SCORE JEE (Advanced)

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1 SLUTIN. ns. (D) L : x + y 0 S L : x + y 0 L : x + y 7 0 Point of intersection of L 0 & L 0 is (,9) Point of intersection of L 0 & L 0 is (0,) line perpendiculr to L nd pssing through (, 9) isx y (i) Line perpendiculr to L nd pssing through (0, ) is x y (ii) Point of intersection of (i) & (ii) is the focus of the prbol which is ( 6, ).. ns. (C) tn æ 4 ö ç è 7 ø 4 æ m -- ( / 4) ö 7 ç + - è m( / 4) ø ( 4,0) m 4 \ eution of the line L 0 is L : x 4y + 0 Now, incentre of the DC is L 0 L : x + 4y + 0 æ 0+ 6 (- 4) (-) ö ç, è 6 6 ø æ ö º ç -,0 º 4 l,k è ø \ ( ) l -, k 0 8 0k 4l 9 4. ns. () + 6 SCRE JEE (dvnced) HME SSIGNMENT # 0 6 cos+ + sin 4 r P(,) Q(+rcos,+rsin ) r 6 C (0, ) 6 (0, ) cos + sin MTHEMTICS 6 sin+ cos sin( + 4 ) sin or or 7 6. ns. () - (, ) ' (, 0) C : tn tn (6, 4) ' (, 0) sin & cos ' º (cos, sin) º (, ) Similrly ' º ( cos, sin) º (6, 4) It cn be checked tht C & C touches ech other. Let the point of contct be C. æ C º + 6, + 4 C º ö ç è ø æ 0 ö ç, Reuired Rdicl xis is line perpendiculr to '' nd pssing through point C æ 0ö ç y- - x - 7. ns. () + ( ) tn + tn tn -tn tn / y7x y x

2 tn -tn -4 tn tn 0 (tn +)(tn ) 0 tn or / \ Slope of longer digonl is. 8. ns. () x + y 6x y (x ) + (y 6) co-ordintes of P re ( + cos, 6 + sin) where tn 6 Q (,6) æ ö \ Pç +., 6 +. è ø P(4, 8) fter rottion of 90 in nti clockwise sense co-ordintes of Q re ( + cos(90 + ), 6 + sin(90 + )) ( - sin, 6+ cos). ns. (C) æ ö ç -, 6 +. è ø Cse-I : when lines re concurrent \ b - (, 7) ( + 6) ( + b) ( b) b + b 0 b 4 b 4 Cse-II : when L 0 & L 0 re prllel b b 6 Cse-III : when L 0 & L 0 re prllel b b \ Sum of vlues of b is P 4. ns. (C) Tngent t P is y (x + ) Putting x 0, we get y Slope of HP slope of SH ÐSHP p SP is dimeter so SP r p []. ns. (C) - (0,) H P (,) S(,0) The eution of the norml to this curve t (sec, tn ) is x cos + y cot...(i) Let the foot of perpendiculr from centre upon the norml be (h, k). eution of norml cn lso be written s hx + ky h + k...(ii) compring eution (i) & (ii) cos cot h k h + k h + k sec, tn h (h + k ) (h + k ) - 9h 9k (x + y ) (y x ) 9x y k, k, k k + k + k 6. ns. () Let the perpendiculr lines re xy xis nd centre is (h, k) then lies on its director circle it mens P is constnt + k (h k ) h + k c x + y c is circle. P (h, k)

3 4. ns. () P 4 sin, p 6 D P is euilterl tringle P re ( D P ) 9. ns. () 4 (, ) P Let yt x + t be the tngent pssing through (, ) \ t + t t t 0 t + t \ 4t + 4t 6 0. ns. (D) y x(y ) (y ) 0 (x )(y ) 0 isectors (y ) (x ) x y (y ) (x ) x + y (0, ) x (, ) y y Reuired circle is x(x ) + (y )(y ) 0 (x ) + (y ). ns. (D) l n (, 0) lies on the given line focus lies on directrix conic s pir of stright lines.. ns. () 4 L. R. l l x 4y 0 l P(, 0) æ ö ç 6 4 \ Distnce between focus & directrix.6 4. ns. () xy x y (i) (x )(y ) 0 l 4x + y 0 x The given pir of lines of which (i) is bisector is eully inclined to co-ordinte xis. h 0.. ns. (C) Put y 0, we get x + 4x + c 0, which gives eul roots if 6 4c 0 c 4. \ Eution becomes x + 4xy y + 4x + ƒy which represents pir of stright lines if bc + ƒgh ƒ bg ch 0 ƒ 4 6. ns. () S : (x ) + (y 4) S : x + y 6x 8y 0 Q rigin lies on the circle. & re points on the circle t distnce from origin. Eution of nother circle whose centre is (0, 0) nd rdius is, is S' : x + y 9 \ chord is, S S' 0 6x + 8y 9 4. ns. () + d c + b + d + d c + b + bc + d c + b + d + d b + c + bc + d b + c Now the given line is ( + d )x + b y + c 0 (b + c )x + b y + c 0 b (x + y) + c (x + ) 0 hence given fmily of lines will pss through the point of intersection of the lines x + y 0 nd x + 0 i.e. (, ) 4. ns. (C) Coordintes of ny points lying on the line æ r r ö y x will be, ç If the given line intersects the curve x 4 + x y + bxy + cx + dy + 6 0, then 4 r r br cr dr

4 r 4 + r + r 4b + r8(c + d ) \ r r r r 4 96 hence.. C. D ns. (D) Fmily of lines x + y 0 nd x + y 0 will psses through the point of intersection of x + y 0 nd x + y 0 i.e. (, 0) ccording to given condition the second fmily of line will lso pss through the point (, 0) 0 b 0 b centroid of the given tringle is (, 7) hence centroid of the tringle is (b, + b) 47. ns. () Let the eution y mx + c be the common tngent to the curve y 8x nd x + y then c /m nd c ( + m ) 4 m ( + m ) m 4 + m 0 (m + )(m ) 0 m m ± hence there will be two common tngents which re perpendiculr to ech other. 49. ns () Let the curve y x + px + cuts the x-xis t (, 0) nd (, 0) then + p, lso y x + px + cuts y-xis t C (0, ) Now eution of fmily of circles pssing through nd is (x ) (x ) + y + ly 0 this circle lso psses through point C(0, ) so l ( + ) so the circle is x + y + px + ( + ) y 0 which lwys psses through (0, ) 4. ns. (D) Given hyperbol is Eution of its director circle will be x + y 4 which is not possible, so no point exist from which two perpendiculr tngents cn be drwn to the hyperbol.. ns. (D) The norml t (cos, bsin) will be x by b cos - sin - If psses through (cosb, bsinb) cosb b sinb - -b cos sin cosb b æ sinb ö - - cos ç sin cosb-cos b æsinb-sin ö ç cos è sin +b -b +b b- sin sin cos sin b cos sin tn tn 8. ns. (C) æ+bö b ç - ny tngent to y x cn be written s y mx m 4 - if it touches nd, then m 4 mx - x - x + D 0 æ m ö x -(+ m)x+ ç + 0 è 4 ø æ m ö (m+ ) - 4ç + è 4 ø 4m 4 0 m 4x 4y 0 9. ns. (,C,D) (x ) + (y ) 4 W (,) (7,) ø (7,) 0 (,) ption () : ( C) 4 ption () : C(7,) º ( ) + ( ) S 0x 6y 0 l - y - 4 0

5 bviously rdicl xis is x y4 ption () is incorrect ption (C) : (x )(x 7) + (y )(y ) 0 {The reuired circles dimeter is } x + y x 4y ption (C) is correct ption (D) is obviously correct 6. ns. (,C) S' (, ) (x, y) P S 0 (, 7) xis (x - ) + (y - 7) + (x + ) + (y + ) PS + PS' ellipse e e xis (y 7) 7 (x -) (y 7) 4(x ) 4x y ns. (,D) S : (x ) + (y + ) 0 \ S is point circle which represents point (, ) nd this point (, ) lso lies on the line L : x + y + 0 Eution of tngents from (, ) to the circle x + y is y + m(x ) 9 mx y + m pplying p r + m + m 9 9(9 + m + 4m ) ( + m ) m 48m 40 0 \ m + m 48 & m m ns. (,C) P N (,4) C(,0) d(p, C) min. {d(p, ), d(p, C)} The region represented is shown by shded region, where P nd CP re ngle bisectors Clerly mximum of d(p, C) occurs, when P is incentre of DC \ Mximum of d(p, C) PN ordinte of incentre P 67. ns. (,C) For concurrency - m- m -7-0 m- m- 0 (m ) (m ) [(m )(m ) (m )(m 7)] 0 0m m + 0 m + 7m + m m 7m m 4m + m 6 0 (m )(m m + ) 0 m is the only rel root ut for m the given line becomes prllel. Thus the given lines cn only be prllel for m. 68. ns. (,) s the given tringle is isosceles, so the ngle bisector of two eul sides will be perpendiculr to third side. Now for ngle bisector 7x + y - 0 x + 7y -0 ± 8 8 (7x + y 0) ± (x + 7y 0) so the eution of ngle bisectors re x y 0, x + y 4 0, so the slopes of the lines

6 perpendiculr to these lines will be, so eution of third line will be y (x + ) or y (x + ) x y or x + y ns.(,,c,d) Let ƒ (x) x 6x nd s (, 0) lies on the dimeter of the circle so mximum vlue of cn be 4 Now ƒ ( ) + 6 > 0 ƒ (0) < 0 ƒ () 9 8 < 0 ƒ (7) 49 4 > so the eution x 6x 0 hs one root between nd 0 & one root betwen nd ns.(,) ny point on the line x + y cn be tken s (l, l) so eution of the chord tking this point s mid point will be lx (y + l) l 4 ( l) It psses through (, ) l ( + l) l 4( l) l + 0 It hs two distinct root then ( + ) > 0 < 0 0 < <. length of ltus rectum cn be between 0 nd ns. (,C) Intersection point of circle x + y 9 nd line y x will be æ 6 ö ç, è ø nd æ 6 ö ç, è ø Coordintes of the point of intersection of perpendiculr drwn from these points to the mjor xis of ellipse nd ellipse will be æ 4 ö ç, - è ø, æ 4 ö ç, - è ø, æ 4 ö ç -, è ø, æ 4 ö ç -,- è ø 6 Eution of tngents t these points will be x + y 0,x y 0 x y + 0 nd x + y ns. (,C,D) It is locus of P(x, y) moving such tht (PS PS') 4 < SS', where S(4, ) nd S'( 4, ). So the locus is hyperbol with S, S' s foci. Centre is mid point of SS'. 4, e 4 e uxiliry circle is concentric circle with rdius '' so x + y ns.(,,c) Let P(t. t ), Q(t. t ) m P m Q t t -4 Let centroid G(h, k) t t. - t + t t + t + so h,k k (t + t ) t t k æh ö ç + 8 è ø (k ). 4 9h x 4(y ) so vertex (0, ), LR 4, xis x 0, tngent t verte Prgrph for Question 80 to 8 4x 4x + 4y 48 (x ) 4(y ) x- 6( y-) æ ö ç è ø 80. Directri Focus æ ö ç,+ P - y LR 6 Q æ ö ç, 8. Let mid point is (h, k) so chord is T S 4hx (x + h) (y + k) h 4h 4k + 49 s it psses through origin it will stisfy (0, 0)

7 so h k 4h 4h 4k 4x x y 0 or 6y x x which is prbol. 8. L L must be in the form of y mx solving it with S 0 4x 4x 4mx x 4x( + 6m) by D 0 6(6m + ) Prgrph for Question 86 to ns. (C) Eution of line is (x - y)(x + y) 0 x+ Ö y0 x Öy0 l l P (h,k) 6m + ±7 m or 4 m + m - Prgrph for Question 8 to 8 (, ) L : x y L 4 0 (, 0) r L 0 D (, ) L 0 x C (4, ) 8. fter I st reflection, eution of L 0 is Slope / nd point (, 0) y 0 (x ) x + y point of incidence on the mirror x 4 0 is (4, ). Eution of L 0 is y + (x 4) x y 7 gin point of incidence on the line mirror y + 0 is (, ) eution of L 4 0 is y + (x ) x + y Closed figure is rhombus re 4 6 s. units 8. Centre of the circle is (, ) nd -- r \ eution of circle is (x ) + (y + ) Let point P is (h, k) æh - k ö æh + k ö + \ ç ç h + 6k 48 x y + 4, b 8 b e 8 4 e 87. ns. () C : bx + y b + b + ( + b) b( + b) Eution of is x cos + y sin...(i) let R º (h, k) is chord of contct to curve C. xh yk eution of is +...(ii) ( + b) b( + b) from (i) & (ii) h cos k sin, ( + b) b( + b) b h + k ( + b) R ( + b) ns. (D) Stndrd property of ellipse. Prgrph for Question 89 to 9 Let the circle be x + y + gx + ƒ y + c 0 If psses through (, 0) \ + g + c 0... (i) R b P

8 Let (t, t) be the point of intersection of two curves \ eution of tngent is Q yt x + t Circle & prbol cuts orthogonly... (ii) \ (ii) psses through the centre of the circle \ ƒt t g ƒt g t (t, t) lso stisfies eution of circle \ t 4 + 4t + gt + 4ƒt + c 0... (iii) Putting the vlue of c & ƒt in bove eution t 4 + 4t + gt + 4g 4t g 0 t 4 + gt + g 0 (t 4 ) + g(t + ) 0 t g suring eution (iii) we get ƒ t (g t ) ƒ ( g) (g ) {Putting the vlue of t } replcing g by x & ƒ by y we get, y ( + x) ( + x) 89. ns. (D), b, c + b + c ns. () y ( + x) ( + x) Consider y + x + x x+ y' / ( + x) ( + x) y ± + x > 0 " x > / + x ò dx -/ + x Put + x t dx tdt 8 Solving we get ns. (C) Prgrph for Question 9 to 94 Let (h, k) be the centre of the circle C k 4h 9... (i) lso P Q Rdius of the circle P Q or h + 8 ±(h 4) h + 8 h + 4 h ( 8,)S ( 8, )P HG -, Putting in (i), k y K J y R(4,) Q(4, ) \ Rdius P \ PR In DPQR, PR PQ + QR QR QR lso co-ordintes of R & S re respectively (4,) & ( 8, ) 9. ns. (C) re of Rectngle PQRS 60 s.units 9. ns. () Now C : (x 4)(x + 8) + (y )(y ) 0 or x + y + 4x 6y 0 centre (, ) & Rdius 6 Eution of tngent through P to C is y + m (x + 8) 6m - + m 6 m 60 Similrly, eution of tngent through Q to C is y + m (x 4) -6m - + m 6 m 60 \ m + m ns. (D) The line through (7, ) t mximum distnce from centre of circle C will be the line

9 perpendiculr to line joining (7, ) & point. - slope of reuired line - -/ ns. () Prgrph for Question 98 to 00 Clerly L will pss through intersection point of &. So its eution will be (x + 4y 0) + l(4x y ) 0 s L psses through origin so l hence eution of L x + 0y 0 x y 99. ns. () s is eully inclined to L & L. Let slope of L is m then 4 4 m - - m - 4 4m 4 + 4m + + 6m 8 + 4m m lso L will pss through point of intersection of &. i.e (, ) so eution of L is y (x - ) y x x y ns. (C) Joint eution of the lines joining & to origin is y 4x æx - 0y ö ç 0 0 0y 44x + 80xy 0 s subtends 90 t origin. So Prgrph for Question 0 to 0 Let eution of stright line through is r cos sin 9 Let L r nd M r, then the coordintes of L nd M re given by L(r cos, r sin) nd m(r cos, r sin). Let N(h, k) be the vrible point such tht N r, then h r cos, k r sin, since L nd M lie on x + 4y 0 nd x + y 0 so r (cos + 4sin) nd r (cos + sin) r r h + 4k nd r r h + k 0. ns. () Now N is the.m. of L nd M so r + r r r r + r r 4 + h + 4k h + k (h + k) + (h + 4k) (h + 4k)(h + k) 6h + 6k + 0hk 4h k 0 hence locus of point N is x + 8y + 0xy 7x y 0 0. ns. (C) r Now r (cos+ sin ) (cos+ 4sin ) cos + sin cos + 4sin cos + sin 0 tn \ eution of L will be x + y 0 0. ns. () x + 4y - x + y - ± Now () + 4() > 0 nd + () > 0 so '+' sign bisector will contin the point (, ) x + 4y x + y - ( )x + ( )y + ( ) 0 Prgrph for Question 07 to 09 Coordintes of, nd C re (,), (, ), C(, ). Now eution of the circles tking, C nd C s dimeter respectively will be S º (x ) (x ) + (y ) (y + ) 0 x + y 6x + 4 0

10 S º (x ) (x + ) + (y ) (y ) 0 x + y 6y S º (x ) (x + ) + (y + ) (y ) 0 x + y 4x 4y ns.(c) Eution of rdicl xis of S nd S will be L º S S 0 x y ns. () Vertex of the prbol is (, ) nd directrix is x + 0 Focus of this prbol will be (, ), it will lie on side C. 09. ns.(c) Eution of rdicl xis of S nd S will be L º 4x y x y clerly rdicl centre will be the point of intersection of L 0 & L 0 i.e., ( 7, 7). Prgrph for Question 0 to 0. ns.() Eution of fmily of circles pssing through & is given by (x )(x 6) + (y 7)(y ) + l(x+y 7) 0 x + y +(l 9)x+(l )y+ 7l0 Eution of the common chord of (i) nd C is (l 9+4)x + (l +6)y + 7l + 0 (x + 6y 6) l(x + y 7) 0 which psses through the point of intersection of x + 6y 60 nd x + y 7 0 i.e. (, /). ns. (C) Centre (,) will lie on the common chord so 8 + 4l 0 l so eution of reuired circle will be x + y x 6y 0. ns. (C) Difference of the sure of the length of the tngents from nd to the circle C is ( ) ( ), P, P 7, P 0 0 Prgrph for Question to. ns. () ' H' S' ' Foci of the ellipse + nd b S(e, 0) nd S'( e, 0) nd its centre is t origin. Now there is nother ellipse whose xis in inclined t n ngle with the xis of ellipse +. b Clerly is the mid point of SS' nd HH'. Thus digonls of the udrilterl H'S' bisects ech other. So its is prllelogrm \ SH S'H' nd ' H'S' Since sum of the focl distnces of point on n ellipse is eul to length of mjor xis so, + ' H'S' + ' (Q H'S') 4. ns. () In D' re hve (') (H) + (S' ) H. S' cos(80 ) (') e + e + e e cos H S (') e + e + ee cos Q ' will be mximum when cos \ (') mx (e + e ). ns. (C) In D we hve S + H S. H cos e + e -(e )(e )cos e + e -ee cos will be minimum when cos \ () min (e e )

11 Prgrph for Question 6 to 8 6. ns. (D) Eution of hyperbol whose symptotes re x 4y nd 4x + y + 0 will be (x 4y + 7)(4x + y + ) + l 0 If it psses though (, ), then l 8 So hyperbol will be (x 4y + 7)(4x + y + ) 8 0 x y 7xy + x + 7y 0 7. ns. () Eution of lines prllel to the lines will be x + y l nd x y µ respectively. If these lines represents symptotes then these lines must pss through the centre of the hyperbol i.e. (, ) so symptotes will be x + y 0 nd x y ns. (D) symptotes of the hyperbol be will 6 æ öæö ç ç - 4 Now ny point on the hyperbol will be (4sec, tn) length of the perpendiculr from this point to the symptotes will be & p (0sec + 0 tn ) p 4 (0 sec - 0 tn ) \ p 4 p Prgrph for Question to 4 Line C will be prllel to ngle bisectors of ngle (, ) so x + 4y - 4x -y - ± (,) C (, ) by +ve sign x 7y ve sign 7x + y So C cn be x 7y l l or 7x + y l l 9 so x 7y or 7x + y 9 Distnce of x 7y from (, ) is nd distnce of 7x + y 9 0 > from (, ) is < so C is 7x + y 9. ns. () DC is right isosceles tringle so re of DC l. l l æ ö ç 0. ns. () Now for '' x + 4y 7x + y 9 x x æ 8 ö ç, 8 y Now for 'C' 4x y 7x + y 9 So æ4 69 ö C ç, - x 4 C l l l x y - Circumcentre will be mid point of C æ7 6 ö ç, - 4. ns. (C) s bisector is such tht point nd C shows opposite sign so it is internl bisector. Prgrph for Question to 7 y 4x Let the centroid is (h, k) nd vertices of D re ( t, t ), ( t, t ), C( t, t) So h ( t t t) k (t + t + t ) (i)... (ii)

12 Let r is the rdius of circle nd t, t, t re roots of eution (t - h) + (t- k) r t 4 + t (4 h) 4kt+ h + k r 0 t + t + t + t 4 0, St t by (ii) t 4 æk ö - ç by (t + t + t + t 4 ) 0 S t -Stt æ ö æ ö + ç - ç è4 ø è ø h 9k 4 -h h + 9k h 4 - h 9k + 4h 9y 4(x 8) here 9 4 so locus is y (x 8). ns. () So LR 6. ns. () Locus of point of intersection of perpendiculr tngents is directrix x 8 4x ns. (D) Chord of contct is T y x x 6, so (6,0) Prgrph for Question 8 to 0 8. ns. () Ellipse x 4 + y Let P(h, k) nd eution of ny tngent is y mx ± 4m + (k mh) 4m + m (h 4) mkh + (k ) 0 + tn( + ) tn m + m -mm tn hk h -4 k - - h - 4 tn xy tn(x y ) x xy cot y D (cot) ¹ 0 H > so hyperbol 9. ns. () m - m tn 4 + mm ( + m m ) (m + m ) 4m m æ k -ö æ hk ö k - ç + ç -4 h -4 h -4 h -4 (x + y ) 4[x y (y )(x 4)] (x + y ) 4[4y + x 4] 0. ns. (D) Let mid point is M(h, k), T S hx + 4ky h + 4k lso let P(x, y ) which lies on director circle. so x + y xx + 4yy 4 4 h k h + 4k (4h) + (4k) (h + 4k ) 6(x + y ) (x + 4y ). ns. () (R), () (R), (C) (Q), (D) (Q) () Length of the LR is 4 4 (, 6) (6, ) (4, 4) \ eution of directrix x + y + l 0 whose distnce from (4, 4) is \ l l± 4-8 l -4 or - x + y 4 or x + y l +l P Q R M

13 () y mx m + 6 which is tngent to the ellipse +, whose foci re 6 (±, 0) \ Product of the perpendiculrs drwn from (±, 0) upon the given line (tngent) is 6 ( b ). (C) co-ordintes of t y 8, x 8 \ (8, 8) fter reflection ry psses through focus (, 0) L : 4x y 8 0 L 0 (, 0) y 8 point of intersection of L 0 with the prbol y 8x is æ ö ç, - è ø fter nd reflection ry moves long the line which is prllel to the xis of the prbol. \ eution of L is y y + 0 \ y , c 4 + c (D) ƒ '(x) 6x 6x 0 (x )(x + ) 0 x, ƒ ''(x) x 6 ƒ ''( ) 8 < 0 x is pt. of mxim ƒ ''() 8 > 0 x is pt. of minim 9 8 / ƒ ( ) ƒ () æ ö ƒç ƒ ( ) \ Mximum vlue on é ù ê-, ú ë û is 8. ns. () (S); () (P); (C) (Q); (D) (R) () Coordintes of point lying t distnce r units from P on line pssing through P nd with slope l will be æ r r ö ç +, + è ø Now this line intersects the curve x + xy 0 so æ(r + ) ö (r + ) ç è ø r + 6r + 0 r + r, r r Now r -r / (s point P lies inside the x + xy 0) 4. ns. () (Q), () (S), (C) (R), (D) (P,Q,R,S) () If x is perfect sure, then Px will be perfect sure only if P is perfect sure, which is not possible s P is prime number. Hence y cnnot be perfect sure. So number of such points will be only one i.e. (0, 0) () H.M. of the lengths of the segments of focl chord is eul to length of semiltus rectum so / \ length of ltus rectum 4 6

14 (C) Eution of the tngent to the prbol y x + 7x + prllel to y x will be y x + l this line touches the prbol so x + 7x + x + l x + 4x + l 0 4C l 0 l hence eution of tngent will be y x the point of contct will be the nerest point to the given stright line i.e. (, 8), b 8 \ b (D) ny tngent to the prbol is ty x + t t the point (t, t) If this is norml to the circle then it will pss through the centre of the circle which is (, ) so t + t t t + 0 t for ny vlue of so the condition stisfies " rel vlue of.. ns. () (S), () (R), (C) (P), (D) (Q) () Eution of common chord 0x + 4y b 0 this chord should pss through the centre of the circle x + y + 6x + 6y b 0 i.e. (, ) 0 b 0 + b 4 + b 4 () Eution of chord of contct drwn from point (, b) will be x + by 0 compring it with given chord x + y, b so + b 0 4 (C) Eution of the fmily of circles pssing through will be (D) x + y x + l (x y) 0 x + y + x (l ) ly 0 centre æ (l-) lö ç -, è ø If is dimeter then centre must lie on, so l (l ) l 4l + 0 l \ reuired circle will be x + y x y 0, b \ + 9b 0 (/, /) P 60 (h,k) Let mid point of the chord is (h, k) In DP cos60 P (h - / ) + (k + / ) (h /) + (k + /) / h + k h + k + 0 \ Reuired locus x + y x + y + 0, b So + b ns. () (Q), () (S), (C) (P), (D) (S) () The given symptotes re perpendiculr so hyperbol will be rectngulr hyperbol so eccentricity is () Eution of norml t ny point P s hyperbol x y - will be b xcos + bycot + b

15 It intersection point with trnsverse nd æ conjugte xes re L + b ö ç sec,0 nd è ø æ + b ö ç 0, tn è ø Locus of mid point of L nd M will be h + b sec, k h sec, + b + b tn bk tn 4h 4bk - ( + b) ( + b) \ Reuired locus x y - æ + b ö æ + b ö ç ç è ø è b ø It eccentricity will be e + b b + Now eccentricity of the given hyperbol is \ \ b e + (C) s the given point lies outside the ellipse so rel tngents cn be drwn. (D) Eution of the symptote of the hyperbol is x - y 0 b b tn0 b nd eccentricity e b ns. () (Q), () (S), (C) (P), (D) (R) () Eution of tngent to the ellipse + prllel to the line x + y 7 6 will y x ± 6+ x+ y ± Clerly x + y is t the shortest distnce from x + y 7 point of contct is (, ) so, b \ + b () Two perpendiculr tngents cn be drwn to the ellipse + from ny 6 9 point lying on the circle x + y i.e. (cos, sin) Eution of chord of contct will be cos + sin cos + 6sin sin + If will be mximum when sin 0 \ mximum distnce \ 6, b \ b (C) Eution of norml to the hyperbol - will be xcos + ycot 4 It hs eul intercepts on positive x nd y so cos - - cot sin, p 6 hence eution of norml is x + y x + y It touches the ellipse so b + x y + b (D) Eution of the hyperbol cn be written s (x )(y ) It is rectngulr hyperbol so b nd b So length of the ltus rectum 4

16 40. ns. 8 The curve is (y + ) 4(x ) eution of the norml to the given curve is 4. ns.00 y Q (7,6) y + m(x ) m m y mx m m which psses through (h, k) - HG 6, 4 K J (,4) P (,4) (,4) (7,4) (7,) R (8,4) x m + m( h) + + k 0 m m m ( + k) m ( + k) Q m m m m m ( + k) + ( + k)( h) + ( + k) 0 ( + k) + h + 0 (y + ) x 4 \ locus of (h, k) re C : (y + ) x 4 k - k, l 4 \ l k 8 x + 4y 0 (8, 6) x + y 6x 8y < 0 (x ) + (y 4) < 0 Point tlest distnce from (, 4) is P(, b) º P(, 4) Points which re gretest distnce from (, 4) re Q(c, d) & R(e, ƒ ) º Q(7, 6) & R(7, ) DPQR is n isosceles tringle & internl bisector of ÐP is y 4 Eution of tngent t origin is x + 4y 0 eution of tngent t æc+ e ö ç +,b º (8, 4) is x 8 re of the right ngled tringle formed by bove three lines is D \ D

PARABOLA EXERCISE 3(B)

PARABOLA EXERCISE 3(B) PARABOLA EXERCISE (B). Find eqution of the tngent nd norml to the prbol y = 6x t the positive end of the ltus rectum. Eqution of prbol y = 6x 4 = 6 = / Positive end of the Ltus rectum is(, ) =, Eqution