THIS FILE CONTAINS TWO DIMENSIONAL GEOMETRY (COLLECTION # 2) Very Important Guessing Questions For IIT JEE 2011 With Detail Solution
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1 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) THIS FILE CONTAINS TWO DIMENSIONAL GEOMETRY (COLLECTION # ) Ver Iportnt Guessing Questions For IIT JEE 0 Wit Detil Solution Junior Students Cn Keep It Sfe For Future IIT-JEEs Pge 0 Two Diensionl Geoetr (D) Te Point Strigt Lines Circles Prbol Ellipse Hperbol Inde For Collection # Question (Pge to 9) Single Correct Answer Tpe Question Copreension Tpe Questions Assertion Reson Tpe Question More Correct Answers Tpe Questions Subjective (Up to Digits) Detiil Solution B Genuine Metod (But In) Clssroo I Will Give Sort Tricks ) Se As Above For Collection # (Pge 9 to 5)
2 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Single Correct Tpe Q. Tere re ectl two points on te ellipse b + = centre of te ellipse re gretest nd equl to ellipse is equl to (A) (B) (C) (codevtpaq6) (D) wose distnce fro te Pge + b. Eccentricit of tis Q. Lengt of te ltus rectu of te prbol 5 ( ) + ( ) = ( + 7) is : (A) (B) (C) / 5 (D) / 5 (codevtpaq7) Q. If te eccentricit of te perbol sec α = 5 is ties te eccentricit of te ellipse sec α + = 5, ten vlue of α is (codevtpaq) (A) π / 6 (B) π / (C) π / (D) π / Q. Ec eber of te fil of prbols = + + s iu or iniu point depending upon te vlue of. Te eqution to te locus of te i or ini for ll possible vlues of is (codevtpaq) (A) strigt line wit slope nd intercept. (B) strigt line wit slope nd intercept. (C) strigt line wit slope nd intercept. Q. 5 Te cute ngle t wic te line = (A) 0 (B) 5 (C) 60 (D) 75 (codevtpaq5) (D) circle + = 5 is intersects te circle ( ) π Q. 6 Let ABC be tringle wit BAC = nd AB = suc tt ( AB)( AC) =. If vries ten te longest possible lengt of te ngle bisector AD equls (codevtpaq7) (A) / (B) / (C) / (D) / + = c c 0 ten te lest vlue of + is equl to Q. 7 If ( ) (codevtpaq8) c c c (A) (B) (C) (D) c Q. 8 Te re enclosed b te prbol = nd its ltus rectu is (codevtpaq) (A) 6 (B) (C) 8 (D), ;, R w, w re colliner ten Q. 9 Tree distinct point P ( u u ) Q ( v v ) nd ( ) (codevtpaq5) (A) uv + vw + wu = 0 (B) uv + vw + wu = (C) uv + vw + wu = (D) uv + vw + wu = Q. 0 At te end points A, B of te fied segent of lengt L, line re drwn eeting in C nd king ngles θ nd θ respectivel wit te given segent. Let D be te foot of te ltitude CD nd let represents te lengt of AD. Te vlue of s θ tends to zero i.e. li equls (codevtpaq) (A) L θ 0 (B) L (C) L (D) L
3 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge Q. Nuber of positive integrl vlues of for wic te curve = intersects te line = is (A) 0 (B) (C) (D)More tn (codevtpaq7) Q. Te re enclosed b te curve = & =, te circle te -is, is (A) π (B) π Q. Fro te point (, ) (C) π (D) π (codevtpaq9) tngent line re drwn to te prbol + = bove =. Te re of te tringle fored b te cord of contct nd te tngents is (codevt5paq) (A) (B) (C) 8 (D) 8 Q. Te eqution to te locus of te iddle point of te portion of te tngent to te ellipse + = included between te co-ordinte es is te curve : 6 9 (codevt5paq) (A) = (B) = (C) + = (D) Q. 5 Let P be vrible point on te ellipse ( b) = + = > wit foci F & F. If A is b te re of te tringle PFF, ten te iu vlue of A is (codevt5paq) (A) b (B) b b (C) b b (D) b Q. 6 All points on te curve = + sin t wic te tngent t is prllel to - is lie on (A) circle (B) prbol (C) n ellipse (D) line (codevt5paq5) Q. 7 Te nuber of possible tngents wic cn be drwn to te curve 9 = 6, wic re perpendiculr to te strigt line = 0 is : (codevt5paq6) (A) zero (B) (C) (D) Q. 8 Let S (, ) nd S' ( 9, ) be two foci of n ellipse. If te coordintes of te foot of te perpendiculr for focus S to tngent of te ellipse is (, ) ten te eccentricit of te ellipse is (A) 5 (B) (C) (codevt5paq8) Q. 9 Te strigt line joining n point P on te prbol (D) 7 = to te verte nd perpendiculr fro te focus to te tngent t P, intersect t R, ten te eqution of te locus of R is (codevt5paq9) (A) + = 0 (B) + = 0 (C) + = 0 (D) + = 0 Q. 0 Te re of te qudrilterl wit its vertices t te foci of te conics = 0 nd = 0, is (codevt5paq0) (A) 5/6 (B) 8/9 (C) 5/ (D) 6/9
4 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge Q. Locus of te feet of te perpendiculrs drwn fro eiter foci on vrible tngent to te perbol 6 9 = is (codevt5paq) (A) + = 9 (B) + = / 9 (C) + = 7 / (D) + = /6 Q. A line pssing troug te point (, 0 ) nd norl to te curve = cn ve te slope equl to (A) (B) (C) (D) 5 (codevt5paq) Q. Te gnitude of te grdient of te tngent t n etreit of lter rect of te perbol = is equl to (were e is te eccentricit of te perbol) b (codevt5paq) (A) be (B) e (C) b (D) e Q. If s, s re te lengt of te perpendiculr on tngent fro te foci, re tose fro te vertices, c is tt fro te centre nd e is te eccentricit of te ss' c ellipse, + =, ten = b ' c (A) e (B) /e (C) /e (D) e (codevt5paq5) Q. 5 TP & TQ re tngents to te prbol, = t P & Q. If te cord PQ ten te locus of T is : psses troug te fied point (, b) (codevt5paq6) b b b (A) = ( ) (B) = ( ) (C) b = ( ) (D) = b( b) Q. 6 Eccentricit of te perbol conjugte to te perbol (codevt5paq7) (A) (B) (C) (D) Q. 7 Consider te sste of equtioins ( p ) + = 0 nd ( ) = is + 7 p = 0. If te sste s te solution oter tn = = 0 ten te rtio : cn be (codevt6paq8) (for onl tis que. More tn one Ans. M correct.) (A) / (B) / (C) (D) - Q. 8 An equilterl tringle s side lengt 8. Te re of te region contining ll points outside te tringle but not ore tn units fro point on te tringle is : (codevt8paq) π π (A) 9( 8 + π ) (B) 8( 9 + π ) (C) (D) Q. 9 An object oves 8 c in strigt line fro A to B, turns t n ngle α, 0, π nd oves 5 esured in rdins cosen t rndo fro te intervl ( ) c in strigt line to C. Te probbilit tt AC < 7 is : (codevt8paq) (A) 6 (B) (C) (D)
5 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Copreesion Tpe # Prgrp for Q. to Q. Consider line : + =, / B, 5. P is point on te line L. L nd te points A ( ) nd ( ) Pge Q. Te bsciss of te point P for wic te re of te PAB is unit (Given P does not lie on te -is) is (codevtpaq9) (A) (B) 6 (C) (D) None Q. A circle psses troug A nd B nd s its centre on te -is. Te -coordinte of te centre is (A) 5 (B) 55 7 (C) 7 (D) None (codevtpaq0) AC + BC is Q. If C is soe point on L ten te iniu distnce ( ) (codevtpaq) (A) 7 (B) (C) (D) # Prgrp for Q. to Q. 6 Consider te lines represented pretricll s L : = = z = + L : = = 5 z = Find Q. cute ngle between te line L nd L, is (codevtpaq) (A) cos 8 (B) cos 6 (C) cos (D) cos 6 L, is (codevtpaq5) Q. 5 eqution of plne P contining te line L nd prllel to te line (A) z 7 = 0 (B) + z 5 = 0(C) = 0 z (D) 9 5 z = 0 Q. 6 distnce between te plne P nd te line L is (codevtpaq6) (A) 7 (B) (C) (D) # Prgrp for Q. 7 to Q. 9 Consider te circles S : = 0; S : = 0 And vrible circle S: + + g + f + c = 0 Q. 7 Nuber of coon tngents to S nd S is (codevt5paq9) (A) (B) (C) (D) Q. 8 Lengt of trnsverse coon tngent to S nd S is (codevt5paq0) (A) 6 (B) (C) 5 (D) Q. 9 If te vrible circle S = 0 wit centre C oves in suc w tt it is lws toucing eternll te circles S = 0 nd S = 0 ten te locus of te centre C of te vrible circle is (A) circle (B) prbol (C) n ellipse (D) perbol (codevt5paq) # Prgrp for Q. 0 to Q. Fro te point P (, k ) tree norls re drwn to te prbol, nd re te slopes of tree norls = 8 nd
6 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge 5 Q. 0 Algebric su of te slopes of tese tree norls is (codevt5paq) (A) zero (B) k (C) k (D) k Q. If two of te tree norls re t rigt ngles ten te locus of point P is conic wose ltus rectu is (A) (B) (C) (D) (codevt5paq) Q. If te two norls fro P re suc tt te ke copleentr ngles wit te is ten te locus of point P is conic ten directri of conic is (codevt5paq) (A) = 0 (B) + = 0 (C) 5 = 0 (D) + 5 = 0 # 5 Prgrp for Q. to Q. 5 + d d = 0 nd psses A conic C stisfies te differentil eqution, ( ) troug te point (, 0 ). An ellipse E wic is confocl wit C ving its eccentricit equl to / Q. Lengt of te ltus rectuof te conic C, is (codevt5paq5) (A) (B) (C) (D) Q. Eqution of te ellipse E is (codevt5paq6) (A) + = (B) + = (C) + = (D) + = 9 9 Q. 5 Locus of te point of intersection of te perpendiculr tngents to te ellipse E, is (codevt5paq7) (A) + = (B) + = 0 (C) + = 8 (D) # 6 Prgrp for Q. 6 to Q. 8 Two equl prbol P nd P ve teir vertices t ( ) + = V 0, nd V ( 6, 0 ) respectivel. P nd P re tngent to ec oter nd ve verticl es of setr. Q. 6 Te su of te bsciss nd ordinte of teir point of contct is (codevt5paq8) (A) (B) 5 (C) 6 (D) 7 Q. 7 Lengt of ltus rectus is (codevt5paq9) (A) 6 (B) 5 (C) 9/ (D) Q. 8 Are of te region enclosed b P, P nd te -is is (codevt5paq0) (A) (B) (C) (D) ( ) # 7 Prgrp for Q. 9 to Q. Four A, B, C nd D in order lie on te prbol = + b + c nd te coordintes, 7 respectivel. of A, B nd D re (, ); (, ) nd ( ) Q. 9 ( + b + c) s te vlue equl to (codevt8paq) (A) (B) (C) (D)
7 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge 6 Q. 0 If roots of te eqution + b + c = 0 nd α re β ten te eqution wose roots re α nd β is (codevt8paq5) (A) + = 0 (B) + + = 0 (C) + + = 0 (D) + = 0 Q. If te re of te qudrilterl ABCD is gretest, ten te su of te bsciss nd ordinte of te point C is (codevt8paq6) (A) 9 (B) 7 (C) 5 (D) # 8 Prgrp for Q. to Q. Consider line pir c = 0 representing perpendiculr lines intersecting ec oter t C nd foring tringle ABC wit te -is. Q. If nd re intercepts on te -is nd nd re te intercepts on is equl to te -is te su ( ) (codevt8paq7) (A) 6 (B) 5 (C) (D) Q. Distnce between te ortocenter nd circucentre of te tringle ABC is (codevt8paq8) (A) (B) (C) 7/ (D) 9/ Q. If te circle + + k = 0 is ortogonl wit te circucircle of te tringle ABC ten k equls (A) / (B) (C) (D) / (codevt8paq9) # 9 Prgrp for Q. 5 to Q. 7 = f, wic intersects te -is Tngent is drwn t te point ( i, i ) on te curve ( ) t ( i+, 0 ). Now gin tngent is drwn t ( i, i ) te -is t (, 0) + + on te curve wic intersects i+ nd te prcess is reputed in ties i.e., I =,,, n. Q. 5 If,,,... n for n ritetic progression wit coon difference equl to log e nd curve psses troug ( 0, ), te eqution of te curve is (codevt9paq) (A) = e (B) = e (C) = + (D) = Q. 6 If,,,..., n for geoetric progression wit coon rtio equl to nd te curve psses toug(, ), ten te curve is (codevt9paq) (A) prbol (B) n ellipse (C) rectngulr perbol (D) perbol wic is not rectngulr Q. 7 Te rdius of te circle toucing te curve obtined in question no. -6 t, 0 is (, ) nd pssing troug te point ( ) (codevt9paq) (A) 5 (B) (C) (D) # 0 Prgrp for Q. 8 to Q. 0 Consider te two qudrtic polnoils C : = + + nd C : =
8 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge 7 Q. 8 If te origin lies between te zeroes of te polnoil C ten te nuber of integrl vlue(s) of is (A) (B) (C) (D) ore tn (codevt0paq9) Q. 9 If vries ten te eqution of te locus of te verte of C, is (codevt0paq0) (A) = 0 (B) = 0 (C) + = 0 (D) + = 0 Q. 0 For =, if te lines = + c nd = + c re coon tngents to te grp of C nd C ten te vlue of ( + ) is equl to (codevt0paq) (A) 6 (B) (C) / (D) none # Prgrp for Q. to Q. Two fied points A nd B re units prt, nd re on te se side of oving line L. If perpendiculr distnces of A nd B s Pnd P fro te line L re suc tt P + P = k, k being constnt, ten te line L lws touced fied circle C. Q. Te centre of te circle C lies on (codevt0paq) (A) line segent joining AB (B) perpendiculr bisector of AB (C) one of A or B (D) noting definite cn be sid Q. If k = ten te rdius of te circle is (codevt0paq) (A) (B) (C) (D) 8, 0,,0 respectivel, ten te centre of te circle C is Q. If A nd B re ( ) ( ) (A) ( 0, ) (B) (, 0 ) (C) ( /, 0 ) (D) cn not be found (codevt0paq) Assertion & Reson Tpe In tis section ec que. contins STATEMENT- (Assertion) & STATEMENT-(Reson). Ec question s coices (A), (B), (C) nd (D), out of wic onl one is correct. Bubble (A) STATEMENT- is true, STATEMENT- is True; STATEMENT- is correct eplntion for STATEMENT-. Bubble (B) STATEMENT- is True, STATEMENT- is True; STATEMENT- is NOT correct eplntion for STATEMENT-. Bubble (C) STATEMENT- is True, STATEMENT- is Flse. Bubble (D) STATEMENT- is Flse, STATEMENT- is True. Q. Consider n epression f (, k ) = + k k were nd k re non zero rel nubers. Stteent-: f (, k ) is lws positive non zero nd rel nd k. (codevtpaq6) Stteent- : A qudrtic epression + b + c is lws positive if ' ' > 0 nd b c < 0. Q. Let C be circle wit centre O nd HK is te cord of contct of pir of te tngents fro point A. OA intersects te circle C t P nd Q nd B is te idpoint of HK, ten Stteent-: AB is te ronic en of AP nd AQ. (codevtpaq8)
9 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Pge 8 Stteent- : AK is te Geoetric en of AB & AO nd OA is te ritetic en of AP nd AQ. Q. Stteent-: Te cuves = nd = re not ortogonl. Stteent- : Te ngle of intersection between ten t teir point of intersection oter tn origin is tn ( ). (codevtpaq) Q. Consider te curve C : = + Stteent- : Te tngent to C t its point P(, 0 ) nd t its point Q(, 0) is prllel to -is. d Stteent - : At te point P(, 0 ) nd Q(, 0) d (codevt8paq0) Q. 5 Consider tringle wose vertices re A (, ), B(, ) nd C(, ) were is rel nuber. Stteent- : Te re of te tringle ABC is independent of (codevt0paq7) Stteent - : Te verte C of te tringle ABC lws oves on line prllel to te bse AB. More tn One Correct Tpe f is circle suc tt te eqution f (, 0 ) = 0 s coincident root Q. Suppose (, ) = 0 equl to, nd te eqution ( 0, ) = 0 g (, ) = 0 is circle centred t ( 0, ) nd tngent to te circle ( ) f lso s coincident roots equl to. Also, f, = 0. Te possible rdii of te circle is (codevtpa9) (A) sin5 (B) cos5 (C) sin8 (D) cos 6 Q. Eqution of strigt line on te cople plne pssing troug point P denoting te cople nuber α nd perpendiculr to te vector OP were O in te origin cn be written s (codevtpaq) z α z α (A) l = 0 (B) Re = 0 (C) Re( α Z ) = 0 (D) αz + α z α = 0 α α Mtc Mtri Tpe Q. Consider te conic C : + + = 0 C : = 0 C : 7 = 0 Colun-II (codevtpbq) Colun-I (P) (A) Lengt of te ltus rectu of C (Q). (B) Lengt of te ltus rectu of C (R). (C) Lengt of te ltus rectu of C (S). SOLUTION (COLLECTION # ) Single Correct Tpe Q. C [Sol. Te given distnce is clerl te lengt of sei jor is + b Tus, = b = ( e ) = e = e = Q. D [Hint: Note tt focus is (, ) nd directri is + 7 = 0 nd distnce fro S to directri is lf te ltus rectu] b 5cos α Q. D [Sol. = e = + = + = + cos α ; 5 5cos α 5 l eccentricit of te ellipse
10 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) cos α 5 + = is 5cos α e = = sin α ; put 5 e = e e = e Pge 9 + cos α = sin α = sin α sin α = ] Q. A [Hint. Mi/ini occurs t =, f ( ) = + b + c s i or ini if = b = nd k = ; eliinting we get k = + Locus is Q. 5 B [Sol. Solving ( ) ( ) + ( ) = 5 i.e. ( 0, ) nd (, ) = nd ( ) + = = 0 = 0 or bc A b b + c b + but b = b = = = + + Q. 7 B [Sol. put = ccos θ ; = csin θ Q. 6 B [Sol. AD = = cos = ( s c = ) = since iniu vlue of te denointor is if > ( ) E = + = c ( sin θ + cos θ) c ( sin θcos θ ) c = c sin θ = E = wen sin θ = ]. = Ans. u u Q. 8 B [Sol. ( ) Q. 9 A [Sol. v v = 0 w w R R R nd R R R 0 B ] u v u v 0 u v u v vu = = v w v w 0 0 v w v w vw 0 0 w w w w R R R ( ) ( ) A ] = + ( ) u w u w + v u w 0 u + w + v = = v w v w vw 0 0 v w v w vw 0 0 w w w w ( ) ( ) ( ) v + w + vw v + w v + w + u = 0 ( ) ( ) v + w + vw = v + w + u v + w uv vw wu = Ans.]
11 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Q. 0 B [Sol. CD=; tn θ = ; Now tn ( L ) tn tn θ = L tn θ + tn θ = L tn θ θ = θ ( ) tn θ = L tn θ + tn θ L li = Ans. ] θ 0 Q. B [Sol. for 0 < te line Alws cuts = for > s e f '( ) e f ( ) tn θ θ θ = L tn θ tn θ θ + θ θ θ = consider ( ) f = e = f '( ) > 0 for > 0 nd ( ) is incresing ( ) nd decresing ( ) for < 0 e for for > 0 f ' < 0 for < 0 = e lws lies bove > Hence never cuts = = ( ] ( ) 0, B ] = i.e. Pge 50 0 π Q. D [Hint. A = d + d = 0 note tt te re is equl to te sector AOB wit centrl ngle 90 ¼ (te re of te circle) required re π π π = Ans. / ( ) ( + ) Q. D [Sol. A = = = 8 Ans. ] Q. A Q. 5 B Q. 6 B = 5/ + 5 = / 5 b = 9. / 5 = 6 00 / 5 < 0 / Q. 7 A [Hint. ( ) ( ) Note tt te slope of te tngent ( /5 ) is less tn te slope of te sptote wic is / wic is not possible] Q. 8 A [Sol. SS' = e, were nd e re lengt of sei-jor is nd eccentricit respectivel ( ) ( ) 9 + = e e = 5 centre is id-point of SS 6 = 8 = Centre =(6, 8), Let te eqution of uilir circle be ( ) ( ) We know tt te foot of te perpendiculr fro te focus on n tngent lies on te uilir circle (, ) lies on uilir circle i.e. ( ) ( ) e = 5 e = 5/ Ans.] Q. 9 B [Sol. T : t t = +.() Line perpendiculr to ( ) troug (, 0 ) t + = t.() = =
12 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) eqution of OP : = 0.() t & eliinting t we get locus] fro ( ) ( ) Q.0 B [Sol. st is perbol ( ) ( ) 9 6 = 6 wit e = 5/ nd nd is n ellipse ( ) ( ) = wit e = / 5 Pge 5 Wit = X nd = Y re = d d =.. = 5 9 Note tt e E.e H = ] Q. D [Sol. = /6 / 9 Locus will be te uilir circle + = /6, ; 9, 6 were tere is no Q. D [Sol. Te point wit slope nd re norl t ( ) ( ) curve, point of norl ( ), ].e.b Q. B [Sol. T : ; = or b.b e = = e Ans.] e = or Q. D [Sol. Let te eqution of tngent b = + + b ±, 0, C 0, 0 Foci ( e, 0 ), ± vertices ( ) ( ) e + + b e + + b s =, s' = b + + b =, ' = + + c = + b + e ss' c = + ' c = e + Ans.]
13 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Q. 5 C [Hint. Cord of contct of (, k ) k = ( + ). It psses troug (, b) bk = ( + ) Locus is b = ( ) ] b Q. 6 A [Sol. e = + = + = e = ; now + = e e = = e = e = ] e Q. 7 B, D [Sol. for non tribil solution p 7 p = 0 ( )( ) p p + = 0 p = or p = 8 if p, 0 if p = 8, + = 0 = ( B ) ] π Q. 8 [A] [Sol. Are =. ( 8.) +. r θ = 7 + 9π 9( 8 + π ) Q. 9 [C] [Sol. If AC = 7, ten = + = = ( D) cosα = = Hence cosα = = α = 60.0 AC < 7 α 0, 60 Now, ( ) Hence p = 60 = [ C ] ] 80 Copreesion Tpe Q. C Q. A P, Q. D [Sol. Let ( ) Hence / = 5 () 9 = 0 rejected = ( ) 8 9 = = Ans. 9 5 () Midpoint of AB is M, ; 7 AB = Eqution of perpendiculr bisector = 7 put = 0 = + = Ans. A ', / () Ige of A in te line L is ( ) now 8 8 AC + BC = A'C + BC A'B = 5 + = Ans.] Pge 5
14 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) z + 5 z + Q. C Q. 5 A Q. 6 C [Sol. L : = = ;L : = = (i) V = i + j + k; V = i + j k + θ = = θ = (ii) Eqution of te plne contining te line L is cos cos C A( ) + B( 5) + C( z + ) = 0 () Were A + B C = 0 since () is prllel to L ence A + B + C = 0 Hence eqution of plne P ( ) ( ) (iii) distnce between P nd L is Q. 7 D Q. 8 A ( ) A B C = = A = 5k; B = k; C = 9k z + = z 7 = 0 (A) d = = Q. 9 D [Sol. (iii) ( ) ( ) r = ; r = ; C = 0, ; C = 6, 0 ; C C = 5 clerl te circle wit centre C nd C re seprted CC = r + r CC = r + r CC CC = r r = constnt] Q. 0 B Q. B Q. A [Sol. Eqution of norl in ters of slope is 0 P, k stisfies tis eqution = ( ) + = point ( ) ( ) Ans.] + k + = 0..() () lgebric su of slopes is k + + = Ans. () If two norls re perpendiculr ten = nd = substituting = in () we get 8 ( k) + + = 0 + ( k) + = 0 + ( ) ltusrectu = Ans. () If te slopes re copleentr ten = = using = in () we get 8 ( k) 0 ( k) 0 k 0 = = Y let Y = = / ( ) directri is Y = or = = 0 Ans. ] Pge 5
15 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Q. B Q. A Q. 5 A d d [Sol. ( + ) d = d = ln = ln ( + ) + C + given =, = 0 C = 0 ence eqution of C is Pge 5 = wic is rectngulr perbol wit eccentricit e =. (i) lengt of te ltus rectu of rectngulr perbol = = (ii) Now for ellipse, e = e =. = = nd b = ( e ) = =. Hence eqution of ellipse is + = Ans. (ii) Locus of te point of intersection of te perpendiculr tngents is te director circle of te ellipse eqution is + =. ] Q. 6 B Q. 7 C Q 8 D [Sol. Te prbols will ve teir concvities in opposite direction oterwise te cn not touc P : = λ...() λ > 0 Let ( ) ( ) P : 6 = λ.() nd ( ) Solving te two eqution ( 6) = λ = ( 6) + λ λ ( ) + 6 λ = λ = 0 ( ) ( ) b c 0; = = λ = λ 9 λ = 8 λ = Hence te prbol re 9 ( ) ( ) 9 9 = ; 6 = Ltus rectu = Ans (ii) d gin, d ust be se for bot ( ) P, d d = λ = d d λ, (were ( ) 9 λ = ) d d 6 9 nd ( 6) = λ d d = (were λ = ), λ ( 6) = = = wen = ten = point of contct = (, ) su = 5 Ans. (i) (iii) Metod 6 6 ( 6) A = ( 6) d. = = 0 = A Ar.PCV ( ) ( ) A = d =..7 A Ar. PCQ 9 7 = 7 7 ( )
16 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) = ( ) ( ) = 0 = 8 0 required re = 8 0 = 8 = ( ) Metod - 6 d 0 / / = 6.. ( ) = (. ) ( 0) + ( 8) = ( ) + ( 8 ) = 8 = ( ) Ans. (iii)] Q. 9 [C] Q. 0 [B] Q. [A] [Sol. () () α = w = w β = w = ence eqution is ( ) w w + w + = 0 Ans. (iii) + + (B) ] Q. [B] Q. [C] Q. [D] [Sol. () As line re perpendiculr = 0 = (coefficient of + coefficient of = 0) using = 0 c = ( D bc + fg f bg c ) ence te two lines re + = 0 nd + = 0 5 int ercepts ; / int ercepts / ; = = = = = Ans () ( CM) = + = = + = = = CM = () Circucircle of ABC + ( ) + = 0 ( + )( ) + = 0 5 ( + ) 5 = 0 + = 0.() Given k 0 using te condition of ortogonlit + + = wic is ortogonl to ( ) we get, = k k = Ans.] Q. 5 D Q. 6 C Q. 7 A f [Sol. Eqution of tngent to = ( ) t (, ) Y = ( X ) i i i i Pge 55 (i) i Yi put Y = 0, X = i = i+ i+ i = (ii) i i d d lies on te curve, d is te coon difference of i i d d A.P.) = = ln ln d d log e = ln ln C 0, C = ln Curve pssing troug ( ) ln = ( ).ln = ln = Ans.
17 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) (ii) gin if,,,..., n in G.P. i i Divide eqution () b i + = i i r = ( i i lies on curve, r is te coon rtio of G.P.) = = d = d + d = d 0 ln = ln c = c s curve psses troug (, ) c = = wic is rectngle perbol Ans., on =, (iii) Eqution of tngent t ( ) + = ; + = + =..() circle toucing () t (, ) is + + λ + = 0..() ( ) ( ) ( ) it pssé troug ( ) ( ), 0 + λ = 0 λ = = 0 r = + + = 5 r = 5 Ans.] + + = 0 Q. 8 B Q. 9 A Q. 0 B [Sol. ( ) () for zeroes to be on eiter side of origin f 0 < 0 = f = + + ( ) + < ( + )( ) < < < integers i.e. {, 0} ( B) C is (, ) 0 0 () Verte of Hence = nd k = = k + Locus = + = 0 Ans. ( ) () Let = + c is coon tngent to = + 0.(0) (for = ) nd = () were = or nd c = c or c solving = = c wit () + c = + 0 or ( ) 0 c = D 0 = gives ( ) 0 c c 0 ( ) + = = +.() l + c = + + c = 0 D = 0 gives = c c = +... Fro () nd () ( ) ( ) 0 + = = = = Ans ] Pge 56
18 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Q. A Q. A Q. B [Sol. () Let A = ( 0, 0) nd B = (, 0) And te line be + b = p = ; p = + b + b + = k + b + b p + p = k now ( 0, 0 ) nd (, 0 ) ust give te se sign i.e. ve wit te line L( < 0) ( ) + = k + b + b ( ) + b = k; k = + b ence centre of te fied circle is (, 0 ) wic lies on te line segent AB ( A) () If k = r = Ans. () p = ; + b p = + b ence wit te se rguent, < 0 nd < 0 p + p = k ( ) + + = k + b + b k = + b Q. A Q. A [Sol. ( ) + b = k ence center is (, 0 ) Ans.] Assertion & Reson Tpe ( AK) ( OA) AB = cosθ = AK ( ) ( )( ) ( )( ) AK = AB OA = AP AQ [AK = AP.AQ using power of point A] Also OA AP + AQ = [ AQ AO = r = AO AP AO = AQ + QP] AP + AQ ( AP)( AQ) = AB Q. C [Hint. ngle of intersection is circle wit centre ( +, 0) nd rdius ( ) + = ] ( )( ) ( AP + AQ) AP AQ AB = ] tn ] Q. [A] [Sol. C is Pge 57
19 Teko Clsses IITJEE/AIEEE Mts b SUHAAG SIR, Bopl, P (0755) Q. 5 A [Sol. l l ( ) ( ) = = = A = = 8] ( ) More tn One Correct Tpe Q. C, D [Sol. f (, ) = 0 will ve centre t ( ) ( ) ( ) = + = CC = 5 Hence rdius r of g(, ) = 0 is 5 + nd 5, nd rdius unit seefigur, one circle eternl nd oter internll tngent Q. B, D [Sol. Required line is pssing troug ( ) te vector OQ Hence z = α + iλ, λ R z α = α z α α purel iginr Re = 0 ( B ) (( ) ) ( ) Re z α α = 0 Re zα α = 0 Also Q. (A) Q; (B) S; (C) P z α z α + = 0 α α ( z ) ( z ) 0 P α nd prllel to α α + α α = z z 0 ( ) α + α α = D ] Mtc Mtri Tpe [Sol. (A) C : + + = 0 ( ) + = + = = (B) (C) C : = ( ) ( ) ( ) ( ) = 0 ( ) ( + ) = = ( ) ( ) + = Let = X; + = Y Y X b. + = L L ( b, ) C : 7 0 ( ) ( + ) = = = = LL = = ( ) ( ) Hence LL = Ans = 0 X Y.6 = Let = X; + = Y = L L = = Pge 58
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