d)ml = m 2 62. Two straight lines having slopes ml and m2 are perpendicular if 1 (d)ml = m 2 63. The straight lines Alx +BIy +CI = 0 and A2x +B2y +C2 = 0 are identicals (i.e. they represent same straight line) if Al BI CI Al BJ C1 Al BI (a) A2 = B2 "j; C2 (b) A2 = B;"" = C2 (c) A2 "j; B2 (d) None of these 64. The straight lines Alx + Bly +C1 = 0 and A2x + B2y +C2 = 0 are parallel if A1 BI CI A1 BI C1 Al B] (a) A2 = B2 "j; C2 (b) A2 =B;""= C2 (c) A2 "j; B2 (d) None of these 65. The straight lines Alx +BIy +CI = 0 and A2x +BzY +C2 = 0 are intersecting line if Al B1 C1 A1 B1 C1 A1 Bl (a) A2 = B2 "j; C2 (b) A2 =B;""= C2 (c) A2 "j; B2 (d) None of these 66. The angle between two lines having slopes ml and m2 is given by (a) tan 8 = ( m1 - m2 ) (b) tan 8 = ( m1 + m2 ) I-mIm2 1+m1m2 ( m -m ) (d) tan8= I 2 1+mIm2 67 The acute angle between two lines having slopes ml and m2 is given by m -m (a) tan 8 = I 2 I-mIm2 m +m (c) tan8= I 2 I-mIm2 68. The straight lines 2x-3y+7 = 0 and 3x+2y+9 = 0 are (a) parallel to each other (b) perpendicular to each other (c) have acute angle between them (d) coincident 69. The straight lines 2x-3y+7 = 0 and 2x-3y+9 = 0 are (a) parallel to each other (b) perpendicular to each other (c) have acute angle between them (d) coincident /83/ 70. The angles between the lines 3x-2y+7 = 0 and 7x+2y+3 = 0 are (a) tan-l ( (b) tan-1( (c) tan-1( (d) tan-l ( 71. The angles between the straight lines having slopes 3 and 4 are given by (a) tan-1(13) (b) (c) tan-l ( 112) (d) tan-l ( (13 -1) 72. If tane = 13 + 1 ' where e is the angle between two straight lines, then e = (a) 150 (b) 3000 (c) 1450 (d) 600 73. The equation of the line passing through the point (-4,-5) and perpendicular to the line joining (1,2) and (5,6) is (a) x+y-9 = 0 (b) x-y-9 = 0 (c) x+y+9 = 0 (d) x-y+9 = 0 74. If P be the perpendicular distance of origin from the line whose intercepts on the axes are a and b, then 111 111 111 (a) p2 b2 (b) p2 b2 (c) p2 = b2 - a2 (d) p2 = a2b2 75. The equations of the straight lines bisecting the angles between the straight lines 3x-4y+5 = 0 and 5x+12y-8 = 0 (a) 2x-16y+15 = 0; 64x+8y+25 = 0 (c) 2x+16y+15 = 0, 64x+8y+25 = 0 (b) 2x+16y+15 = 0; 64x-8y+25 = 0 (d) 2x-16y-15 = 0; 64x+8y+25 = 0 - ANSWERS-(1) b (2) c (3) a (4) b (5) c (6) a (7) b (8) a (9) b (10) a (11) c (12) d (13) a (14) d (15) c (16) c (17) b (18) d (19) a (20) c (21) d (22) a (23) b ,24) b (25) d (26) c (27) b (28) b (29) d (30) c (31) a (32) c (33) c (34) b (35) a (36) c (37) a (38) c (39) d (40) b (41) c (42) d (43) d (44) b (45) a (46) c (47) b (48) c (49) d (50) a (51) b (52) b (53) b (54) c (55) d (56) a (57) b (58) b (59) c (60) d (61) a (62) c (63) b (64) a (65) c (66) d (67) d (68) b (69) a (70) c (71) d (72) a (73) c (74) b (75) a Chapter 3 THE CIRCLE Circle is the locus of a point which moves on a plane such that its distances from a fixed point is always constant. The fixed point is called center of the circle and the constant distance is called radius of the circle. Central Form Of Circle Let C (h, k) be the center of a circle with radius 'r' and let P (x,y) be an arbitray point on the circle. Therefore by definition of circle CP= r o y (?\(X/Y) ~ or ~ ( x - h)2 + (y - k)2 = r, [By distance formula] -0-+-------+ X or (x - h)2 + (y - k)2 = r2 - - (1) This form of circle is called Central form of circle. Standard Form Of Circle Every circle can be shifted in x - y plane, so that its centre concide with the origin without having any change in its radius. Thus radius is the main parameter. Hence on shifting the circle (I), so that its center coincide with origin, i.e. (h,k) = (0,0) Equation (1) becomes, x2 + y2 = r2 This form of circle is called Standard form of circle, as it contains only single parameter 'r'. General Form Of Circle The equation of circle in central form is given by (x _h)2 +(y _k)2 = r2 or x2 -2xh +h2 +y2 -2yk +k2 = r2 or x2 +y2 -2hx -2ky + h2 +k2 _r2 =0 Let -h = g, c = h2+ k2_r2 and -k = f The equation of circle becomes x2 +y2 + 2gx + 2fy + c = 0 This form of circle is called General form of circle. Center and Radius of a Circle given in General Form. Equation of circle in general form is x2 + y2+ 2gx + 2fy + c = 0; Whereg=-h and f=-k .. Center (h,k) = (-g,-f) = of of y) ] RADIUS c = h2 + k2 _r2 or r2 = h2 + k2 -c or r = + k2 - C = +(-f)2- c 22' = of x) + of y) - constant term /85/ Note Every 2nd degree equation in x and y involving no term containing xy and in which coefficient of x2 = coefficient of y2 represent a circle. Diameter Form of Circle-Let A and B be two end-points of a diameter AB of a circle having co-ordinates (Xl, Yl) and (x2, respectively. Let P(x, y) be any arbitrary point on the circle. We know that angle in a semicircle is one right angle. P(x,y) .. LAPB=90 .. BP => (Slope of AP ) . (Slope of BP) = -1 (xlly,) (x y) V 1 or (y - Yl) (y - Y2) = -(x -Xl) (x -x2) or (X - Xl) (x - x2) + (y - Yl) (y -Y2) = 0, is the required equation of circle in diameter form /86/ Example 1 Solution or or Example 2 Solution Find the equatin of circle having center at (1,-5) and radius equal to 3 Given center of circle (h,k) = (1,-5) and radius r = 3 By central form of circle (x -hf + (y _k)2 = r2, we get (x -If + (y +5)2 = (3)2 x2 -2x +1 +y2 +10y +25 -9 = 0 x2 + y2 -2x +10y +17 = 0 is the required equation of circle. Find the eq. of circle having center (4,5) and passing through the point (7,9). Given center of circle C( 4,5) and the circle passes through the point A (7,9). radius (r) = AC = = ..)9+ 16 =5 rx,9) :. By central form of circle or or Example 3 (i) (ii) Solution (i) and or or (x- h)2 +(y _k)2 = r2 (x _4)2 +(y -5f = (5) 2 x2 -8x +16 +y2 -lOy +25 = 25 x2 +y2-8x -lOy +16 = 0, is the required equation. Find the equation of circle having center at (3, 4) and touching x - axis y-axis Since the circle touches the x-axis, x - axis is tangent to the circle at the point of contact L (say). We know that the .ir through the point of contact passes through the center of y circle. 0 radius (r) = CL = 4 center of circle is (h, k) = (3, 4) By central form of circle, (x - h)2 + (y - kf = r2, we get (x -3) 2 + (y - 4)2 = (4f x2 -6x +9 +y2 -8y +16 = 16 x2 +y2 -6x -8y +9 = 0 is the required equation of circle. x (ii) or or Example 4 Solution and or or or Example 5 (i) (ii) (iii) Solution (i) /87/ Now, the circle touches y -axis, therefore y -axis is tangent to the circle at the point of contact L. radius (r) = CL = 3 and center of circle is ( h, k) = (3,4) By Central form of circle, (x -h)2 + (y _k)2 = r2, we get (x _3)2 + (y _4)2 = (3)2 x2 -6x +9 +y2 -8y +16 = 9 x2 +y2 -6x -8y +16 = 0 is the required eq. of circle y o x Find the equation of circle having center at (-1,2) and passing through the point of intersection of st. lines 2x +y -4 = 0 and x -y -5 = 0 Equation of given st. lines are 2x +y -4 = 0 - - - (1) x -y -5 = 0 - - - (2) Solving (1) and (2) for x and y, we get x = 3 and y = -2 The circle passes through point A(3,-2), the point of intersection of straight lines (1) and (2). center of circle is (h, k) = (-1, 2) radius (r) = +(-2-2)2 = .J16+16 = 132 By central form of circle (x -h)2 +(y _k)2 = r2, we get (x +1)2+(y-2)2 = (132)2 x2 +2x +1 +y2_4y + 4 = 32 x2 +y2 +2x -4y +5 -32 = 0 x2 +y2 +2x -4y -27 =0 is the required eq. of circle. Find the center and radius of the circle x2 +y2 +6x -8y +10 = 0 (x +5)2 +(y -9)2 = 10 (x -1)2 +(y _7)2 = 25 Equation of given circle is x2 + y2 + 6x -8y + 10 = 0 ------ (1) center of circle (1) is given by (-g, -f) = of x), - of y)] f88I = = (-3,4) and radius of circle (1) is given by r = = (%coefficientofxr +(%coefficientofY r -constantterm = = +(_4)2 = .J9 + 16 = J2s = 5 units Center = (-3, 4), radius = 5 units (ii) Equation of given circle is (x +5f + (y _9)2 = 10 - - (2) comparing equation (2) with central form of circle (x -hf + (y -kf = r2 we have, h = -5, k = 9, r = .JIO .. Center is (-5,9) and radius (r) = .JIO units (iii) Equation of given circle is (x -If + (y - 7f = 25 comparing eq. (3) with central form of circle (x -hf + (y -kf = r2 We have, h =1, k=7 and r = 5 Center is (1, 7) and radius = 5 units - -(3) -- Example 6 Find the equatin of circle concentric with circle 2X2 +2y2 +8x -12y -14 = 0 Solution:-and passing through the point (2,5). The eq. of given circle is 2X2 + 2y2 + 8x -12y -14 = 0 Dividing the eq. through by 2 we get, x2 +y2 +4x -6y -7 = 0 - -(1) Center of circle (1) is The required circle is concentric with circle (1) or or Example 7 Solution or or Example 8 Solution /89/ [i.e. both having same center], therefore the center of required circTe is C(h,k) = (-2, 3) and it passes through the point A(2,5) its radius will be r = AC = J( -2 _-2-)-2 -+-(3---5-)-2 = Jl6+4 =m =-J4x5 = 2.J5 units By central from of circle (x -h)2 +(y -k)2 = r2 [x +2]2 +[y -3]2 = 20 x2 +4x +4 +y2_6y +9 -20 = 0 X2+y2 + 4x -6y -7 = 0, is the required equation of circle. Find the equation of circle concentric with circle x2 +y2 -4x +10y -7 =0 and passing through the center of circle x2 +y2 +8x +4y +12 = 0 Equation of given circle are x2 +y2 -4x +10y -7 = 0 and x2 +y2 +8x +4y + 12 = 0 The required circle is concentric with circle (1) (i.e. both having same center) (1) is - -(1) - -(2) (-1(-4),-1(10)) = (2,-5) The center of reqd. circle is C (h, k) = (2, -5) The required circle passes through the center of circle (2) i.e. C (-4, -2) The radius of required circle is r = CCI = --5-+-2)-2 = -J36+9 r = J45 = 3.J5 By central form (x _h)2 +(y - kF = r2, we have (x _2)2 +(y +5) 2 = (3.J5) 2 x2 -4x +4 +y2 +10y +25 -45 = 0 x2 + y2 -4x +10y -16 = 0, is the required equation of circle. Find the equation of circle having co-ordinates of end points of one of its diameters as (2,5) and (7,-3). Let A (2, 5) and B (7, -3) be the end points of a diameter of required circle. Using diameter form of circle, we have /90/ i.e. or or (x -2) (x -7) +(y -5) (y -(-3)) = 0 (x -2) (x -7) +(y -5) (y +3) = 0 x2 -7x -2x +14 +y2 +3y -5y -15 = 0 x2 +y2 -9x -2y -1 = 0, is required equation of circle. A(2,5) B(7,-3) Example 9 Find the equation of circle having its center at (4, 5) and one of the extreme points of a diameter as (3, -2). Solution Center of required circle is (h, k) = (4, 5) and one end point of a diameter is A(3, -2) = (Xl Yl) (say) Let B (X2' Y2) be the other end point of given diameter. .. Center C (h, k) is mid point of Diameter AB Xl +X2 Thus by mid-point formula, h = 2 and k = Yl +Y2 2 -2+Y2 A(3,-2)1----------1 B(X2'Y2) i.e. ' or or i.e. or 3+X2 4= --,-2 3 +x'-=8 2 and 5= ---'--=-2 and -2 +Y2 =10 x2 = 5 and y 2 = 12 B (5, 12) is the other end point of given diameter. The equation of required circle (diameter form) is (x - 3) (x -5) + (y +2) (y - 12) = 0 x2-5x -3x + 15 + y2 -12y +2y -24 = 0 x2 + y2 -8x -lOy -9 = 0 is the required equation of circle. Example 10 Find the eq. of circle passing through three points (2, 4) , (-3, 3), & (5, -2). Solution \ i.e. i.e. and i.e. Suppose that the equation of required Circle is x2' + y2 + .2gx_ + 2fy + c = 0 - - (1) Since the circle.passes through the points A (2, 4), B (-3,3) and C(5,-2). Hence these points would satisfy eq. (1) 4 +16 +2g(2)-+2f(4) +c = 0 4g +8f,+-c = -20 - -(2) 9 +9 +2g(-3) +2f(3) +c = 0 + 6f + c = -18 - - (3) 25 +4 +2g (5) +2f(-2) +c = 0 109 +c = -29 - - (4) (2} - (3) gives A(2,4) C(5,-2) 10 g + 2 = -2 or 5g + = -1 (3) - (4) gives -16g + 10 = 11 (5) x 10 - (6) gives or 50g + 10 = -10 -16g + 10 = 11 + 66g = -21 g 7 g= - 22 21 66 --(5) --(6) Putting this value o g in equation (5), we get or or 5(-;2) +=-1 35 = -1 + 22 -22+35 22 13 22 Putting the values of f and g in equation (2), we get 4 ( -;2) + 8 ( ~ ~ ) + c = -20 14 52 c= -20+---11 11 c= -220+ 14 -52 11 -258 c=--11 Putting the values of f, g and c in equation (1), we get ( 7) ( 13 ) - 258 x2 +y2 +2 - 22 x +2 22 Y - U = 0 or 11x2 +l1y2 -7x +13y -258 = 0, is the required equation o circle. /91/ /92/ MULTIPLE CHOICE QUESTIONS 1. The locus of a point which is moving on a plane such that its distance from a fixed point is always constant is a (a) Ellipse (b) Straight line (c) Circle (d) Parabola 2. The distance of each point on the circle from its center is called (a) diameter (b) chord (c) line segment (d) radius 3. Radius of a circle is the distance between any point on circle and ___ ' (a) any other point on circle (b) centre (c) origin (d) x-axis 4. The line segment joining any two arbitrary points on a circle is called a (a) diameter (b) chord (c) line segment (d) radius 5. The largest chord of a circle is called its (a) diameter (b) semi-circle (c) line segment (d) radius 6. A line touching a circle at a point on circle is called (a) tangent ,(b) secant (c) chord (d) diameter 7. A line passing through any two points on the circle is called (a) tangent (b) secant (c) chord (d) diameter 8. The point on the circle at which tangent touches the circle is called (a) centre (b) origin (c) point of contact (d) pole 9. Diameter is ___ . __ of radius (a) half (b) one-fourth (c) twice (d) one-third 10. Number of diameter of a circle (all having same length) (a) One (b) Two (c) Three (d) Infinite 11. Diameter of a circle passes through (a) centre (b) origin (c) point of contact (d) pole 12. Angle in a semi-circle is (a) 1800 (b) 900 (c) 2700 (d) 1500 13. The length of boundary of a circle is called its (a) circumference (b) area (c) diameter (d) radius 14. The circumference of a circle of radius 'r' is 1 (a) 1tr2 (b) 21tr (c) -nr2 2 (d)41tp 15. The area (interior region) of a circle of radius 'r' is 1 (a) 1tr2 (b) 21tr (c) -1tr2 2 (d)41tp 16. The central form of circle is /93/ (a) X2+y2 = a2 (b) x2+y2+2gx+2fy+c = (c) (x_h)2+(y_k)2 = r2 (d) (x-x1)(X-X2)+(Y-Y1)(Y-Y2) = 17. The centre of circle (X+2)2+(y-3)2 = 25 is (a) (2,3) (b) (2,-3) (c) (-2,-3) (d) (-2,3) 18. The radius of circle (X-3)2 + (y_5)2 = 16 is (a) 3 (b) 5 (c) 4 (d)16 19. The radius of circle (x+3)2+(y+5)2 = 6 is (a) 3 (b) 5 (c) 6 (d) .J6 20. The standard form of circle is (a) X2+y2 = a2 (b) x2+y2+2gx+2fy+c = (c) (x_h)2+(y_k)2 = r2 (d) (x-x1)(X-X2)+(Y-Y1)(Y-Y2) = 21. The general form of equation of circle is (a) X2+y2 = a2 (b) X2+y2+2gx+2fy+c = (c) (x_h)2+(y_k)2 = r2 (d) (x-x1)(X-X2)+(Y-Y1)(Y-Y2) = 22. The diameter form of equation of circle is (a) X2+y2 = a2 (b) x2+y2+2gx+2fy+c = (c) (x_h)2+(y_k)2 = r2 (d) (x-x1)(X-X2)+(Y-Y1)(Y-Y2) = 23. The centre of circle x2+y2=a2 is (a) (a,O) (b) (O,a) (c) (0,0) (d)( -a,O) 24 The radius of circle x2+y2=a2 is (a) a (b) a2 (c) .fa (d)2a 25. The centre of circle X2+y2=4 is (a) (2,0) (b) (0,2) (c) 0,0) (d) (-2,0) 26. The radius of circle X2+y2=4 is (a) 2 (b) 4 (c) J2 (d)8 27. The centre of circle (x-h)2+(y-k)2 = r2 is (a) (-h,-k) (b) (-h,k) (c) (h,-k) (d) (h,k) 28. The radius of circle (x-h)2+(y-k)2 = r2 s (a) r2 (b) r (c) Jr (d)2r 29. The centre of circle x2+y2+2gx+2fy+c = is (a) (g,f) (b) (-g,-f) (c) (-g,f) (d) (g,-f) 30. The radius of circle X2+y2+2gx+2fy+c = is (a) (b) (c) (d) g2+f2-c 31. The centre of circle X2+y2+5x-4y+7 = is (a) (%,-2) (b) G,-2) (c) (-; ,-2) (d) ( -; ,2) /94/ 32. The centre of circle 2x2+2y2+16x+8y+10 = 0 is (a) (-4,-2) (b) (-4,2) (c) (4,-2) (d) (4,2) 33. The radius of circle X2+y2+6x+4y-8 = 0 is (a) 3Fs (b) J2i (c) Fs (d) 4Fs 34. The radius of circle 2X2+2y2+ 16x+8y-40 = 0 is (a) 2m (b) 2J5 (c) 2.J10 (d) m 35. The equation circle having (3,5) and (-4,2) as the end points of one of its diameters is (a) X2+y2_X+7y-2 = 0 (c) X2+y2+X-7y-2 = 0 (b) X2+y2-x-7y-2 = 0 (d) X2+y2+X+7y+2 = 0 36. The equation of circle having(3,5) as one end point of diameter and (4,2) as its centre is (a) x2+y2+8x+4y+10 = 0 (b) x2+y2-8x-4y+10 = 0 (c) x2+y2-8x+4y+10 = 0 (d) x2+y2+8x-4y+10 = 0 37. In the figure, if ~ and ~ are the slopes of AP and BP then (a) m1=m2 (b) ml=-m2 (c) m1.m2=1 (d) ~ m 2 = - 1 A 1"----4o;-:'---'lB 38 The equation of circle having its diameter as the line segment joining the points (0,0) and (1,1) is (a) x2+y2-x-2y+1 = 0 (b) x2+y2-x+2y+1 = 0 (c) x2+y2+x-2y+1 = 0 (d) x2+y2+x+2y+1 = 0 39. The equation of circle having its centre at (1,3) and radius 4 units is (a) x2+y2+2x+6y+6 = 0 (b) X2+y2_2x-6y-6 = 0 (c) X2+y2+2x-6y+6 = 0 (d) X2+y2_2x+6y+6 = 0 40. The equation of circle having its centre at (3,1) and passing through the point (2,5) is (a) X2+y2_6x+2y-7 = 0 (b) X2+y2+6x-2y-7 = 0 (c) X2+y2+6x+2y-7 = 0 (d) X2+y2_6x-2y-7 = 0 41. The equation of a circle having its centre at (4,5) and passing through the centre of circle X2+y2+4x-6y = 12 is (a) x2+y2+8x+10y+1 = 0 (b) x2+y2+8x-10y+1 = 0 (c) x2+y2-8x+10y+1 = 0 (d) x2+y2-8x-10y+1 = 0 42. The equation of a circle passing through the points (0,0); (a,O) and (O,b) is (a) x2+yi+ax+by = 0 (b) X2+y2_ax+by = 0 (c) X2+y2_ax-by = 0 (d) X2+y2+ax-by = 0 , /95/ 43. The equation of circle concentric with circle x2+y2+2x-4y+l0:::::: and having radius 3 is (a) X2+y2+2x-4y-4 :::::: (b) X2+y2+2x-4y+4 :::::: (c) X2+y2+2x-4y+6 :::::: (d) X2+y2+2x-4y-6 :::::: 44. The equation of circle concentric with circle x2+y2+2x-4y+l0 :::::: and passing through the point (0,2) is (a) X2+y2+2x-4y-4 :::::: (b) X2+y2+2x-4y+4 :::::: (c) X2+y2+2x-4y+6 :::::: (d) X2+y2+2x-4y-6 :::::: 45. The equation of circle touching the x-axis and y-axis at points A(2,0) and B(0,2) is (a) X2+y2_4x-4y+8 :::::: (b) X2+y2_4x-4y+4 :::::: (c) X2+y2_4x-4y-4 :::::: (d) X2+y2_4x-4y-8 :::::: 46. The equation of circle having its centre at (-1,2) and area equal to 81t square units is (a) X2+y2+2x-4y+3 :::::: (c) X2+y2+2x-4y-3 :::::: (b) X2+y2_2x+4y-3 :::::: 0 (d) X2+y2+2x-4y+5 :::::: 47. The equation of circle having its centre on x-axis and passing through the points (0,0) and (2,4) is (a) X2+y2_5x :::::: (b) x2+y2+10x :::::: (c) X2+y2_20x :::::: (d) x2+y2-10x :::::: 48. The intercepts made on y-axis by the circle X2+y2+4x-5y+6 :::::: are (a) 3,4 . (b),4,.8 (c) 3,2 ' (d)4,5 49. The intercepts made on x-axis by the circle X2+y2+5x+2y+4 :::::: are (a) -1,4 - (b) 1,-4 (c) 1,4 '(d)-1,-4 50. The circle X2+y2+5x+2y+4 :::::: cuts x-axis at points (a) (-1,0); (4,0)'" . (b) (1,0); (-4,0) - (c) (1,0); (4,0) (d) (-1,0); (-4,0) 51. The circle :::::: cuts y-axis at the points (a) (0,3); (0,2) - '(b) (0,4); (0,8) (c) (0,3); (0,4) (d) (0,4); (0,5) 52. The radius of the passing through the points (0,0); (4,0) and (0,3) is (a) 1.5 units (b) 2.5 units (c) 3.5 units " (d)4.5 units 53 The equation of circle having centre at point (1,0) and passing through the point of intersection of lines x+y::::::8 and 2x-y :::::: 7 is . (a) X2+y2+2x-24 :::::: (b) X2+y2_2x+24 :::::: (c) X2+y2_2x-24 :::::: (d) X2+y2_2x-25 :::::: 54. The area of circle passing through the points (0,0); (4,0) and (0,3) is (a) 6.251t sq. units (b) 12.5 1t sq. units (c) 25 1t sq. units (d)51t sq. units 55. The circumference of circle passing through the points (0,0); (4,0) and (0.3) is (a) 6.25 1t units (b) 2.5 IT. units (c) 10 1t units (d)5 1t units /96/ - ANSWERS-(1) c (2) d (3) b (4) b (5) a (6) a (7) b (8) c (9) c (10) d (11) a (12) b (13) a (14) b (15) a (16) c (17) d (18) c (19) d (20) a (21) b (22) d (23) c (24) a (25) c (26) a (27) d (28) b (29) b (30) c (31) d (32) a (33) b (34) c (35) c (36) b (37) d (38) a (39) b (40) d (41) d (42) c (43) a (44) b (45) b (46) c (47) d (48) c (49) d (50) d (51) a (52) b (53) c (54) a (55) d Chapter 4 CONIC-SECTION A conic is the locus of a point moving on a plane so that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line. The fixed point is called focus of conic and the fixed straight line is called Directrix of conic. Important Terms: Focus - The fixed point is called focus of conic; denoted by S. Directrix - The fixed straight line is called directrix of conic. Eccentricity - The constant ratio is called eccentricity of conic and is denoted by Ie'. Axis of Conic - A straight line through the focus and perpendicular to directrix is called axis of conic. Vertex - The point of intersection of axis and conic is called vertex of conic. The value of eccentricity (e) decides the nature of conic, as given below-i) If el, then thE" conic is a Hyperbola. PARABOLA -A parabola is the locus of a point moving on a plane such that its distance from a fixed point is always equal to its distance from a fixed straight line. i.e. PS = PM The fixed straight line is called Directrix of Parabola and the fixed point is called focus of parabola. Standard Forms of Parabola We have four types of parabolas-i) Rieht handed Parabola [y2 = 4ax] ii) Left handed Parabola [y2 = -4ax] iii) Upward Parabola [X2 = 4ay] iv) Downwards Parabola [X2 = -4ay] fixed straight line (Directrix) p (fixed pont) S(Focus) /98/ Their shapes are as given below -Y Y X' X X' X Y' Y' (i) Right handed Parabola [y2 = 4ax] (ii) Left handed Parabola [y2 = -4ax] Y A X' X' X Y' Y' (iii) Upward Parabola [X2 = 4ay] (iv) Downwards Parabola [x2 = -4ay] We shall derive the equation of right handed parabola. The derivation of equations of other parabolas can be done on same steps. To find the equation of Right Handed Parabola [y2 = 4ax] Let MN be a fixed straight line i.e. Directrix and 5 be the fixed point i.e. Focus. Then Axis of parabola is a straight line perpendicular to Directrix MN and passing through the focus 5, let it be X-axis. The point of intersection of axis of parabola and the parabola i.e. A is called vertex of parabola and X' N Y' X X /99/ let it be origin. Let the line perpendicular to axis at point A be y-axis. Since point A lies on the curve, therefore by definition of parabola, AS=AZ=a (say). Thus co-ordinates of focus S are (a,o) and equation of Directrix is x = -a or x+a = 0 Let P(x,y) be an arbitrary point on the parabola. Hence by definition of parabola, distance of point P from focus S. i.e. or = Perpendicular distance of point P from Directrix PS=PM [By distance formula and .lr distance of a point from a given straight line] or =Ix+al Squaring both sides we get, (x-a)2+y2 = 1 x+a 12 or (x_a)2+y2 = (x+a)2 [.: 1 x 12 = X2] or x2+a2-2ax+y2 = x2+a2+2ax or y2 = 2ax+2ax [on canceling terms x2, a2 from both sides] or y2 = 4ax Hence y2 = 4ax is the equation of Right Handed Parabola. It is called Right Handed Parabola as it opens towards Right Hand Side. Latus Rectum -The chord of parabola passing through focus S and perpendicular to the axis of parabola i.e. LL' is called Latus Rectum of parabola. Points Land L' lies on Parabola y2 = 4ax y L(a,Za) Both Land L' have x-eo-ordinate = a . y2 = 4a(a) or y2 = 4a2 Za or y = 2a Thus Land L' have their y-co-ordinates x' z X 2a and -2a respectively. Za Hence co-ordinates of Land L' are (a,2a) and (a,-2a) respectively. L'(a,-Za) . LL' = LS + SL' = 2a+2a = 4a . LL' = 4a Y' /1001 Thus length of Latus Rectum is 4a and Equation of Latus Rectum is x = a or x -a = 0 Focal Distance - The distance of a point P on the parabola from its focus(5) is called Focal Distance of point P Let P(x1,y 1) be any point on the parabola y2 = 4ax, having focus 5(a,o) . . Focal Distance of point P is PS = - a)2 + (Y1 - 0)2 = 1 2 - 2ax 1 + a 2 + Y / X' [.: point P(xl,y 1) lies on parabola Y' y2 = 4ax i.e. Y/= 4axl] = -+-a-=-2-+-4-a-x-1 = +2axl +a2 = + a)2 = Xl + a Thus focal distance of point P(X1'YI) on parabola y2 = 4ax is xl+a Symmetry of Parabola Right handed parabola y2 = 4ax is symmetrical about x-axis. Fundamental Results for the standard forms of Parabolas. Form of Equation of Vertex Focus Axis of Length of Parabola Parabola (A) (5) Parabola Latus Rectum Right handed y2 = 4ax (0,0) (a,O) x-axis, y=O 4a Left handed y2 = -4ax (0,0) (-a,O) x-axis, y=O 4a Upwards x2 = 4ay (0,0) (O,a) y-axis, x=O 4a Downwards x2 = -4ay (0,0) (O,-a) y-axis, x=O 4a x Equation of Directrix x+a = a x-a = 0 y+a = a y-a = 0 Eccentricity: Eccentricity of a conic is the ratio of distance of any point on the conic from its focus to the perpendicular distance of that point from the Directrix. PS i.e. eccentricity (e) = PM But for parabola PS = PM (By Def.) e=l Thus eccentricity of parabola is 1 independent of its type. Example 1 (i) (iii) Solution X' Find the focus, vertex, length of Latus Rectum, equation of Directrix, axis of parabola, equation of Latus Rectum for the following parabolas-y2 = 8x (ii) x2 = 12y (iv) y2 = -lOx x2 = -6y Y Y' (i) y2 = 8x is the equation of right Handed Parabola of the form y2 = 4ax .. 4a = 8 or a = 2 Focus (S) : Co-ordinates of focus are given by (a,O) = (2,0) Vertex (A): Co-ordinates of vertex are given by (0,0) Length of Latus Rectum = 4a = 4(2) = 8 units Equation of Directrix is given by x = -a i.e. x = -2 or x+2 = 0 Axis of Parabola: It is x-axis whose equation is y = 0 Equation of Latus Rectum: It is given by x = a i.e. x = 2 or x-2 = 0 (ii) y2 = -lOx is the equation of left handed parabola of the form y2 = -4ax .. 4a = 10 5 a= -2 Focus (S) : Co-ordinates of focus are given by (-a, 0) = ( -% ,0 ) Vertex (A): Coordinates of vertex are given by (0,0) Length of Latus Rectum = 4a = 4(%) = 10 units /101/ X /10'1/ Equation of Directrix is given by x = a 5 i.e. x = 2 or 2x = 5 or 2x - 5 = Axis of Parabola: It is x-axis whose equation is y = Fquation of Latus Rectum: It is given by x = -a 5 i.e. x = -2 or 2x+5 = , (iii) x2 = 12y is the equation of Upwards parabola of the form x2 = 4ay .. 4a = 12 or a = 3 Focus (5) : Co-ordinates of focus are given by (O,a) = (0,3) Vertex (A): Coordinates of vertex are given by (0,0) Length of Latus Rectum = 4a = 4(3) = 12units Equation of Directrix: It is given by y = -a i.e. y =-3 or y+3 = Axis of Parabola: It is y-axis whose equation is x = Equation of Latus Rectum: It is given by y = a i.e. y = 3 or y-3 = (iv) x2 = -6y is the equation of Downwards parabola of the form x2 = -4ay .. 4a = 6 3 or a =-2 Focus (5) : Co-ordinates of focus are given by (O,-a) = (0,-%) Vertex (A): Co-ordinates of vertex are given by (0,0) Length of Latus Rectum = 4a = 4(%) = 6 units Equation of Directrix is given by y=a 3 i.e. y= 2 /103/ or 2y -3 = 0 Axis of Parabola: It is y-axis whose equation is x = 0 Equation of Latus Rectum: It is given by y = -a Example 2 Solution 3 i.e. y =-2 or 2y+3 = 0 Find the equation of parabola having focus at point (3,5) and equation of directrix 3x-4y+l0 = 0 Let P(x,y) be an arbitrary point on the parabola and let S(3,5) by the focus of parabola and let PM be length of perpendicular from point P(x,y) on the directrix 3x-4y+l0 = 0 By definition of parabola 13x - 4y + 101 +(y_5)2 = +(_4)2 = 13x - 4y + 101 5 Squaring both sides we get (3x - 4y + 10)2 (x_3)2+(y_5)2 = 25 25[x2-6x+9+y2-10y+25] = (3X)2+( -4y)+2( -4y)(10)+2(10) (3x) or 25x2+25y2-150x-250y+850 = 9x2+16y2+100-24xy-80y+60 or 16x2+9y2-210x-170y+24xy+750 = 0 is the required equation of parabola. Example 3 Find the equation of parabola having focus at(5,2) and vertex at (3,2) Solution Let SZ be the perpendicular drawn from focus S on the Directrix MN. Y Let A be the vertex of parabola. Then we know that A is the mid point of SZ x' X i.e. ZA = AS Let Co-ordinates of Z be (Xl'Yl)' then by mid-point formula N Y' /104/ or or or or or or or Xl+5_'"l d Yl+2_2 -,-, an -2 2 xl+5 = 6 and YI+2 = 4 Xl = 1 and Y I = 2 (1,2) are the co-ordinates of point Z, lying on the Directrix MN. -1 Now slope of MN = slope of SZ [.: MN 1. SZ] -1 Slope of MN = [2-2] 5-3 -1 o Equation of MN (By slope point form) is given by -1 (y-2) = 0 (x-I) x-I = 0 Equation of Directrix is x-I = 0 Now by Definition of Parabola, PS = PM , (Where P(x,y) is an arbitrary point on the parabola) lx-II j(x - 5)2 + (y - 2)2 = )(1)2 + (0)2 )(X-5-)2 +(y_2)2 = IX-II Squaring both sides, (x-5f+(y-2)2 = (x-l)2 x2-10x+25+y2-4y+4 = x2-2x+l or y2-4y-8x+28 = 0 is the required equation of parabola. Example 4 Solution or Find the co-ordinates of focus, vertex, equation of latus rectum, length of latus rectum, axis and equation of Directrix of the parabola 2x2+5y-3x+4 = a Equation of given parabola is 2X2+5y-3x+4 = a - (1) 2 5 3 x + - Y - - x + 2 = a [Making the coefficient of x2 = 1] 2 2 or or or or .. or or or or 2 3 5 x --x=--y-2 2 2 2 3 995 x --x+---=--y-2 2 16 16 2 2 3 9 5 9 x -"2x+16=-"2y-2+16 (x- :r =-iy-

4 2 16 5 Compare it with X2 = -4a Y we get -(2) -(3) -(4) 5 3 23 4a= -2 ' X = x - - and Y = Y +-4 40 5 a= -8 Fo,:us : Focus of Parabola (4) is (O,-a) = (0,-%) or i.e. or Vertex: i.e. or 5 X = 0 and Y = -8 3 d 23 5 x--=Oan Y+-=--4 40 8 3 48-6 x = - and Y = -- or-4 40 5 (3 -6) Co-ordinates of focus of parabola (1) are '4'5 Vertex of parabola (4) is (0,0) X = 0 and Y = 0 3 23 x--=O and v+-=O 4 - 40 /105/ /106/ or 3 -23 x=- and y=-4 40 (3 -23) Co-ordinates of vertex of parabola (1) are "4' 40 Equation of Latus Rectum: Equation of Latus Rectum of Parabola (4) is Y =-a 23 5 i.e. y+-=--40 8 5 or 40y +23 = --x40 8 or 40y +23 = -25 or 40 -y +48 = 0 or 5y +6 = 0 , is the equation of Latus Rectum of Parabola(1). Length of Latus Rectum of Parabola(l) = Length of Latus Rectum of Parabola( 4) = 4a = 4(%) = % units Axis of Parabola (4) is Y-axis i.e. X = 0 or or x - ~ = O 4 4x - 3 = 0, is the equation of axis of parabola (1) Equation of Directrix: Equation of directrix of Parabola (4) is Y=a i.e. 23 5 y+-=-40 8 or 40y + 23 = 25 or 40y - 2 = 0 or 20y - 1 = 0 is the equation of directrix. THE ELLIPSE An ellipse is the locus of a point moving on a plane such that its distance from a fixed point bears a constant ratio 'e' (less than one) to its distance from a fixed straight line. i.e. or PS - =e SA' = eA'Z A'Z 1 Adding (i) and (ii) we get, SA + SA' = e(AZ + A'Z) or AA' = e(AZ + A'Z) or 2a = e[(CA+CZ)+(CZ-CA')] or 2a = e[a+CZ+CZ-a] or 2a = e(2CZ) or a = e.(CZ) a or CZ = -e Subtracting (ii) from (i) SA - SA' = e(AZ-A'Z) or (CA+CS)-(CA' -CS) = e[AA'] or (a+CS) - (a-CS) = e(2a) or 2CS = 2ae or CS = ae - -(ii) [Measuring the distance from C] - -(iii) - - (iv) a a From (iii) we have CZ = --, i.e. distance of directrix from centre C(origin) is - ,. e e a Since directrix is parallel to y-axis, its equation is x = --e a or x+-= e and From (iv) we have CS = ae, i.e. distance of focus(S) from centre C(origin) is ae. Since focus S lies on x-axis, Co-ordinates of S are (-ae,O) Let P(x,y) be any point on the ellipse. Then hy definition of ellipse flO9j PS -=e PM or PS= ePM or +(y_O)2 = e e +(0)2 [By using distance formula and .ir distance of a point from a straight line] Squaring both sides, (x+ae)2+y2 = e21x + or or or or or x'+2aex +a'e'+y' = e'( x + :: + 2:) x2+2aex+a2e2+y2 = e2x2+a2+2aex x2(1_e2)+y2 = a2(1-e2) x2 y2 -+ -1 a2 a2(1- e2) --- (v) Again let the Ellipse cuts Y-axis at points Band B' having Co-ordinates (o,b) and (o,-b) respectively. Then by definition of Ellipse BS' BM' =e or BS' = eBM' +(b-0)2 =e[;] a2e2 + b2 = a2 b2 = a2(1-e2) From (v) and (vi) we have x2 y2 -+-=1 a2 b2 [.: BM'= CZ = frOm(iii)] - - (vi) WhIch is the required equation of standard form of Ellipse of first type /110/ Latus Rectum of Ellipse: The chord of ellipse though foci Sand S' and perpendicular to Axis of Ellipse are called Latus Rectum of Ellipse. Thus ellipse has two Latus Rectum Ll L' 1 and L2L' 2' Abscissa of Ll and L'l is -ae and that of L2 and L' 2 is ae Now L1, L'1,L2 and L' 2 lie on the Ellipse or or (ae)2 y2 ---=:-'--+-= 1 a2 b2 X' Z Y A' A z' or , multiplying and dividing by a2 or y a Co-ordinates of L1, L'1,L2 and L' 2 are ( b2) ( _b2) (b2) (_b2) -ae,-;- ; -ae'-a-; ae,-;- and ae'-a-2b2 Length of Latus Recta = --a a Equation of Latus Rectum Ll L'l and L2 L' 2 are x = -ae and x = ae respectively i.e. x = ae for L2L' 2 and Ll L'l Focal Distances respectively. X The distance of a point on Ellipse from its focus is called its focal Distance since an ellipse has two foci, there are two focal distances of every point P on the ellipse SP Now by definition of Ellipse, if P(Xl'Yl) is any point on ellipse then PM = e or SP = ePM y = e[ZN] = e[CZ+CN] B = e[;+xl] = a+exl Also S'P = ePM' x' = e [Z'N] = e[CZ'-CN] B' =e[;-x1 ] = a - aexl Y' Thus for any point P(X1'Yl) on ellipse there are two focal distances SP = a +exl and S'P = a-exl Now, SP + S'P = (a+exl) + (a-exl) Thus SP + S'P = 2a /111/ X This is a very important property of ellipse, that the sum of focal distance of any point on an ellipse is constant and is equal to the length of major axis i.e. 2a. Major and Minor Axis Length of major axis = AA' = 2a Length of minor axis = BB' = 2b Equation of Major axis is y = 0 and Equation of Minor axis is x = 0 2 2 Symmetry - Equation of Ellipse is ;- + .;.. = 1 a b Equation of ellipse remains unchanged if x is replaced by -x, which means curve is symmetric about y-axis. Similarly, equation of ellipse remains unchanged if y is replaced by -y, which means curve is also symmetric about x-axis. Thus ellipse is symmetric about both the axes. Equation Directrices - Like first type of ellipse, second type of ellipse also has two a directrices having equation y = - , both parallel to x-axis. e flU/ Latus Rectum - Second type of ellipse also has a pair of Latus rectum, having equation 2b2 y = tae, both pcuallel to x-axis and each having length = -;-Focal Distances PS = a-eYl and PS' = eYl +a Thus PS + PS' = 2a (length of major axis) Major and Minor axis :- y-axis is major axis and x-axis is minor axis in second type of ellipse having length 2a and 2b. i.e. Length of major axis = AA' = 2a Length of minor axis = BB' = 2b Equation of major axis (y-axis) is x = and Equation of minor axis (x-axis) is y = Symmetry Second type of ellipse is also symmetric about both the axes. Fundamental Results for the Standard Forms of Ellipse Type I Equation of Ellipse x2 y2 1 -+-'-=1 . a>b a2 b2 ' 2 Centre (0,0) :1 Vertices (a,O) 4 Foci (ae,O) a 5 Equations of Directrices x=-e 6 Equation of Latus Recta x = ae 7 Length of Latus Recta 2b2 a 8 Relation lJ2 = a2(1-e2) 9 Major Axis x-axis (y=O) 10 Minor Axis y-axis (x=O) 11 Length of Major Axis 2a 12 Length of Minor Axis 2b 13 Eccentricity (e) eb b2 a2 (0,0) (0, a) (0, ae) a y=-e y= ae 2b2 a lJ2 = a2(1-e2) y-axis (x=O) x-axis (y=O) 2a 2b e Denominator of y2 (1) is an ellipse of first type x2 y2 -+--1 a2 b2-comparing (1) and (2) we get a2 = 9 and b2 = 4 a = 3 and b = 2 -- (2) Co-ordinates of vertices of (1) are given by ( a,O) = ( 3,2) Co-ordinates of foci of (1) are given by ( ae, 0) We know, b2 = a2 (1-e2) or 4 = 9 (1-e2) 4 - = 1-e2 9 or or 4 5 e2= 1--=-9 9 $ or e=-3 Co-ordinates of foci are ( ae,O) = ( 3. ~ ,0 1 = ( ~ , O ) 2b2 Length of Latus Recta is given by -a 2(4) 8 . = --=- unIts 3 3 Equation of Latus Recta are given by x = ae /114/ l.e. or x= J5 a Equation of Directrices are given by x = -e i.e. or Length of Major axis Length of Minor axis x=( ~ J = 2a = 2(3) = 6 units = 2b = 2(2) = 4 units Centre of Ellipse: Co-ordinates of centre of ellipse are (0,0) (ii) Equation of given ellipse is 16x2 + 9y2 = 144 or Vertices: Foci: or or or 16x2 + 9y2 = 1 144 144 - - (1) or Denominator of x2 < Denominator of y2 (1) represents equation of 2nd type of ellipse x2 y2 -+-=1 b2 a2 Comparing (1) and (2) we get a2 = 16 and b2 = 9 a = 4 and b = 3 -- (2) Co-ordinates of vertices are (0, a) = (0, 4) Co-ordinates of foci are (0, ae) We know, b2 = a2 (1-e2) 9 = 16 (1-e2) 9 - = 1-e2 16 or 9 7 e2= 1--=-16 16 J7 or e=-4 Co-ordinates of foci will be (0, ae) -- [0,-_1- 4J747 J = (0,J7) Length of Latus Rectum : a 2(9) 9 . = --=- umts 4 2 Equation of Latus Recta : are y = ae i.e. y ~ 4( '7J or y=J7 a Equation of Directrices : are y = -e i.e. or Length of Major axis Length of Minor axis 4 y= J7 4 16 y= J7 = 2a = 2(4) = 8 units = 2b = 2(3) = 6 units Centre of Ellipse: Co-ordinates of centre of ellipse are given by (0,0) 1 /115/ Example 6 Find the equation of ellipse having eccentricitY"2' focus at (2,4) and equation of directrix x+2y-9 = 0 Solution Let P(x,y) be any arbitrary point on the ellipse PS By definition of ellipse PM = e /116/ or PS = ePM Co-ordinates of focus (S) are (2,4) 1 and e = -2 1 = -PM x' 2 [By distance formula] 1 jx+2y-9j = 2 + (2)2 11x +2y-91 ="2 .J5 Squaring both sides we get, 11x + 2y - 912 (X_2)2+(y-4)2 = -------4 (5) y Y' [.-ir distance of a point from a straight line] or 20[X2_4x+4+y2-8y+ 16 J = x2+4y2+81 +4xy-36y-18x or 19x2+16y2-62x-124y-4xy+319 = 0 is the required equation of ellipse. x Example 7 Find the standard equation of ellipse which passes through the points (2,3) and (-3,-1). ? 1 Solution Let the equation of required ellipse in standard form be + "'- 1 a- b-Also Since ellipse (1) passes through the point (2,3) 4 9 -+-=1 a2 b2 ellipse (1) passes through the point (-3,-1) 9 1 -+-=1 a2 b2 (2) - (3) x 9, we get 81 = 1-9 a2 b2 - - (1) - - (2) -- (3) f117f or From (3) , or ~ = 1 - 9 ( ~ ) b2 77 1 5 or b2 = 77 b2 = 77 5 or 2 2 E . (1) b __ x -+-Y--1 quation ecomes, 77 / 8 77 / 5 -or 8x2 +5y2 = 77, is the required equation of ellipse. Example 8 Find the centre, eccentricity, foci, vertices, equation of directrices, equations of Latus-Recta of Ellipse 3x2+4y2-12x-8y+4 = 0 Solution or or or or or or or Equation of given ellipse is 3x2+4y2-12x-8y+4 =0 (3x2-12x)+(4y2-8y) +4 =0 3(X2_4x)+4(y2_2y) +4 = 0 3[x2-4x+4-4]+4[y2-2y+1-1] +4 = 0 3[(x-2)2-4]+4[(y-1)2-1]+4 = 0 3(x-2)2+4(y-1)2-12-4+4 = 0 3(x-2)2+4(y-1)2 = 12 (X_2)2 + (y_1)2 =1 4 3 Compare it with X2 y2 -+-=1 a2 b2 a2 = 4 and b2 = 3, x = x-2, Y = y-1 . . a = 2 and b = .J3 Centre: Co-ordinates of centre of (3) are (0,0) i.e x = 0, y = 0 or x-2 = 0, y-1 = 0 --(1) -- (2) - - (3), we get /118/ or. x = 2, Y = 1 Co-ordinates of centre of (1) are (2,1) Eccentricity: or or or We know, b2 = a2(1-e2) 3 = 4(1-e2) 3 - = 1-e2 4 e2 = 4 4 1 e =-2 Foci: Co-ordinates of foci of (3) are (ae,O) i.e. X = ae, Y = 0 or x-2 = 2(%) , y-1 = 0 or x-2 = 1, Y = 1 or x = 3, 1 and y = 1 ., Co-ordinates of vertices of (1) are (3,1) and (1,1) Vertices: Co-ordinates of vertices of (3) are (:+.-a,O) i.e. X = fa, Y = 0 or x-2 = 2, y-1 = 0 or x = 4, Y = 1 or x = 0, y = 1 .. Co-ordinates of vertices of (1) are (4,1) and (0,1) Equations of Directrices Equations of Directrices of Ellipse (3) are a 2 X = ;- = 1/2 = 4 or x - 2 = 4 or x-6 = 0 or x+2 = 0, are the required equation of directrices of (1). Equation of Latus Recta: of (3) are X = ae or or or x = = 1 x-2 = 1 x-3 = 0 or x-1 = 0, are the equations of Latus Recta of equation (1) Length of Latus Recta = 2b2 = 2(3) = 3 units a 2 Major Axis: of (3) of X-axis I.e. Y = 0 or y -1 = 0, is the equation of major axis of ellipse (1). Length of Major Axis = 2a = 2(2) = 4units Minor Axis of(3) is Y-axis i.e. X = 0 or x-2 = 0, is the equation of minor axis of ellipse (1) Length of Minor axis = 2b = 2.J3 units THE HYPERBOLA /119/ A hyperbola is the locus of a point which moves on a plane such that its distance from a fixed point bears a constant ratio to its distance form a fixed straight line. PS i.e. PM = e > 1 p or PS> PM The fixed point is called focus of Hyperbola and the fixed straight line is called Directrix of hyperbola Standard Forms of Hyperbola A' Z' (i) First type of Hyperbola [x2_L = 1] a2 b2 X' A' z z' (ii) Second type of Hyperbola y x-[ 2 ~ l -? --? =lJ a- b-X /120/ We shall derive the equation of first type of hyperbola. The derivation of equation of second type of hyperbola can be done on same steps. To find the Equation of standard form of hyperbola of first type. Let MN be a fixed straight line i.e. Directrix and S be a fixed point i.e. Focus, The axis of hyperbola is a straight line perpendicular to directrix and passing through the focus(s), let it be x-axis. Let Z be the point of intersection of directrix and axis of hyperbola. Let Ie' be the eccentricity of hyperbola. Divide SZ internally and externally in the ratio e:l at points A and A' respectively. SA e and SA' e i.e. -------- ----AZ 1 A'Z 1 or SA = eAZ and SA' = eA'Z Y N' B' N Y' - - (i) - - (ii) Then by definition of hyperbola, A and A' lie on the hyperbola. Let AA' = 2a and C be the mid point of AA' Thus AT = AC = a Take C as origin and a line lx to AA' at point C as y-axis Adding (i) and (ii) we get, SA + SA' = e(AZ+A'Z) = e(AA') (CS-CA)+(CS+CA') = e(2a) or 2CS - CA + CA' = 2ae or 2CS - a + a = 2ae or 2CS = 2ae or CS = ae Subtracting (i) from (ii) we get, SA' -SA = e{A'Z-AZ) or AA' = e[(A'+CZ)-(CA-CZ)] = e[a+CZ-a+CZ] = e[2CZ] or 2a = c2(CZ) or CZ a e x /121/ a Thus co-ordinates of focus 5 are (ae,O) and the equation of Directrix MN is x = -, e Let P(x,y) be any arbitrary point on the Hyperbola and Let :RM be the perpendicular distance of point P from the directrix MZN. Then by definition of hyperbola PS -=e PM or PS=ePM or +(y-O)' = +-;1 +(0)2 or Squaring both sides, (x-ae)2+y2 = e2( x - -; r or or or or x2-2aex + a2e2+y2 = e2x2-2aex+a2 x2(1-e2)+y2+a2(e2..,.1) = 0 _x2(e2_1)+y2 = -a2(e2-1) x2 y2 or -- -1 a2 a2(e2 -1) -- - (iii) Take a2(e2-1) = b2 (say) i.e. b= Let Band B' be two points on Y axis having Co-ordinates (O,b) and (O,-b) respectively. Then BB' = 2b Thus (iii) becomes x2 y2 ---=1 a2 b2 Which is the required equation of hyperbola of first type. /122/ Latus Return of Hyperbola -The chord of hyperbola through S' and S and perpendicular to axis of hyperbola are called Latus-Rectum of hyperbola. Thus hyperbola has two Latus rectum L1L' and L2 L'2 Abscissa of Ll and L1' are -ae and that of L2 and L'2 are ae. x2 2 Now Lv L1', L2 and L2' lie on the hyperbola 2 - Y 2 = 1 a b or or or or = = = y2 a2 2 b2 = a2 (e -1) b2 ~ b2 -a Co-ordinates of L1, L1', L2 and L2' are y ( -ae, b:): ( -ae, _ ~ 2 ) : (a< ): ( ae, - ~ 2 ) respectively Lengths of Latus Recta L1L'1 (=L2L'2) = L1S + SL1' b2 b2 = --+-a a =--a x ?b2 length 10 Latus Recta of Hyperbola = -=--a /123/ Equation of Latus -Recta LILI' and L2L2' are x=-ae and x = ae respectively. Thus equations of latus - Recta of Hyperbola are x= ae FOCAL DIST ANCE:-The distance of a point on the hyperbola from its focus is called its focal distance. Since hyperbola has two foci, there are two focal distance of each point P on the hyperbola. y By definition of hyperbola, if P(Xl'Yl) is any point on Hyperbola, then or or Also or or SP PM =e SP=ePM = e[ZQJ = e[ CQ - CZJ a = e[x - - J I e SP = exCa Sip --=e PMI S'P = e PM' = e [Z'QJ = e[CZ' + CQJ a = e[- + x J e I S' P = a +exI (ii) - (i) gives S' P -SP = 2a P(x,y) N' N - - (i) [ distance is measured from C] -,- (ii) Thus for any point P(XI'YI) on hyperbola there are two focal distances, SP = exl-a and S'P = a +ex1 and S'P-SP = 2a x This is an important property of hyperbola that the difference of focal distance of any point on it is constant equal to the length of transverse axis i.e. 2a. Transverse and Conjugate Axis- x-axis is called transverse axis and y-axis is called conjugate axis of first type of standard hyperbola. Length of Transverse axis = A' A = A'C + CA = a +a = 2a /124/ and length of conjugate axis Equation of Transverse axis is y = a = BB' = BC + B' C =b+b = 2b And Equation of Conjugate axis is x = a x2 y2 Symmetry: - Equation of hyperbola is 2 - 2 = 1 a b Equation of hyperbola remains unchanged if x is replaced by -x. which means curve is symmetrical about y-axis. Similarly, Equation of hyperbola remains unchanged if y is replaced by - y, which means curve is symmetrical about x - axis. Hence hyperbola is symmetric about both the axes. FUNDAMENTAL RESULTS FOR THE STANDARD FROM OF HYPERBOLA. Type - I Type - II (1) Equation of Hyperbola x2 l y2 x2 ---=1 ---=1 a2 b2 a2 b2 (2) Co-ordinates of Center (0,0) (0,0) (3) Vertices ( a, 0) (0, a) (4) Foci ( ae, 0) (0, ae) a a (5) Equation of directrices x=- y=-e e (6) Equation of Latus Recta. x = ae y= ae (7) Length of Latus Recta 2b2 2b2 a a (8) Relation b2 = a2 (e2 -1) b2 = a2(e2-1) (9) Transverse axis x-axis; eq. y = a y-axis; eq. x= a (10) Conjugate axis y-axis; eq. x = 0 x-axis; eq. y = a (11) Length of Transverse axis 2a 2a (12) Length of conjugate axis 2b 2b (13) Eccentricity (e) e>l e>l Example 9 Find the eccentricity, co-ordinates of foci, center, vertices, the axes, equation of directrices and length of latus rectum of hyperbola, y2 -4x2 = 4 Solution Equation of Hyperbola is y2 -4x2 = 4 y2 x2 or ---=1 4 1 (1) is a Hyperbola of second type 2 2 Compare it with ~ 2 - ~ 2 = 1, we get a2 = 4 and b2 = 1 or a = 2 and b = 1 Eccentricity: ',' b2 = a2(e2-1) or 1 = 4(e2-1) 1 or - = e2-1 4 5 or e2= -4 J5 or e=-2 Foci: Co-ordinates of Foci are (0, ae) - (O,2. V;) -(o,Fs) -(1) Centre: Vertices: Axes: Co-ordinates of centre are (0,0) Co-ordinates of vertices are (0, a) = (0, 2) Equation of transverse axis is x = 0 Equation of conjugate axis is y = 0 Length of transverse axis = 2a = 2(2) = 4 units Length of conjugate axis = 2b = 2(1) = 2 units a Equation of Directrices : are given by y = -e i.e, 2 4 y=- or y = -J5 J5 2 2b2 Length of Latus Rectum = -a = 2(1) 2 = 1 unit /125/ /126/ Example 10 Solution or Find the equation of hyperbola having one of the foci at (2,3) and a directrix x-y+7 = 0 and having eccentricity equal to 2. Let P(x,y) be an arbitrary point on the required hyperbola and 5(2,3) is focus; equation of directrix is x-y+7 = 0 and e = 2 By definition of hyperbola, SP -- = e or SP = ePM PM Squaring both sides we get (x-2)2+(y-3f = 21 x-y+712 or x2-4x+4+y2_6y+9 = 2x2+2y2+98-4xy-28y+28x or X2+y2_22y + 32x-4xy+85=O is the required equation of hyperbola. Example 11 Solution or or or or or or or Find the eccentricity, Foci, centre, vertices, axes, equations of directrices ands length of latus - Rectum of the hyperbola 16x2-9y2-32x-54y-79 = 0 Equation of given hyperbola is 16x2-9y2-32x-54y-79 = 0 - - (1) 16x2-32x-9y2-54y= 79 16(x2-2x)-9(y2+6y) = 79 I6[x2-2x+l-I]-9[y2+6y+9-9] = 79 I6(x-I)2-9(y+3)2 = 79 +16 - 81 16(x-l)2 -9(y+3)2 = 14 8 9 - (x-I? - - (y+3? = 1 7 14 (x -If _ (y + 3)2 = 1 7 14 - - (2) - ---8 9 X2 y2 Compare it with - - - = 1 we get a2 b2 ' -- (3) 7 14 a2== - b2 = -8 ' 9. X = x-I and Y = y+3 Eccentricity: By relation b2 = a2( e2-I) 14 7 or - = - (e2-1) 9 8 or 16 -+1 = e2 9 5 or e =-3 Foci of hyperbola (3) are (ae, 0) ../7 5 i.e. X = 2../2'3' Y = 0 5../7 x-1 = 6../2' Y + 3 = 0 i.e. 5M x = 1 12 ' y = -3 or Hence foci of hyperbola (1) are [1 5: ,-3 J Centre of hyperbola (3) is (0,0) i.e. X = 0, Y = 0 or x-1 = 0, y+3 = 0 or x = 1, Y =-3 Hence centre of hyperbola (1) is (1,-3) Vertices of hyperbola (3) are (a,O) ../7 i.e. X = 2.J2 ' Y = 0 M x-1 = -4- , y+3 = 0 or M x = 14 ' y=-3 or Hence, vertices of hyperbola (1) are [1 ~ ,-3] Axes: Equation of transverse axis of hyperbola (3) is Y = 0 i.e. y--3= 0 Hence, equation of transverse axis of hyperbola (1) is y-3 = 0 Equation of conjugate axis of hyperbola (3) is X = 0 Hence equation of conjugate axis of hyperbola (1) is x-1 = 0 /127/ /128/ Length of transverse axis = 2a = {:;) = tUnits L gth f' '2b 2M , en 0 conjugate aXIS = = -3-umts Equation of directrices of hyperbola (3) are or or a X=-e ..fi 3 x-I = 2J2'S = 3M x=l +--- 20 3M 20 Hence, equation of directrices of hyperbola (1) are X = 1 Length of Latus Rectum 2b2 2(14) 2J2 = -;- = -9-' ..fi Example 12 (i) (ii) Solution = 56 fI = 8M units 9 9 Find the equation of the hyperbola for which The transverse and conjugate axis are 6 and 5 The between the foci is 8 and e=2 (i) Length of transverse axis = 6 units " 2a = 6 or a =3 and, length of conjugate axis = 5 units 5 2b = 5 or b = "2 Let the equation of hyperbola be x2 y2 ---=1 a2 b2 or or or x2 4y2 ---=1 9 25 25x2 - 36y2 = 225, is the reqd. equation of hyperbola (ii) Distance between the foci = 8 1. 2. 3. or 2ae = 8 or or or ae = 4 a(2) = 4 a=2 or a2 = 4 [ .,' given e=2) By relation b2 = a2(e2-1), we have or b2 = 4(e2-1) or b2 = 4(4-1) or b2 = 12 Let the equation of hyperbola be i.e. x2 y2 ------= 1 a2 b2 x2 / ---=1 4 12 or 3x2- y2 = 12, is the reqd. equation of hyperbola MULTIPLE CHOICE QUESTIONS Eccentricity of parabola is a. 1 Eccentricity of ellipse is a. 1 Eccentricity of hyperbola is a. 1 d. d. d. 0 0 0 4. Eccentricity of a conic is the ratio of focal distance of a point on conic to the distance of point from a. x-axis b. y-axis c. Latus rectum d. Dirictrix 5. Curve of which conic is unbounded /129/ a. circle b. parabola c. ellipse d. none of these b. Curve of which conic is bounded a. parabola b. hyperbola c. ellipse d. none of these /130/ 7. Sum of focal distances of any arbih'ary point on an ellipse is equal to a. 2a b. 2b c. a d. b 8. Difference of focal distances ef any arbitrary point on a hyperbola is equal to a. 2a b. 2b c. a d. b 9. Relation between a,b and e for an ellipse is a. b2=a2 (1 +e2) b. b2=a2 (1-e2) c. bba2 (e2-1) d. a2=b2 (1-e2) 10. Relation between a, band e for a hyperbola is a. b2=a2 (1 +e2) b. b"=a2 (1-e2) c. b2=a2 (e2-1) d. a2=b2 (1-e2) 11. Standard parabola is a symmetric curve about a. x-axis b. y-axis c. both x and y-axes d. Either x-axis or y-axis depending upon its type. 12. Standard ellipse and hyperbola are symmetric curve about a. x-axis b. y-axis c. both x and y-axes d. Either x-axis or y-axis 13. Equation of ellipse having semi-major axis 8 and e=lj2 is a. 4x2+3y2=192 b. 3x2+4y2=192 c. 16x2+9y2=144 d. 9x2+16y2=144 14. Eccentricity of ellipse for which the distance between foci is equal to the length of the Latus rectum is J5+1 a ---. 2 b. -1s 2 c. J5-1 2 13-1 d. 2 15. Eccentricity of ellipse for which the length of the Latus rectum is equal to semi major axis is 1 a. 2 b. 1 3 c. 1 13 1 d. J2 16. Eccentricity of ellipse for which the length of latus rectum is equal to semi minor axis is a. 1 2 b. 13 2 c. 1 J2 1 d. 13 17. Eccentricity of an ellipse for which the distance between foci is equal to minor axis IS a. 1 2 b. 13 2 1 c. J2 1 d. 13 18. If the distance between foci of an ellipse is 4 units and e = 1/2, then, length of major axis is a. 8 b. 6 c. 4 d. 2 19. The equation of dlipse having major axis 12 units and eccentricity equal to 1/2 is a. 3x2+4y2=108 b. 4x2+3y2=108 c. 4x2+5y2=108 d. 5y2+4x2=108 /131/ 20. Length of latus rectum of parabola y2=4ax which passes through the point (1,4) s a. 4 b. 8 c. 16 d. 32 21. If the focus of a parabola is at (0,-3) and its dirictrix is y=3, then its equation is a. x2 = -12y b. x2 = 12y c. y2 = -12x d. y2 = 12x 22. The co-ordinates of a point on the parabola y2=8x whose focal distance is 4 are b . (1,2J2") c. (2,4) d. none of these 23. If the parabola y2=4ax passes through (3,2) then the length of its latus rectum is a. 2/3 b. 4/3 c. 1/3 d. 4 24. The co-ordinates of focus of the parabola y2 = 20x are a. (4,0) b. (5,0) c. (0,5) d. none of these 25. The equation of dirictrix of parabola y2 = -12x is a. x-3=0 b. x+3=0 c. y-3=0 d. y+3=0 26. The length of latus rectum of parabola "X2=12y is a. 16 b. 20 c. 12 d. none of these 27. Vertex of all standard parabolas is a. (a,O) b. (0, a) c. (0,0) d. none of these 28. x2 y2 Equations of major and minor axis of ellipse ~ + b2 =1 respectively are a. y=O,x=O b. x=O,y=O c. x=a,y=a d. none of these 2 2 29. Equations of major and minor axis of ellipse ~ 2 +.;- =1 respectively are a b-a y=O,x=O b. x=O,y=O c x=a,y=a e. none of these 2 2 30. Co-ordinates of foci of ellipse ~ + L = 1 are 4 9 a. (0,J3) b. (J3,0) c. ( ~ , O ) d. (0,J5) 31. Vertices of ellipse 9x2+4y2=36 are a. (0, 3) b. (3,0) c. (0,2) d. (2,0) 32. Equations of dirictries of ellipse 16x2+25y2=400 a. 3x 25 = 0 b. 3y 25 = 0 c. 3x 16 = 0 d. none of these 33. Length of a latus rectum of ellipse 16x2+25y2=400 is a. 1.6 units b. 3.2 units c. 6.4 units d. none of these 34. Eccentricity of ellipse 9X2+5y2-30y=0 is a. 1/3 b. 2/3 c. 3/4 d. none of these /132/ 35. The length of latus rectum of ellipse 3x2+y2=12 is 36. 37. 38. a. 4 b. 3 c. 8 x2 y2 . . The equation ----- + ----- =1 represents an ellIpse If 10-a 4-a 4 d . .J3 a. a4 c. 4 0) Lt x ~ O (1+X)1/X =e or (1+1y =e x ~ c ( ) x Lt 10g(1 + x) = 1 x ~ O x Lt aX-1 --=loge a x ~ O x Lt eX-1 --=1 x ~ O x L -Hospital's Rule: -Lt f(x) . . 0 C() If x ~ a g(x) be the lImIt of the form 0 or C() Lt f(x) f' (a) Then x ~ a g(x) = g' (a) ; Provided f' (a), g'(a) "# 0 f" (a) g" (a) if f'(a), g'(a) = 0 and so on. Lt sin x -x Example 10. Evaluate x o--x-' -Solution: Lt sinx -x x = Lt [SinX -1] x Lt [Sinx]_ Lt (1) = X-40 x x-)O = 1 -1 = 0 Lt xsinx Example 11. Evaluate x 0 (1- cos2x) Solution: Lt xsinx x 0 (1 - cos 2x) Lt xsinx = Lt x = x 1 = 2(1) Lt xsinx Example 12. Evaluate x 0 (1- cosx) Solution: Lt xsinx x 01-cosx Lt xsinx = 0 2 = 1 2 [.: Lt sinx = 1] x [.: 1- cos2x = 2sin 2 x] /147/ /148/ Lt X2 sinx sm -2 1 L t sin x x [ ]2 = i x O-x- 1 _.!. Lt ( sin x )[ 2 x J2 - x . x sm-2 _.!. Lt (SinX)4[ J2 - 2x 0 x . x sm-2 = .!.(1)( 4)(1)2 2 = (i)X4 =2 Lt 1-cosx Example 13. Evaluate x 0 x2 Solution: Lt (1- cosx) x2 = 2 . 2 X Lt sm-2 _ Lt 2[Sin - X Lt = 2 [Multiplying Num. & Deno. by xl l ]2 . X Lt 1 sm2 2.---= 4 Lt ( Example 14. Evaluate x 0 1 + 2" Solution: Lt ( 1+- 2 1 1 1 = (e)2 =Fe Lt x.(3r -x Example 15. Evaluate 0 1 -cosx Solution: Lt x(3r -x l-cosx Lt x[(3y -1] sm -2 Lt x.x[(3r-11._1_ = x 2' 2 X sm -2 /149/ .,' (f(x)t = f(x) [ Lt I I t ]11 1 Lx->a [ Lt 1 1 .,' (1+x)x =e I [Multiplying Num. and Den. hy xl /150/ Lt .!.[_x 12[ (3r -1) x-)02 "x x 51112 _ Lt .!.[ 2 " ~ 12 Lt ((3r -1) - x-)02 "x x-)O x 51n-2 = 2 log 3 [ aX-1 ] ": Lim - ;--- = log a Lim eax _ ebx Example 16. Evaluate 0 x-) x Solution: Lt eax -1 + 1- ebx x-)O x Lt (eax _1)_(ebx -1) Jjm = Lt [eax -1]_ Lt [ebX -1] x-)O x x-)O X ": [f(x) f g(x)] -x -- > a Lim Lim f(x)t- g(x) x -) a x - > a Lt [ (eax -1)] Lt [ (ebX -1)] = x-)O a ~ - - x-->O b - - ~ = a Lt (eaX-l)_b Lt (ebX_l) x ~ O ax x ~ O bx = a(l) - b(l) = a - b Lt 2x _ 3x Example 17. Evaluate x ~ 0 x Solution: = x ~ O x = x ~ O x = Lt ((2X -1))_ Lt ((3X -1)) x ~ O x x ~ O x = log 2 -log 3 Lt tan x - xcosx Example 18. Evaluate x ~ 0 x Lt tan x - xcosx Solution: x ~ O x Lt tanx xcosx = x ~ O x x Lt tanx Lt = --- cosx x ~ O x x ~ O = 1-1 = 0 [.: Lt eX -1 = 1] x ~ O x [ Lt (aX -1) 1 .: x ~ 0 -x- = log a [.: log(:) = logm-logn ] /151/ /152/ Lt (X)5 - 32 Example 19. Evaluate x 2 (x _ 2) Solution: Lt (X)5 - 32 x-2 4 = = Lt (X)5 _ (2)5 x-2 Lt (X)5 _ (2)5 (x-2) .: = n(a)n-1 [ Lt (x)n - (a)n ] x--a = 5(2)5-1 = 5(2)4 = 5(16) = 80 Lt X4 -16 Example 20. Evaluate 2 5 32 x -Solution: Lt (X)4 -16 -32 Lt (X)4 _ (2)4 = X 2 (X)5 - (2)5 Dividing Num. and Den. by (x-2) [X4 -(2)4] Lt x-2 = x-2 Lt [X4 - (2)4] x-2 = Lt [(X)5 - (2)5] x-2 (4)(2)4-1 = 5(2)5-1 4(2)3 4x8 2 = 5(2)4 = --= -5 x 16 5 [. . Lt f(x) = x /(x) I . Lt g(x) [ Lt n n ] X - a ( )n-l .: =n a x-a MULTIPLE CHOICE QUESTIONS Q.1 If f is a function from Set A to Set B, then Set A is called /153/ (a) Range (b) Domain (c) Co-domain (d) None pf Q.2 If f is a function from Set A to Set B then Set B is called (a) Range (b) Domain (c) Co-domain (d) None of these Q.3 If f is a function from Set A to Set B then Set f(A) = {f(x):x EA} is called (a) Range (b) Domain (c) Co-domain (d) None of these Q.4 Range of a function f is a subset of (a) Domain (c) R=Set of Real Numbers (b) Co-domain (d) None of these Q.5 If a function f from Set A to Set B is a real valued function, then (a) A c;;;; R (b) B c;;;; R (c) f(A) c;;;; R (d) 'Both (b) and (c) Q.6 If a function f is defined by, f(x) = _ i; then, f(10) = (a) .J8 (b) 98 (c) 7.J2 (d) 8 Q.7 If f(x) = _x_. x ;;j:. -1 0 then f(2) = x+l' " x 1 (a) x + 1 x (b) x + 1 -x (c) x + 1 Q.8 If f(x) = .Jx2 -4; x E R is a real valued function, then, (a) -2 $; x$;2 (b) x 2 x+l (d) x (c) x$; -2 (d) Either x $; -2 or x 2 Q.9 =? \ (a) x x-3 1 (b) -x (c) f(x) Q.l0 If f(x) = -3-' then f(x) is not defined for -x (a) x = 2 (b) x = -3 (c) x = 3 1 (d) x--x (d) x = 4 Q.l1 If for a function f(x), f(x+y) = f(x)+f(y) for all reals x and y, then f(O) is equal to (a) 0 (b) 1 (c) 2 (d) None of these Q.12 If f is a function from Set A to Set B then for xEA f(x), (a) can have two values (b) is unique (c) can have multiple values (d) none of these /154/ 1 Q.13 If f(x) = x, g(x) = --, then f(g(x))= x (a) f(x) (b) g(x) (c) x (d) None of these Q.14 If f is a function defined from Set A = {2,3,4} to Set B = {4,9,16,25} by f(x) = x2, then Range of function f will be (a) {2,3,4} (b) {4,9,25} (c) {4,9,16} (d) {9,16,25} {4X + 5 for 0 ~ x ~ 4 Q.15 If f(x) is a function defined by f(x) = 2x -1 for x> 4 then f(3) = ? (a) 17 (b) 5 (c) 7 (d) 4 Q.16 If f(x) = x2+1; then the value of f(f(x)) = ? (a) x4-1 (b) x4+2x2+2 (c) x4-x2+1 (d) None of these Q.17 The Set A = {xER :2f(x) = f(2x)}, where f(x) = x2+3x is (a) {} (b) {O} (c) {I) (d) {O,4} Q.18 If f(x) = x2+1, then f(a+l) -f(a-l), aER is (a) 2a+4 (b) 2a-4 (c) 2a x Q.19 If f(x) = - -1-; x;t:O,1 then f(l-x) = ? x-1 (a) f(x) (b) ---f(x) (c) 2f(x) Q.20 (2X + 1 ) If f(x) = 3x --2" ' then f(f(2)) = ? (a) 1 (b) 3 (c) 2 Q.21 If f(x) = 2x+7 and g(x) = (x+l}2 then, f(g(x)) = ? (a) x2+4x+9 (b) x2+4x+12 (c) 2x2+4x+9 1 Q.22 If f(y) = sin21ty and y(x) = -;2" then f(y(2)) = ? (a) 1 (b) 0 (c) 2 I-x Q.23 For f(x) = -1-; then f(f(tan8)) = ? +x e e (a) cot 8 (b) coti (c) tan i Q.24 If f(x) = l-x2, then f(sin8)+f(cos8) = ? (a) 2 (b) 0 (c) 1 (d) 4a (d) None of these (d) 4 (d) x2-4x+9 (d) 1/4 (d) tan 8 (d) -1 /155/ Q.25 If f(x) = x2-1 then f(secS)-f(tanS) = ? (a) 2 (b) 0 (c) 1 (d) -1 l-x Q.26 If f(x) = -1-; then f(cosS) = ? +x e e e (a) cot2- (b) tan2- (c) tan-- (d) tan S 2 2 2 Q.27 1[ If f(x) = log x then find f(2sin x) + f(cos x) at x = 4" (a) 1 (b) 2 (c) 0 (d) -1 x Q.28 The values of ~ ; x :f- 0 are (a) 1,0 (b) 1,2 (c) -1,0 (d) -1,1 Q.29 Period of sin x, cos x, sec x and cosec x is (a) 1t (b) 21t (c) 1t/2 (d) None of these Q.30 Period of tan x and cot x is (a) 1t (b) 2 1t (c) 1t/2 (d) None of these Lt x2 -9 Q.31 3 --3 is equal to x---) x-(a) 2 (b) 3 (c) -6 (d) 6 Lt x2 -5x +6 Q.32 x---)2 (x-2) is equal to (a) 1 (b) -1 (c) 2 (d) -2 Lt x2 -6x+9 Q.33 x---)3 (x- 3) is equal to (a) 1 (b) 0 (c) 3 (d) None of these Lt x-l Q.34 x ---) 1 x3 -1 is equal to (a) 0 (b) 1/2 (c) 1/3 (d) None of these Lt xn _an Q.35 --- is equal to x-)a x-a (a) nan-1 (b) nxn-1 (c) ann-1 (d) nan Lt x3 -8 Q.36 ---- is equal to x ---) 2 x2 - 4 /156/ (a) 3/2 (b) 3 2 2 Lt (x)3 - (3)3 Q.37 x --> 3 --r--! (x)3_(3)3 Q.38 Q.39 Q.40 Q.41 Q.42 Q.43 Q.44 Q.45 (a) ifj (b) 2?f3 Lt I .! (1 + x)x is equal to x -) 0 (a) 2 (b) 1 Lt aX-1 For a > 0 -- =? ' x ~ O x Lt eX-1 0-- is equal to x - ~ x (a) 1 Lt a2x_1 --=? x ~ O x -(a) logea Lt 1 (1+2x)x x ~ O (a) Je 1 (b) log 1 (b) 1f2log a (b) e2 1+- =? Lt ( 3X)3X x ~ O 2 (a) Je (b) e2 Lt ( 1)X 1+- =? x ~ o o x (a) Je Lt X Oa for a>O x ~ (a) a (b) e2 ,(b) 0,. (c) 2/3 (d) None of these (c) J3 (d) 2/3 (c) e (d) None of these (c) 1 (d) None of these (c) 2 (d) None of these (c) Joga2 (d) a2 (c) e (d) None of these (c) e (d) None of these (c) e (d) None of these (c) 1 (d) None of these /157/ Lt esinx -1 Q.46 ---=? x ~ O sinx . (a) 0 (b) 2 (c) 3 (d) 1 Lt sin3x Q.47 --=? x ~ O x . (a) 0 (b) 1 (c) 1/3 (d) 3 Lt tan x Q.48 -- =? x - ~ 0 3x . (a) 0 (b) 1 (c) 1/3 (d) 3 Lt 1- cosx Q.49 ------- =? X -) 0 x2 . (a) 0 (b) 1 (c) 1/2 (d) 2 Lt 1- cos2x Q.SO ---- =? X -) 0 x2(1 + cos2x) . (a) 0 (b) 1 (c) 1/2 (d) 2 Lt 3x _2x Q.S1 -----= ? x - - ~ O x 2 3 (a) log6 (b) log) (c) log2 (d) None of these Lt atanx -1 Q.S2 For a > 0,- - ---- = ? x ---) 0 tan x (a) logae (b) logea (c) 1 (d) None of these Lt tan2x -- x Q.S3 --- -- - -------- - = ? X - ~ 0 4x - sin x . (a) 0 (b) 1 (c) 1/3 (d) 3 Lt sin x - cosx ----Q.S4 x ~ ~ ( x-1) 1 (a) 0 (b) -./Z (c) -12-(d) 1 /158/ Lt e3x _ e2x Q.55 sin x (a) 0 (b) 2 (c) 3 (d) 1 Lt eax _ ebx Q.56 sin x (a) b-a (b) a+b (c) a-b (d) 0 Lt x3 +8 Q.57 --=? x+2 . (a) 8 (b) -8 (c) 12 (d) None of these Lt 3-x Q.58 - = ? . (a) -J5 (b) 15-(c) 2 J5 (d) None of these Lt x-2 Q.59 --- ---------- - ? x--)2.J3x-2-.JX+2 -. (a) 2 (b) -2 (c) 3 (d) None of these Lt (a+ht-(a)n Q.60 =? h (a) an (b) al1-1 (c) na (d) nan-1 Lt Ix -11 Q.61 ----- =? x-1 . (a) 1 (b) -1 (c) 0 (d) Does not exist Lt Ixl Q.62 -- =? x . (a) 1 (b) -1 (c) 0 (d) Does not exist Lt 4x3- 3x2 + 2x I- 5 Q.63 ----- - ----- --x --) OC) 2x3, x2 +x+2 (a) 4 (b) 2 (c) 5/2 (d) 1/2 Lt tan x Q.64 --=? sinx -(a) 0 (b) -1 (c) 1 (d) None of these /159/ Lt 2x -1 Q.65 -.-------- -? X --> 0 .J1+x -1 -. (a) log2 (b) 3log2 (c) 2log2 or log 4 (d) 0 Lt JX+I Q.66 (a) -1 (b) 1 (c) 0 (d) None of these {X2 + 2; x 2 Lt Q.67 Iff(x) = 2x+5; x>2 I then =? (a) 6 (b) 9 (c) 0 (d) Does not exist Q.68 Lt (1 + x)n -1 x (a) n-l (b) n (c) 1 (d) None of these Q.69 Lt (1+2+3+4+ ........ +n) x 00 n2 (a) n (b) 1 (c) 1/2 (d) None of these Q.70 If both f(x) and g(x) are defined in a neighbourhood of 0; f(O) = 0 = g(O) Lt f(x) and '(0) = 8 = g'(O) then x 0 g(x) =? (a) 2 (b) 1 (c) 16 (d) None of these Lt ( 4)X Q.71 1 + - is equal to x (a) e1/4 (b) e4 Lt Q.72 , 4 4 x-(a) 2J5 (b) J5 Lt tan eo Q.73 --=? e . 1r (a) 180 (b) 1t (c) 4e 2 (c) J5 (c) 1 (d) None of these (d) None of these (d) None of these /160/ Lt cos5x -cos13x Q.74 x 0 cos5x - cos l1x (a) 5/2 (b) 3/2 (c) 1/2 (d) None of these Lt I-cosmO Q.75 -------- 1-cosnO m2 m3 m (a) - (b) n3 (c) (d) m.n n2 n Lt tan x - sin x Q.76 x3 (a) 1 (b) 0 (c) 2 (d) 1/2 Lt sin x Q.77 J[ (a) 1 (b) 0 (c) 1t (d) -2 Lt cos ax - cos bx Q.78 x2 a-- b b -- a a2 _ b2 b2 _a2 (a) - (b) --(c) (d) -"-2 2 2 2 Lt 2sinx -sin2x Q.79 x3 (a) 2 (b) 0 (c) 1 (d) None of these Lt cos2x Q.80 x (a) 2 (b) 0 (c) 1 (d) None of these Lt Q.81 O(secx +