fo/u fopjr Hh# u] ugh vjehs de] foifr ns[ NsMs rqjr e/;e eu dj ';ea iq#" flg ldyi dj] lgrs foifr vusd] ^cu^ u NsMs /;s; ds] j?qcj j[s VsdAA jfpr% euo /ez iz.sr ln~xq# Jh j.nsmnlh egj STUDY PACKAGE Subjec : Mhemics Topic : HIGHLIGHTS ON PARABOLA, ELLIPSE & HYPERBOLA Avilble Online : www.mhsbysuhg.com R Suden s Nme : Clss Roll No. : : Address : Plo No. 7, III- Floor, Ner Pidr Sudio, Above Bond Clsses, Zone-, M.P. NAGAR, Bhopl : 0 90 90 7779, 9890 5888, WhsApp 9009 60 559 www.teoclsses.com www.mhsbysuhg.com
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com HIGHLIGHTS ON PARABOLA, ELLIPSE & HYPERBOLA.HIGHLIGHTS OF PARABOLA CHORD WITH A GIVEN MIDDLE POINT :Generl mehod for finding he equion of chord of ny conic wih middle poin (h,. Exmple: Le he prbol be y 8x nd (h, (, we hve o find he equion of AB y m(x...( now y 4x nd y 4x Subrcion y y 4(x x or y x y x bu y y 6 nd 4 8 m 4 y y 8 6 m 4 Hence equion of AB y 4 (x 4x y 0 ] Conversely : To find he mid poin of given chord :Le he equion of he line be 4x y 0 given.to find he mid poin (h, of AB 4 4 4 8 here m y y since 4h 0 4h 9 0 h hence M is (, For prbol in priculr Equion of AB y m (x h...( Bu m AB ( lso ( m AB Hence equion of chord whose mid poin is (h, y (x h...( or 4h y (x h 4h (x h y (x h 4h T S EXAMPLES :Ex-Find he locus of he middle poin of chords of he prbol y 4x which (i psses hrough he focus [Ans. y (x ] (ii re norml o he prbol [Ans. y (y x 4 8 4 0 ] (iiisubend consn ngle α he verex(homogeniseans.(8 y x n α 6 (4x y ] (ivre of given lengh (sy l (v re such h he normls heir exremiies mee on he prbol [Ans. y (x ; Hin : use ] Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com[sol. (ih (...( (...( [ ] h h y (x This could lso be spelled s locus of he middle poin of ll focl chords of ll he pricles y 4x. or Locus of he middle poin of ll he chord of conc of he pir of ngens drwn from ny poin on his direcrix. (ii h (...( ; (...( lso from (h [( ] using (, 4...( 4 8 4 4 4 8 h 4 4 8 4 ( h 4 8 4 0 Ans. ] Ex- A series of chords is drwn so h heir projecions on he srigh line which is inclined n ngle α o he xis re of consn lengh c. Prove h he locus of heir middle poin is he curve (y 4x(y cos α sin α c 0 [Sol. Le nˆ cosαî sin α ĵ PQ v ( î ( ĵ;;;h ( ; ( lsoprojecion of v on nˆ c v nˆ c;;; nˆ ( cosα ( sin α c ;;; ( [ ( cos α sin α ] c [( 4 ] [( cos α sin α ] c ] Ex-Through ech poin of he srigh line x my h is drwn he chord of he prbol y 4x which is biseced he poin. Prove h i lwys ouches prbol(y m 8(x h. [Sol. Equion of vr. chord ABy y i (x my h y i yy i y i x my h y i (y my i (x h 0 (y m 8(x h ] DIAMETER : The locus of he middle poins of sysem of prllel chords of 4 Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com Prbol is clled Dimeer. Equion o he dimeer of prbol is y /m, where m slope of prllel chords. Explnion : Slope of AB ism...( lso ( consn m Hence equion of he dimeer is y m i.e. line prllel o 4 he xis of he prbol. Solving y wih y 4x, we hve, m m ( 4x orx m Hence coordines of Q re, Hence he ngen he exremiy of dimeer of m m prbol is prllel o he sysem of chords i bisecs.since poin of inersecion of he wo ngens re A nd B is, ( or, Hence he ngen he ends of ny m chords of prbol mee on he dimeer which bisecs he chord. Noe:A line segmen from poin P on he prbol nd prllel o he sysem of prllel chords is clled he ordine o he dimeer bisecing he sysem of prllel chords nd he chords re clled is double ordine. IMPORTANT HIGHLIGHTS : If he ngen & norml ny poin P of he prbol inersec he xis T & G hen Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 4 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com ST SG SP where S is he focus. In oher words he ngen nd he norml poin P on he prbol re he bisecors of he ngle beween he focl rdius SP & heperpendiculr from P on he direcrix. From his we conclude h ll rys emning from S will become prllel o he xis of he prbol fer reflecion. Deduce h, if Q is ny poin on he ngen nd QN is he perpendiculr from Q on focl rdius nd QL is he perpendiculr on he direcrix hen QL SN. Noe : Circle circumscribing he ringle formed by ny ngen norml nd x-xis, hs is cenre focus. (b The porion of ngen o prbol cu off beween he direcrix & he curve subends righ ngle he focus. (c (d m m ( ; ( m m ] The ngens he exremiies of focl chord inersec righ ngles on he direcrix, nd hence circle on ny focl chord s dimeer ouches he direcrix. Also circle on ny focl rdii of poin P (, s dimeer ouches he ngen he verex nd inerceps chord of lengh on norml he poin P. Noe : ( For compuing p drw perpendiculr from S (, 0 on ngen P. Any ngen o prbol & he perpendiculr on i from he focus mee on he ngen he verex. i.e. locus of he fee of he perpendiculr drwn from focus upon vrible ngen is he Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 5 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com ngen drwn o he prbol is verex. Explnion : Tngen '' is y x...( Pssing hrough (h, hence h...(a A line hrough (, 0 wih slope y (x...( ( lso psses hrough (h, h h...(b (A (B gives 0 ( h x 0 which is he ngen he verex. (e If he ngens P nd Q mee in T, hen : TP nd TQ subend equl ngles he focus S. ST SP SQ & The ringles SPT nd STQ re similr. (i To prove h α β, i will be sufficien o prove h 'T' lies on he ngle bisecor of he ngle PSQ i.e. perpendiculr disnce of 'T' from he line SP is equl o he perpendiculr of T from SQ. equion of SP y (x ly x ( y b 0 p ( ( ( 4 p α β, Hence proved. Now (ii SP SQ ( ( ( (iii (f Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'. ( lso (ST ( ( [ ( ( Hence (ST SP SQ This is conclusive h produc of he focl rdii of wo poins P nd Q is equl o he squre of he disnce of focus from he poin of inersecion of he ngens drwn P nd Q. ST SQ gin, nd α β SP ST hence he wo ringles SPT nd SQT re similr. Tngens nd Normls he exremiies of he lus recum of prbol y 4x consiue ] Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 6 of 0
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com [Hin: squre, heir poins of inersecion being (, 0 & (, 0. figure is Self explnory] Noe : ( The wo ngens he exremiies of focl chord mee on he foo of he direcrix. ( Figure L NL G is squre of side (g Semi lus recum of he prbol y 4x, is he hrmonic men beween segmens of ny focl chord of he prbol is : [Sol. bc b c i.e. b b ( b c...( b c c c from ( nd ( bc b c...( i.e.. b c ] b c b c (h The circle circumscribing he ringle formed by ny hree ngens o prbol psses hrough he focus. TPT α β Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 7 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com nθ ; n α nθ ; nθ nθ n θ nθ nθ n α n(θ θ...( m ( m n (θ θ ly m n (θ θ n β n( θ n( θ nθ n θ θ n( θ θ n( θ n θ n θ θ θ n β n (θ θ...( from ( nd (, we ge α β hence proved ] (i The orhocenre of ny ringle formed by hree ngens o prbol y 4x lies on he direcrix & hs he co ordines, (. (j Find inersecion of BD nd CE o ge 'O'. The re of he ringle formed by hree poins on prbol is wice he re of he ringle formed by he ngens hese poins. Refer figure of poin (h Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 8 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com A A ( ( ( MORE ABOUT NORMALS If norml drwn o prbol psses hrough poin P(h, hen mh m m i.e. m m( h 0. h Then gives m m m 0 ; m m m m m m ; m m m. where m, m, & m re he slopes of he hree concurren normls. Noe h he lgebric sum of he : slopes of he hree concurren normls is zero. ordines of he hree conorml poins on he prbol is zero. Cenroid of he ringle formed by hree co norml poins lies on he x xis. Exmple : Consider of he ringle ABC is (m m m x (m m m y 0 now x [(m m m m m ] ( h (h cenroid is (h, 0 bu (h > 0 h > Hence bsciss of he poin of concurrency of concurren normls >. EXAMPLES : Ex. Find he locus of poin which is such h ( wo of he normls drwn from i o he prbol re righ ngles, (b he hree normls hrough i cu he xis in poins whose disnces from he verex re in rihmeicl progression. [Ans : ( y (h ; (b 7y (x ] [Ex.7, Pg., Loney] [Sol. ( we hve m m lso m m m Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 9 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ex. m pu m is roo of m ( hm 0 (b y mx m m hence m, m, m m m m m m m m (h (m m m ( m m (h m which is roo of m ( hm 0 ] If he normls hree poins P, Q nd R mee in poin O nd S be he focus, prove h SP SQ SR SO. [Ex.8, Pg., Loney] [Sol. SP ( m ; SQ ( m ; SR ( m SP SQ SR ( m ( m ( m ( m m ( ( m mm mm m m (mm zero m Ex. A circle circumscribing he ringle formed by hree co norml poins psses hrough he verex of he prbol nd is equion is, (x y (h x y 0. [Q., Ex-0, Loney] [Sol. Equion of he norml P y x psses hrough (h, ( h 0...( 0 h, Le he circle hrough PQR is x y gx fy c 0 Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge 0 of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ex.4 Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com solving circle x, y 4 4 g f c 0 4 ( g 4f c 0...( 4 0 bu 0 4 0 circle psses hrough he origin hence he equion of he circle x y gx fy 0 now equion ( becomes ( g 4f 0...( ( nd ( mus hve he sme roo ( g h g (h nd 4f f Hence he equion of he circle is x y (h x y 0 (x y (h x y 0 ] Three normls re drwn o he prbol y 4x cosα from ny poin on he srigh line y b sin α. Prove h he locus of he orhocenre of he ringle formed by he corresponding x y ngen is he ellipse, he ngle α being vrible. b [Sol. y 4Ax where A cos α y x A psses hrough λ,b sin α b sin α λ A A A (A λ b sin α 0 Ex. 0 ; lso h A cos α...( nd A ( bsin α A (0 A b sin α...( from ( nd ( locus is x y ] b bsin α A LEVEL PROBLEMS [Q.5, Ex.0, Loney] Locus of poin P when he normls drwn from i re such h re of he ringle Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'. Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge of 0
FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com formed by heir fee is consn. [Hin: Are of ABC consn m m m m m m C (m m (m m (m m C...( consider (m m m mm m [m (m m m m m (m m [( m m h m m m ] ( m m m (h (h (h m m m x l m n hence equion ( becomes (x l (x m (x n consn x (l m nx (lm mn nlx lmn 0 [Q.6, Ex-0, Loney] (h m m m (h 9 7 mm mm mm (h 4 mm m consn ] Ex. The sides of ringle ouch prbol y 4x nd wo of is ngulr poins lie on noher prbol y 4b(x c wih is xis in he sme direcion, prove h he locus of he hird ngulr poin is noher prbol. [Q., Ex-0, Loney] [Sol. consider he equions bined by puing he coordines of A nd B in y 4b(x c ( 4b( c...( ( 4b( c...( his implies h nd re he roos of he equion, ( 4b( c hving nd s is roos i.e. ( 4b 4bc ( b 4bc 0 Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.
Ge Soluion of These Pcges & Lern by Video Tuorils on www.mhsbysuhg.com FREE Downlod Sudy Pcge from websie: www.teoclsses.com & www.mhsbysuhg.com Ex. Ex.4 b 4bc nd (b ; hence (b h 4bc h 4bc hence locus is y (b (x 4bc ] (b b h 4bc Circles re drwn hrough he verex of he prbol o cu he prbol orhogonlly he oher poin of inersecion. Prove h he locus of he cenres of he circles is he curve, y (y x x x(x 4 [Q.6, Ex.8 (Loney] If he norml P nd Q mee on he prbol, prove h he poin of inersecion of he ngens P nd Q lies eiher on cerin srigh line, which is prllel o he ngen he verex, or on he curve whose equion is y (x 4 0. [Q., Ex.9 (Loney] Ex.5 ( Prove h infinie number of ringles cn be consruced in eiher of he prbols y 4x nd x 4by whose sides ouch he oher prbol. (b Prove h he locus of he cenre of he circle, which psses hrough he verex of prbol nd hrough is inersecions wih norml chord, is he prbol y x. [Q.5, Ex.0 (Loney] Teo Clsses, Mhs : Suhg R. Kriy (S. R. K. Sir, Bhopl Phone : 0 90 90 7779, 0 9890 5888. pge of 0 Successful People Replce he words lie; "wish", "ry" & "should" wih "I Will". Ineffecive People don'.