# EXERCISE - 01 CHECK YOUR GRASP

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1 UNIT # 09 PARABOLA, ELLIPSE & HYPERBOLA PARABOLA EXERCISE - 0 CHECK YOUR GRASP. Hin : Disnce beween direcri nd focus is 5. Given (, be one end of focl chord hen oher end be, lengh of focl chord 6. Focus of prbol y 8 is (, 0. Equion of circle wih cenre (, 0 is ( + y r A AB is common chord Q is mid poin i.e. (, 0 AQ y where y 8 8 r AQ + QS so circle is ( + y 9 Q S O (,0 0. Since QR is focl chord so vere of Q is (, nd R is (, re of PQR 0 0 A A [ ]. Le he poin be (h, Now equion of ngen o he prbol y whose slope is m is y m + m s i psses hrough (h, mh + I hs wo roos m, m B m m h m + 0 m + m h, m.m h m h m h... (i... (ii from (i & (ii (h h Thus locus of poin is y 9 7. Le slope of ngen be m So equion of ngen is 9 h. y m + m Now ngen psses hrough (, so m + m 0 m ± equion of ngens re y (...(i y (...(ii inercep of ngen (i & (ii on line is y & y respecively. Now y y is 6 8. Equion of direcri of prbol will be he required locus.. We now h re of ringle so formed / y 6 /. Equion of ngen o y P(, y is yy ( + yy (i Le (h, be mid poin of chord QR. Then equion of QR is y ( + h b (h + b + y + h 0... (ii Clerly (i nd (ii represens sme line. y h y nd h y h h h mid poin of QR is (, y

2 5. Le P( y be poin of conc of wo prbol. Tngens P of he wo prbols re yy ( + nd (y + y yy (... (i nd y (y clerly (i nd (ii represen sme line y y Hence locus of P is y... (ii EXERCISE - 0 BRAIN TEASERS. Le poin P(, on y equion of line joining P & vere y... ( equion of line which is perpendiculr ngen P & pssing S(, 0 is y +... ( from ( & ( elimining we ge he locus of R y +. Le P (, Relion beween & equion of line PR y ( Pu y 0 nd R (( +, 0 R (( +, 0 Lengh of PS ( + So AR is wice of PS. A S(,0, we ge P(, R Q(, 6. Since line pssing hrough focus so Poin of inersecion of ngen P & Q re (, ( + S (,0 P(, Q(, Poin of inersecion of norml P & Q re (( + + +, ( + (, y (, ( + (, y (( +, ( + y y 9. The curve y is he pr of curve y equion of norml P, y ( Since line cu he curve orhogonlly so equion ( will psses (, solving we ge so equion of line which psses (, 6 is y Equion of ngen nd norml P (, on y re y +... ( y ( So T(, 0 & G( +, 0 equion of circle pssing P, T & G is ( + ( ( + + (y 0 (y y ( + 0 equion of ngen on he bove circle P(, is + y (+ ( + 0 slope of line which is ngen o circle P m ( slope of ngen P, m n ( n sin n. Le B (, ; C (, A (, ( + equion of ngen of y (, y +... ( p ( P ( (

3 p ( S : + y y 0 equion of OR will be S S 0... (iii. p ( p p p. Hence p, p, p in G.P. P(, R (0,0 ( + ( y 0 y n from (i co + co n b b. P (. y Q(, P Le P(, & Q(, (/,b (,b P so n & n co + co +... (i (, b/ equion of circle wih (0, 0 & (, s end poins of dimeer is ( + y(y so S : + y y 0 similrly oher circle is... (ii + yb + b 0 Similrly P y by + + b 0 Common chord is P P 0 y 0 ( + y( y 0 slope will be, EXERCISE - 0 MISCELLANEOUS TYPE QUESTIONS Mch he column :. ( A Equion of norml (, on y y + + using homogenizion (y y ( for ming 90, coeff. + coeff. y 0 0 ( B Poin on y whose prmeer re,, (,, (,, (6, 8 Are ( C Equion of norml is y m m m since i psses hrough,. so we ge m m + 0. Vlue of m re, /, /, so normls cn be drwn. ( D Equion of norml (, o y y (i If i gin mee he curve gin (, hen so, &. ( A Required re / S ( B ( + (y / ( y 6 5 y 6 ( (y 5 focus, is (, & direcri is + y 6 0 disnce beween focus nd direcri is Lengh of Lus Recum 5

4 5 ( C y + is focus (0, Le poin on y + is (, + ( ( 6 ( + 0 ± 6 6 so poin re (± 6, & + b ( D (y ( + vere is (, so equion is (y ( + Y X Le poin on Y X is, From fig. n 0 so poin on prbol is (6,. y 0 0 ( /, Bu when vere chnge, disnce (or lengh of side of equilerl ringle remin sme lengh of side (6 (. Asserion & Reson :. Le P (, & Q, on P (, & Q, y equion of P P : ( + y +...(i equion of Q Q ( + y +...(ii dd (i & (ii which is direcri of y Locus of poin of inersecion of ngen is direcri. In cse of prbol direcor circle is direcri Comprehension # Ais of prbol is bisecor of prllel chord A B & CD re prllel chord. so is C(, A (0, equion of prbol is ( y + b I pssing (0, & (, (,D (,B so + b...( + b...( from ( & (. Vere (,. 8 & b ( (y direcri of y is y y 8 y 8 y + 0. Le prmeric poin on y re A(, B(,C( nd D( So + + Equion of circle pssing hrough OAB is + y + g + fy + c 0 fourh poin M( 5 puing he vlue (, in circle we ge four degree equion. In his equion Similrly circle pssing hrough OCD & fourh poin N( 6 we hve I men boh poin M nd N re sme so common poin (, (,

5 EXERCISE - 0[A] CONCEPTUAL SUBJECTIVE EXERCISE. Prbol y P( (, & Q( (, Given K equion of chord PQ ( + y + so ( + y + y [L L Type] so y 0 & fied poin (, 0 8. y... ( Le equion of OP y m... ( 9. equion of OQ y... ( m from ( & ( we ge P(m, m from ( & ( we ge Q, m m equion of PR y m m ( m y + m m +...( equion of QR is y m m( + m Q O R(h, 90 y m +...(5 m Locus of R solving ( & (5 & elimining m we ge y, poin (sin, cos no lie ou side y + 0 cos + sin 0 sin + sin 0 sin ( sin 0 sin 0 or sin, or 5, 6. y m + c ouch y 8 ( + (m + c 8 ( + m + (mc 8 + c (i line ouch he prbol so D 0 of equion (i (mc m (c 6 0 P m c 8mc + 6 m c + 6m 0 m mc + 0 Since m is rel D 0 c 6 0 c (, ] [,. Le poin on y be P(, equion of ngen of P y +... ( I inersec he direcri... ( poin of inersecion of ( & ( is A(, ( Le mid poin of PA is (h, h... ( + (... ( from ( & ( elimining & replce h & y we ge y ( + ( + 7. Le poin on prbol y is P(, Given ± ing posiive P(, equion of ngen P is y + If inersec -is T hen T(, 0 Norml (, mee gin prbol Q(, (using Q(9, 6 Now P(,, T(, 0, Q(9, 6 PT ( ( 80 PQ ( 9 ( 6 5 PT PQ Le poin be (h, Equion of norml (m, m y + m m + m mh m m m + m( h ( Is slope is m, m & m m.m.m m Pu in ( [Given m m ] y (

6 . Equion of norml (m, m on y y + m m + m... ( I cus is y 0 i.e. ( + m, 0 Le middle poin (h, h m + + m h m + & m... ( from ( & ( h + Locus y ( vere (, 0 L.R.. Le P(, & Q(, on y co-ordine of T(, ( + which is poin of inersecion of ngen P & Q equion of PQ which is norml P y ( equion of PQ is ( + y +... ( equion ( & ( re sme Compre slope + Now mid poin of TP ( ( which is direcri Hence TP bisec he direcri. Norml P(m, m on y y + m m + m... ( G ( + m, 0 Equion of QG is + m Solving wih prbol we ge y ± m QG PG ( + m (m m (m which is consn P (m,m G T Q (+m, +m P Q EXERCISE - 0 [B] BRAIN STORMING SUBJECTIVE EXERCISE. Le prbol y Le P(,, & Q(, OP & OQ re perpendiculr. P(, 90 O R(h, 90 (0,0 Q(, Now digonls of recngle bisec ech oher h h ( +...( ( +...( + + h 8 Required locus is y ( 8. Equion of ngen of y in slope form (, y is y m + m... ( equion of norml (b, b on by + y b + b I psses hrough (, y + y b + b... ( ( & ( re sme equion so compre m m(b b m m (b + b m (b + b... ( Pu m b + b in equion ( b + + b 0 will be rel > 8b

7 . Le prbol y poin on prbol P(, & Q(, Poin of inersecion of ngen P & Q is T (, ( + T (,( + P(, Q(, Norml P & Q mee gin in he prbol so relion beween equion of line perpendiculr o TP & pssing hrough mid poin of TP is y ( + ( ( +... ( y + ( + + ( + similr equion of pssing mid poin of TQ nd o TQ y + ( + + ( +... ( from ( & ( & using Elimining & we ge he locus of circumcenre y ( 6. Le P(, & Q(, on y equion of chord of PQ ( + y +... ( Poin on -is is K(, 0 PK ( + + (( + + QK (( + + PK + QK (( + (( (( (( PK + QK ( so fied poin K (, 0 9. Le prbol is y will be independen of K equion of norml (m, m y + m m + m i psses hrough (h, m + m( h 0 is roo re m m & m m 0, m m m m m le equion of circle be h + y + g + fy + C 0 I psses (m, m m + m + gm + fm + C 0 m + m ( + g + fm + C 0 is roos m, m, m & m m + m + m + m 0, m +m + m 0 m 0 circle psses (0, 0 m m +m m + m m + m m + m m + m m g h g h + g g h f m m m + m m m + m m m + m m m f f equion of circle + y (h + + y 0

8 EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS. I is fundmenl heorem.. Given prbols re y y... (i... (ii Puing he vlue of y from (ii in (i, we ge 6 5. y ( 6 0 0, from (ii, y 0,. Le A (0, 0; B (, Since, given line b + cy + d 0 psses hrough A nd B, d 0 nd 8b + c 0 b + c 0. ( 0 Obviously, d + (b + c 0 + y y 6 5 y + 6 Verices will be (, So h or nd Required locus will be y Poin mus be on he direcri of he prbol Hence he poin is (, 0 8. Locus of poin of inersecion of perpendiculr ngen is direcri of he prbol. so 9. ngen of slope m of y is y m 5 m 5 y lso ngen o 5 / 5 / m m m + m m + m 0 m which sisfy m m + 0 which gives y 5 s ngen So I & II boh re rue.

9 EXERCISE - 05 [B]. ( The prbol is y. 8 Puing y Y, 8 X, he equion Y..X JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS. 0,. + or + +, (b The direcri is X + 0, i.e Bu 0 is he direcri. So, 8 8, Any norml is y I is idenicl wih + y if 6 nd 6 9 Alier : y + c [m + m ] c [6( + ( ] c ±9. ( Any ngen o y is y m + m. O Y (,0 X I ouches he circle, if or 9( + m m m or m, m ± m m m For he common ngen o be bove he -is, m Common ngen is, y + y + The locus is y b( b, b Direcri is ( b + b 0 or 0. The given curves re y 8... ( nd y... ( If m is he slope of ngen o (, hen equion of ngen is y m + /m. If his ngen is lso ngen o (, hen m m m + m + 0 m + + m 0 We should ge repeed roos for his equion (condiions of ngency D 0 ( m. m 0 m m Hence required ngen is y Le P be he poin (h,. Then equion of norml o prbol y from poin (h,, if m is he slope of norml, is y m m m 0 As i psses hrough (h,, herefore mh m m 0 or m + ( h m ( which is cubic in m, giving hree vlues of m sy m, m nd m. Then m m m (from equion bu given h m m We ge m Bu m mus sisfy equion ( + ( h + 0 h 0 Locus of P(h, is y + ( Bu ATQ, locus of P is pr of prbol y, herefore compring he wo, we ge nd 0

10 8. The given equion of prbol is y y ( (y ( Any prmeric poin on his prbol is P ( +, + Differeniing ( w.r.., we ge y dy d dy d y dy d 0 Slope of ngen o ( p. P( +, + is m Equion of ngen P( +, + is y ( + ( y y + ( ( Now direcion of given prbol is ( 0 Tngen o ( mees direcri Q 0, Le p. R be (h, ATQ R divides he line joining QP in he rio : i.e. : eernlly. (h, ( 0, h ( + nd h nd Elimining we ge h ( ( h (h ( + 0 locus of R(h, is, ( (y The given curve is y + 6 Equion of ngen (, 7 is Q +50 C ( 8, 6 (y y ( ATQ his ngen ( ouches he circle + y y + C 0 Q. (cenre of circle ( 8, 6. Then equion of CQ which is perpendiculr o ( nd psses hrough ( 8, 6 is y + 6 ( y ( Now Q is p. of inersecion of ( nd ( i.e. 6, y 7 Req. p. is ( 6, 7.. Wihou loss of generliy we cn ssume he squre ABCD wih is verices A(,, B(,, C(,, D(, P o be he poin (0, nd Q s (, 0 Then, PA PB PC PD QA QB QC QD 5 5 [( ] (( Le C' be he sid circle ouching C nd L, so h C nd C' re on he sme side of L. Le us drw line T prllel o L disnce equl o he rdius of circle C, on opposie side of L. Then for N, cenre of circle C', MN NO N is equidisn from line nd poin locus of N is prbol. 5. Since S is equidisn from A nd line BD, i rces prbol. Clerly AC is he is, A(, is he 6. focus nd T, AT. is he vere of prbol, T T lus recum of prbol Are (T T T QP ST ArPQS ST ArPQR TR 8 PQ TR sq. unis.

11 7. For PRS, Ar(PRS SR PT 0 0, PS, b PR 6, c SR 0 y. + r r A( r B( +, + rdius of circumference similrly r is R bc Rdius of incircle re of PQR semi perimeer of PQR s. lso possible (/, P (, A y 8 We hve PR 6, b QP PR 6 c PQ nd PQ TR 6 s 6 6 r C : y C : + y ( 0 y ± 8 (,0 so he curves ouches ech oher wo poins (, & (,. h + h + 9 T (,0 y 9 / ( P(,,0 N(+,0 (,0 A' (, re of PAA'.8. 6 ( (Usi ng proper y : Are of ringle formed by ngens is lwys hlf of originl ringle. Le P be (h, y on using secion formul P, h nd y h nd y (, y lies on y 6 6h h Locus of poin P is y. 5. Equion of norml is y m m m I psses hrough he poin (9, 6 hen 6 9m m m m 7m (m (m (m + 0 m,, Equions of normls re y + 0, y + 0 & y + 0

12 6. Focus of prbol S(,0 poins of inersecion of given curves : (0,0 nd (,. y Q (, (0,0P S(,0 8. Lengh of focl chord P Q ( 9. Le F(, 8 [( + ] [ + ] 5 where y Are (PSQ.. sq. unis (0, E (0, G F(,8 Prgrph for Quesion 7 nd 8 7. Single ngen he erimiies of focl ( + (, 90 M(, ( + O P(, S(,0 Q(, chord will inersec on direcri. M(, ( + lies on y + ( ( n & ( ( EFG ( (6 8 d 0 d + 0 / 8 m 0 (EFG m 6 y 0 8 & y 0 minim mim n 5 bu is obuse becuse O is he inerior poin of he circle for which PQ is dimeer. n 5

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