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Index 1. Theory 2. Short Revision 3. Exercise 4. Assertion & Reason 5. Que. from Compt. Exams 6. 39 Yrs. Que. from IIT-JEE(Advanced) 7. 15 Yrs. Que. from AIEEE (JEE Main) Student’s Name :______Class :______Roll No. :______

Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal : 0 903 903 7779, 98930 58881, WhatsApp 9009 260 559 www.TekoClasses.com www.MathsBySuhag.com FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com (iii) • The constant ratio is called the the called is ratio constant The the called is line straight fixed The the is called point The fixed • • • (iv) (i) Get Solution of These Packages & Learn by Video Tut (ii) .Scin f ih crua cn b dfeet plane different by cone circular right of Section 2.2.2. Sections: Conic 1.1.1. • A point of intersection of a conic with its axis i axis its with conic a of intersection of Apoint & focus the through passing line The • • 2. 1. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful iue- -5 Figure or parallel to the axis of the cone is an ellipse o ellipse an is cone the of axis the to parallel or cone is a parabola as shown in the in as shown parabola a is paralle cone plane a by cone circular right a of Section Section Section of a right cone circular by a plane neither cone circular a right of Section shown in thein shown as the vertex through passing lines straight of is a pair vertex its through passing a by plane the in shown as is cone circular right A d perpendicular its to ratio a constant in is point is or aof the locus w conic point A section, conic parallel to its base is a as shown in the the in shown as circle a is base its to parallel a by plane cone circular a right of Section oi Sections: Sections: Conic Conic eto o rgt iclr oe y ifrn planes planes different different by by cone cone circular circular right right of of Section Section oi Sections: Conic eto o rgt iclr oe y ifrn planes different by cone circular right of Section Parabola Focus. Eccentricity Directrix. denoted by e. denoted try" & "should" try" with "I& "should" Will". Ineffective People figure figure s called a a called s parallel parallel to any generator of the cone nor perpendi hich moves in a plane so that its distance from a f in a from so plane its distance moves hich that istance from a fixed straight line. straight a fixed from istance r hyperbola as shown in the the in as shown hyperbola r l to a generator of the of generator a to l to the directrix is called the the called is directrix to the −−− − 3 Figure -6Figure orials on www.MathsBySuhag.com . Vertex . figure figure s Axis −−− − . 5 & 65 & don't. . cular ixed

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 2 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com .Dfnto ad Terminology and Definition 5.5.5. PARABOLA Get Solution of These Packages & Learn by Video Tut .Dsigihn vros ois : conics prop various directrix Focal conic: Distinguishing a of equation 4.4.4. General 3.3.3. Solution. :Examples 5. 4. 3. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful h² = ab h² < ab h² > ab h² > ab; a + b = 0 = b + a ab; > h² ab > h² ab < h² ab = h² For parabola y parabola For = y² 4ax; = y² are the of parabola forms standard Four (directrix). line straight a fixed from distance to is perpendicula equal (focus) point a fixed from distance whose a point, of locus the is parabola A depends The section the conic upon of nature the po (l & q) (p, focus with a conic of equation general The : View 3D ⇒ ⇒ ⇒ In this case case this In Directrix. The On Lies Focus The When (I) Case case e. Two different eccentricity the of value the eiiin n Terminology Terminology and and Definition Definition itnusig aiu cnc : : conics conics various various Distinguishing Distinguishing (i) a parabola an ellipse a hyperbola rectangular hy rectangular a hyperbola ellipse an 1; = e a parabola Directrix. On Lie Not Does Focus The When (II) Case e < 1 1 < e eea euto o a oi: oa drcrx proper directrix Focal conic: a of equation General e = 1 1 = e e > 1 1 > e if: lines straight (iii) Focal Distance:Focal eea euto o a oi: oa drcrx proper directrix Focal conic: a of equation General Focal Chord : Chord Focal Double Ordinate: Double Latus Rectum:Latus (ii) (i) NOTE : For y² = 4ax.= y² For (L.R.). Rectum the Latus called is parabola of (iii) e Px y b ay on o te aaoa hs focu whose parabola the on point any directrix to be perpendicular PM Draw 0. = 3 + – 2y y) x at P(x, is Let focus whose parabola the of 0. = 3 + 2y – x equation the Find ⇒ eiiin n Terminology and Definition itnusig aiu cnc : conics various Distinguishing eea euto o a oi: oa drcrx proper directrix Focal conic: a of equation General 2 + m + ≡ ≡ ≡ 2 ∆ h h h ) [(x [(x ) 5(x 1) + [(x 5 1) + (x Vertex is (0, 0) (0, is Vertex Axis is y = 0 = y is Axis Vertex is middle point of the focus & the point of of point & the focus the of point middle is Vertex Perpendicular distance from focus on directrix = ha on directrix focus from distance Perpendicular Two are said to be equal if they have the have they if to be equal said are Two parabolas SP PM = SP

≠ 2 2 2 < ab the lines will be imaginary. will lines the ab < intersecting > distinct & real be will lines the ab > 0, 0 < e < 1; 1; < e < 0 0, 2 2 = PM = coincident. will lines the ab − + y + ∆ p)

≡ 2 2 A chord of the parabola, which passes through th through passes which parabola, the of chord A − + (y + 2) + (y + abc + 2fgh 2fgh + abc = 4ax: the passing A through focus or ordinate double a 2 2 4ax; x² = 4ay; x² = = x² 4ay; = x² 4ax; + 2x + 4y + 5) = (x = 5) + 4y + 2x + + (y (y + The distance of a point on the parabola from the the f from on parabola the a of point distance The 2 2 + (y + 2) + (y + A chord of the parabola perpendicular to the axis axis the to perpendicular parabola the Aof chord − q) ⇒ ⇒ 2 = = 2 ] = e = ] 2 − ∆ ] = (x – 2y + 3) + 2y – (x = ]     af

x ≠ 2 2 0; e > 1; 1; > e 0; − ends of the latus rectum are L(a, 2a) & & L’ 2a) (a, L(a, are rectum latus the of ends = 4a. rectum latus the of Length (lx + my + n) + my + (lx − 1 2 bg + y 4 + 2 − 2 + 4y + 3 − 4ay ch     (ii) (iv) 2 2 2 = 0 & the general equation of a conic represents a represents a conic of equation general 0 = the & + 9 – 4xy + 6x – 12y) – 6x + 4xy – 9 + 2 2 try" & "should" try" with "I& "should" Will". Ineffective People ≡ focus is (a, 0) (a, is focus Directrix is x + a = 0 = a + x is Directrix ax ∆ s s arise. r

≠ 2 x – 2y + 3 = 0. Then by definition, by Then 0. 3 = + 2y – x directrix lx + my + n = 0 is: n = + my + lx directrix sition of the focus S w.r.t. the directrix the & S of focus sition also w.r.t.u the directrix 0; e > 1; 1; > e 0; + 2hxy + by + 2hxy + orials on www.MathsBySuhag.com intersection of directrix & & axis. directrix of intersection at S. at same latus rectum.latus same lf the latus rectum. the latus lf – , 2 ad h drcrx h line the directrix the and 2) – 1, (– i S– , 2 ad h directrix the and 2) – 1, S(– is s 2 focal chord perpendicular to the axischord perpendicular focal + 2gx + 2fy + c = 0 = c + 2fy + 2gx + e e focus. ocus. of the symmetry.the of ∆

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Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 3 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Example : Example 5. 3. Get Solution of These Packages & Learn by Video Tut .Prmti Representation: Parametric 6.6.6. 2. Solution. 1. Problems Practice Self Example : Example 4. Solution. 6. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful i.e. the equations x = at² & y = 2at together repre together 2at = y & at² = x equations the i.e. Ans. and latusrectum. (1 is whose focus parabola the of equation the Find Ans. Find the latus rectum & equation of parabola whose Y ⇒ ⇒ The simplest & the best form of representing the co the representing of form best the & simplest The aaerc Representation: Representation: Parametric Parametric Find Find the extremities of latus rectum of the parabol Ans. sketches.roguht So, the equation of the directrix is directrix the of equation the So, Latusrectum -Latusrectum -Directrix 0). = Y 3/4, – = (X are focus the of coordinates the So, i.e. Focus- is axis the of equation the So, Axis: 4y is The equation given This is of the form Y form the of is This Find the equation of the parabola whose th is focus the parabola of the equation Find 3x – 4y + 2 = 0. = 2 + 4y – 3x This is the equation of the required parabola. the required of is equation the This Find the vertex, axis, focus, directrix, latusrectu directrix, focus, axis, vertex, the Find 4y Ans. Using these relations, equation (i) reduces to reduces (i) equation relations, these Using So, the coordinates of the vertex are the vertex of coordinates the So, - Vertex Find the vertex, axis, focus, directrix, latusrectu directrix, focus, axis, vertex, the Find Let x = X – – X = x Let ⇒ aaerc Representation: Parametric 2 – = 3X 2 2 + 12x – 20y + 67 = 0 = 67 + 20y – 12x + 0 = 67 + 20y – 12x +    2 1 Find Find the parametric equation of the parabola (x – (2x – y – 3) – y – (2x 4x x = – – = x (–17/4, 5/2) [Putting X = 3/4 in (ii)]in 3/4 X = [Putting 5/2) (–17/4, Y = = –a, (X 0)are focus the of coordinates The = y Y is = 0. parabola the of axis the of equation The x – 1= 2atx ∵ y       4 , The coordinates of the vertex are (X = 0, = 0, Y (X = are 0) vertex the of coordinates The 2 − y 9 4 – 5y = – 3x – – 3x – = 5y – 2 − 2 + y + 7 2    5 2

, 5 2 , x ,    5 2 11    3 4 2 2 2 2    + 4xy + 4x + 32y + 16 = 0 = 16 + 32y + 4x + 4xy + The length of the latusrectum of the given parabol given the of latusrectum the of length The X = a X is i.e. = directrix the of equation The 4a = – 12 + y + , 7 2 = – 3x – – 3x – = 9 4 , y = Y + + Y = y ,    2 2 = – 20 (x + 2y – 4), Axis 2x – y – 3 = 0. LL 0. = 3 – y – 2x Axis 4), – 2y + (x 20 – = – 2xy + 8x + 8y = 0 = 8y + 8x + –2xy 2 = – 4aX. On comparing, we get 4a = 3 = 4a we get comparing, On – 4aX. = ⇒ 67 4 42 4 5 2 x = 1 – 6t, y = 2–3t = y 1–6t, = x [Putting X = 3/4 in (ii)]in 3/4 X = [Putting (ii)] 0 in = Y [Putting (ii)] 0 in = Y 0, = X [Putting Ans. ⇒ ⇒ ⇒ ⇒ 16x a = 3, y – 2 = at–2 y = 3, a= 2 + 9y + try" & "should" try" with "I& "should" Will". Ineffective People y y    m of the parabola y the parabola m of a y = x m of the parabola, also draw their rough sketches. rough their draw also parabola, the of m 2 2 y 2 sents the parabola y² = 4ax, t being the parameter. the being t 4ax, = y² parabola the sents – 5y + + 5y – + 5y – 3x + + 24xy – 12x + 16y – 4 = 0 = 4 – 16y + 12x – 24xy + vertex vertex is origin & directrix is x + y = 2. − e point (0, 0)and whose directrix is the straight l is straight the 0)and whose (0, directrix e point 2 , – 1) and whose vertex is (2, 1). Also find its ax its find Also 1). (2, is – whose vertex and 1) , − 5 2 ordinates of a point on the parabola is (at², 2at) (at², is parabola the on a point of ordinates    2 2 – 2x + 3. ....(iii) ....(ii) 1) orials on www.MathsBySuhag.com = – 3 – = 2    2 = –12 (y – 2) 5 2    ⇒ 2 3 4    = – 3x – – 3x – = 67 a 3/4.= . 4 x ′ = 4 = + = 0 = a a is 4a = 3. 7 2 2 – 8y – + x 19 their= draw Also 0.    5 . 67 4 + + ....(i)    5 2    2 don't. ine is Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 4 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Example : Example 4. 3. 2. 1. Problems Practice Self Solution. :Example 1. Problems Practice Self Get Solution of These Packages & Learn by Video Tut Example : Example .Ln & Parabola: a & Line 8.8.8. Parabola: a to Relative point a of Position 7.7.7. Solution. Example : Example Example : Example 1. Problems Practice Self Solution. Solution. Solution. 8. 7. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful through a fixed point. Find the point? Find Find the parametric equation of the parabola x ∴ since P, S & Q are collinear t + (1 a = PS ∵ 3. zero or negative. is positive, (x point The If PSQ is focal chord of parabola y parabola of chord focal is PSQ If Ans. y parabola of chord focal of end one If Ans. Find the of midpoint the chord x + y = 2 of the par Ans. + 3x = y line the If = y Equation of chord joining (at theIf endpoint t ⇒ ⇒ ⇒ ie a Parabola: Parabola: a a & & Line Line Parabola: Parabola: a a to to Relative Relative point point a a of of Position Position Length of the chord intercepted by the parabola on on parabola the by intercepted chord the of Length ∴ so y = x + 1 is to the parabola. the to tangent is 1 + x = y so Prove that focal distance of a point P(at point a of distance focal that Prove If tIf 2. Find the set of value's of of value's of set the Find OE:1. NOTE : Let parabola is y (t Ans. y y 2at– coincident or imaginary according as a according imaginary or coincident ie a Parabola: a & Line Parabola: a to Relative point a of Position 1 This line passes through a fixed point (– ak, 0). + t + 1 , , t t 1 2 2 are end points of a focal chord then show that t ) ) y – 2at + 2 1 = Discuss the position of line y = x + 1 with respect with 1 + x = y line of position the Discuss m at + a = PM = PS Check weather the point (3, 4) lies inside or outsi or inside lies 4) (3, point the weather Check Solving we get (x + 1) + (x we get Solving (1, – 4)(1, (4, – 2) (– t t ∴ ∵ If t If co a a t PS Length of the focal chord making an angle an angle making chord focal the of Length 1 1 t 2 1 t 2 1 PQ ∈ 2 The equation of a chord joining t joining a chord of equation The – 1 = t = 1– − t ∞ = 1= – + ( k ( ak) + (x 2 1 1 (– 4 – 2 – 4 (– & t & ordinates at the extremities of a focal chord can b can chord a focal of extremities the at ordinates = m 1 + + , , 1/3) t + 2 2 y 2 ). 1 t 1 2 = = ) lies outside, on or inside the parabola y² = 4ax y² = 4ax parabola the on or inside outside, ) lies 2 1 2at– SQ 2 , , t (3, 4) lies outside the parabola.the outside lies 4) (3, S y PS 2 2    are the ends of a focal chord of the parabola y² = y² parabola the of chord focal a of ends the are 1 = 4ax 1 (x – at(x 2 1 2 2 m – 4x = 0 = 4x –

of a chord satisfy the relation t t + t 4 λ λ ≡ ≡ 1 2 = = 2 t2 intersect the parabola y parabola the intersect − y 1 1    t 1 3 1 2 1 t 2 a = 2x –= 2x 2at 1 2 – 4x – 1 2 + 4 – , 2 . ) c m a m a α α ) ( ) ( 1 1 for which ( which for 2 1 , , 2at − + = 16 – 12 = 4 > 0 > 4 = – 12 16 = 2 = 4x 4x = The line y = mx + c meets the parabola y² = 4ax in in 4ax = y² parabola the meets c + mx = y line The ∵ 1 1 2 ) ) and (at 2 3 2 = 4ax (a > 0), where S is focus then prove that prove then S focus is where 0), > (a = 4ax )

⇒ 2 α = 16x is (16, 16) then coordinate of other end is. end other of coordinate then 16) (16, is = 16x , – , 2 – t < > 2 1 t , 2at) on parabola y parabola on , 2at) 2 2 c 2 = k) = , , 2at

1 m m & t& 2 2 = 4x at two distinct point's then set of value's o value's of set then point's distinct two at = 4x = 4ay try" & "should" try" with "I& "should" Will". Ineffective People

α 2 . ⇒ ) ) is abola y (x 1)– (x 2 1 ) lies inside the parabola y parabola the inside lies ) is 2x 2x is t condition of tangency is, c = a/m. = c is, tangency of condition 2 = k (const.) then prove that the chord always pass 1 α the line y = m = y line the t 2 with the xthe with = –1. 2 − 2 = 0 = = 4x. Ans. (t orials on www.MathsBySuhag.com de the paabola y paabola de the 1 to parabolas y parabolas to + t+ 2 2 = 4ax (a > 0) is a(1 + + ta(1 is 0) > (a 4ax = e taken as (at² e taken ) y + 2 2 at + y ) according as the expression y as the expression according x = 2at, y = at = y 2at, = x − axis is 4acosec² 4acosec² is axis

x + is:+ c x 4ax then t then 4ax 1 t 2 2 = 0.= = 4x. = 2 2 + 4x 0.= 4x + 4x.= , 2at) & & 2at) 2 . 1 t α. 2 = = 2    ). two points real, points two t − a 2 1. Hence the Hence 1. , don't. − 2 1 f 'f ² t a − λ    4ax ' ' is es 1 Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 5 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Solution. : Example Solution. : Example Solution. : Example Solution. : Example 5. Get Solution of These Packages & Learn by Video Tut .Tnet t te aaoa ² 4ax: = y² Parabola the to 9.9.9. 9. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful but the given tangent is y = mx + mx = y is tangent given the but its equation and its point of contact. of point its and equation its Equation of tangent to parabola y parabola to tangent of Equation y paabola the to tangents the to equation the Find ∵ :NOTE (iii) (i) Find the length of focal chord whose one end point ⇒ y to tangent of Equation pa the of tangents common the to equations the Find ∴ ∴ 10) (4, through passes it Since agns o h Prbl y = 4ax: 4ax: = = y² y² Parabola Parabola the the to to Tangents Tangents ∴ for common tangent, (i) & (ii) must represent same same represent must & (ii) (i) tangent, common for Equation of tangent to x to tangent of Equation y = mx + + mx = y y = m(x + a) + + a) + m(x = y agns o h Prbl y = 4ax: = y² Parabola the to Tangents equation of tangent’s are y = = y are tangent’s of equation + 4m = 10 y = mx + + mx = y tangent’s y = – 2x – 1 at at 1 – 2x – = y tangent’s Equation of tangent of slope m to the parabola y parabola the to m slope of tangent of Equation m = m are tangent’s required of Slope y parabola the to A tangent y parabola the to ‘m’ slope of tangent of Equation the touches c + mx = y line straight the that Prove t y y = = y y = mx + + mx = y x = m = x y = = y m = = m Point of intersection of the tangents at the point the at tangents the of intersection of Point

y = x + a + x = y

y 1 4 =–2 m 2, – = 1 9 = 2 = m m 1 3 1 4 1 1 1 ∓ y + + y ± , ,

a (x + x + (x a x – – x m 3 a 1 2 1 9 4

m m t² at (at², 2at). (at², at t² x + 4 at (8, 8) (8, at 4 + x a a m 4 b ( 9 m 1 m . b 1 1 ) ⇒ ) at the point (x point the at ) 2 2 2 = 4ax is 4ax = = 4by is 4by = 2 = = 2 1 y = mx + a a + mx = y ⇒

   2 c = 9x is9x = 2 1 2 = 8x makes an angel of 45° with the straight line line 45°of straight an angel the makes with = 8x , − 2 16 m 16 ...... (ii) ...... (i) 1 , y ,    1 ) ;)    4 x 2 m – 40 m + 9 = 0 = 9 + m 40 – try" & "should" try" with "I& "should" Will". Ineffective People + 9 & y = = y & 9 + + ∴ is ‘t’. m 1 2 = 9x which goes through the point (4, 10). (4, point the through goes which 9x =    line. rabolas yrabolas (ii) c = am + + am = c orials on www.MathsBySuhag.com 2 t = 4ax is 4ax = 2 = 4a(x + a) isa) + 4a(x = 1 parabola y parabola & t & 2 2 y = mx + mx = y = 4ax and xand 4ax = is [ at [ is m a 9 4 1 x + 1. + x 2 t = 4a (x + a) if c = ma + + ma = c if a) + (x 4a = [Ans. 2, a(t m a 1 (m (m 2 = 4by.= + t+ a    2 t ≠ ) ].) + 0) at at 0) 1 t    2 y = 3x + 5. Find+ 5. y = 3x ]    m a don't. 2 , 2 m a m a   

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 6 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Example :Example Solution. :Example Solution. :Example 4. 3. 2. 1. Problems Practice Self Get Solution of These Packages & Learn by Video Tut 0 oml t te aaoa ² 4x : 4ax = y² parabola the to Normals 10.10. 10.10. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful If the at If normal point ‘t (iii) (ii) (i) NOTE : (iii) (ii) (i) Prove that the area of triangle formed by three tan Prove that image of focus in any tangent to parabol Prove that perpendicular drawn from focus upon any Ans. Find equation tangent to parabola y ∴ ⇒ ∴ Find Find the locus of the point N which from 3 normals t (ii) Hence ∵ ⇒ ∴ Since at normal t If the normals at points t ∴ Slope of at normal P = – t oml t te aaoa ² 4x : : 4ax 4ax = = y² y² parabola parabola the the to to Normals Normals i Two of them are equally inclined to x-axis (i) i t (i) Hence (i) t (i) Hence formed formed by their points of contacts. from from equation (i) & (iii), we get oml t te aaoa ² 4x : 4ax = y² parabola the to Normals ( t t point 't point If the normals to the parabola y² at = to the 4ax the poin parabola the normals If poin the at 4ax = y² parabola the to normals the If t at normals of intersection of Point at + 2at = tx + y mx = y y = y is tangent common of equation t (t t t t– t t y m m 1 1 3 3 1 1 3 2, − a + t + = – t– = t t– = t–= t 1 1 then t then − 2 2a, 2a, 0). ≠ 2 = 1 1 2 (t + t2) + y = = bm – = t t (ii) 2 = =m& m =    2 x 1 1 2 2 1 1 + t + t = – = ,t , − t– = + 1 1 2 t 2 3 t – – 1 = 2 = b a − ' then t then ' 2 – − 2 2 2 + 2 1 1 2 2am 2am =    + t + ) + 2 ) (t+ = t= & t t 2 2 t 2 t y 2 t =2...... (iii) 2 = 2 /1 t 1 2 2 1 2 + − 3 a 2 1 3 1 2 x + a a + x = 0 = (t meet the at curve t (x (x    1 − ’ intersects the parabola again at ‘t 1 t 2 1 2 am 2 1 ⇒ , , t t t– 3 + 2)+ 2 1 − at (at at = 2; t 2; = and slope chord of PQ = 2 x meet at the point t    t 1 3 2 1 ) = 0 ) = 1 − at (am at ) at (x at )    a b . 2, 3    .....(ii) .....(i) ⇒ = m = = 2at). m a /1 3 2 1, = – – = − = 4x whose intercept on y–axis is 2. 2, . y

(t −    1 2am) ) ; ) 1 − + t + 3 t 1 2 b a m + t + t– = b    2 1 2 ) and the line joining t joining line the and ) 1 /1 3 2 & t & 3 on the parabola then prove that + t + 1 try" & "should" try" with "I& "should" Will". Ineffective People – 2 gents to the parabola y 3 are, a (ta are, are drawn to the parabola y = 0 = a lies on its directrix. tangent of a parabola lies on the tangent at the ve t t 2 1 1 + . 2 t 2 2 orials on www.MathsBySuhag.com ’ then show that t ts ts t 1 2 + t + t t t 1 & t 1, meets the parabola again at the point the at again parabola the meets 2 2 2 + t + intersect again on the parabola at the on again the parabola intersect 1 & t & 1 t 2 + 2); 2); + 2 2 = 4ax is half the area of triangle passes through a fixed point fixed a through passes 2 –t= − 2 a = 4ax are such that

1 t – 1 t 2 t 2 (t 1 1 + t+ 2 ). don't. rtex.

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 7 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Sol. : Example Solution. : Example 6. 5. 4. 3. 2. 1. Problems Practice Self Solution. Get Solution of These Packages & Learn by Video Tut 1 ar f Tangents: of Pair 11.11. 11.11. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful Ans. Find the length of normal chord at point ‘t’ to the Ans. the at If normal point P(1, 2) on the parabola y Ans. y parabola the of points the Find ∴ ∴ Let the is normal passes through N(h, k) Two of them are perpendicular Equation of to normal to yeach other (ii) Equation Equation tangent of to y Find the focus of the point P from which tangents a ⇒ ⇒ We know the equation of pair of tangents are given the Write equation of pair of tangents to the parab be can which d tangents of pair the to equation The the If normals at 3 points P, Q & R are concurrent, y parabola the of chord the that Prove chord at a normal If 't' point on ythe parabola (iii) (iii) The centroid of (i) The sum of slopes of normals is zero, Sum (ii) ar f Tangents: Tangents: of of Pair Pair (ii) If If two normal’s are perpendicular two If are nromal equally inclined to th x-axis, (ii) (i) m m S S Let Let m For given value’s of (h, k) it is cubic in ‘m’. t the curve and that its length is 6is length its that and curve the is given by: SSby: given is ∴∴∴ (i) (i) m that ar f Tangents: of Pair ∴ 2 = 2 = 1 1 ≡ m m y² y² 2 2 1 m m+ 1 + m+ , , m − 3 = – rm()m from (3) (9, (9, – 6) ∴ ∴ m k = – mh 2am – am y 8y y = mx – – y 2am mx am= 4x;S ; 4ax y = mx + + mx = y from (2) – 1 – + from (2) rm()m (5)from & (6), we get from (1) (y      2 2 3 a 2 – x– 2 1 = & m 2 m 2 – (4 4x) 4)+ (2y = 2+ – (x 1)) + m + m= – 32x = 4y = – 32x , 4 3 − a k m+ 2 – 2xy – 6x + 2y = 1 = 2y+ 6x – – 2xy 2 3 2 t(a are root’s 0 a 3 + m + 1 m m 2 t = T² where : where T² = 2 3     + 1 3 m m a ∆ m 0= , , 1 PQR PQR lies on the axis of parabola. )1 3     = =0...... (i) 0 = 3 2 2 3 a = –= 1 2 2 + 4x + , 2 = 4ax is 2 y (const) (ii) (ii) (const) – 1 – – = 4ax, is a 2 2 1 a = = – a(x 3a) 1 3 a 3 = y = − + m + = 3 3 2 h     + 4 + 8xy – 8y – 8x – 8xy + 4+ a k 1 k a k a ² ² 2 2 = – = ⇒ − (m = 2 – = 2 4ax = 4ax at which the normal is inclined at 30° at th to inclined is normal the which at 4ax = 3 1 a + m + a k . 1 2 = 4ax, whose equation is y is – x equation whose 4ax, = a h .....(iv) 0 = y ...... (iii) ...... (ii) ⇒ ; T T ; 2 2 ) = ) = θ 2 1 2 + + = 4ax subtends a then angle at pr right the vertex 2 = 4x cuts it again at point Q then Q = ? parabola y a try" & "should" try" with "I& "should" Will". Ineffective People a θ of of ordinates of points P, Q, R is zero la ol y − am then show that re drawn to parabola y 2 = = h by SS ≡ 3 rawn from any point (x point any rawn from y y y + (2a+ k + 0= – m h) en m θ 2 0 = 4x drawn from a point P(–1, 2) 1

.....(vi) .....(v) − orials on www.MathsBySuhag.com 1 1 2 = T²= + m 2a(x + x + 2a(x = 4ax. 2 = 0 (const) 1 ). 2 = 4ax having slopes m 2 1, 1, + 4a+ y 1 ) to the parabola y² ) 4ax= to the parabola 2 = 0, is a normal to a is normal 0, = e axis.e don't. 1 , , m ove thatove 2 s uch

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 8 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Solution. : Example Solution. : Example Solution. : Example 1. Problem Practice Self Get Solution of These Packages & Learn by Video Tut 3 hr o Contact: of Chord 13.13. Circle: Director 12.12. 13.13. 12.12. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful For parabola y parabola For y (y at P & Q. & P at From From (i) & (ii) Let it is chord of contact parabolafor y ∴ But given is ∴ If If the line x – y – 1 = 0 intersect the parabola y ∵ Let tangent at P(t Find the length of chord of contact of the tangents : NOTE drawn tangents of contact of chord the to Equation perpendic the of intersection of point the of Locus Ans. If two tangents to the parabola y ∴ Let passesit through P(h, k) Equation of Equation tangent to of y Find the locus of point whose chord of contact w.r. ⇒ Let (h, k) be point of intersection of tangents the hr o Contact: Contact: of of Chord Chord Circle: Circle: Director Director yy i tan (i) then find the locus of P in each of the following c ∴ locus of P P locus is of ∴ i m (i) i)cos (ii) i)tan (ii) hr o Contact: of Chord Circle: Director 2 = 4ax. = 1 1 ² ² = 2a (x + x + (x 2a = − 4ax m = = m x – y – 1 = 0 y – = 1– x x–y h=0.....(i) 0 4h= yk – + 4x h) + 4(x yk = m at y = = y + h) yk = 2b(x Equation of chord of contact is = = a = PQ = y(i) ⇒ point point 4 = k 1, – = h 4 1 The area of the triangle formed by fr tangents the formed The area the of triangle 1 1 1 2 ) = = t + m + 0= a+ h mk – 2 θ 2 3/2 2 θ = x = θ 0 2 – 2ax = k = = 1 2 ÷ 2a. 1 ≡ b + tan k − cos − b 1 (–1, 4) 2 2 x + + x 1 k 1 ). = m = = 4ax it’s equation is x + a = 0 which is parabola is which 0 = a + x is equation it’s 4ax = 1 y = (x – y a) = (x tan , , at( m y( 1 = = − ) ) & Q(t 1 m 2 a 1 2 θ 1 m 2 t(( − 2 2 1 θ + − 0 = = = 4 − 1 λ k bh = = 2 + 4 m at = h x 1 m ax t λ 2 2 2 2 2 2 2 , x(ii) a(t & h k 2 λ 2 = 4ax is+ y = mx (a constant) ) ) meet at (x ) k bh ) 1 a (a constant) 2 2 )( = 2 + − y 1 2 1 2( θ 2 + − = 0 1 2 = 4ax from a point P make angles tt4 ⇒ ⇒ 4 2 t(a h/k λ a 1 )(( 2 h/a 1 2 − – {(x a) + t + 1 ) t t , , y 1 2 2 = 4bx w.r.t. the point P(h, k) + 2 )) 1 ) ) = y ) t y = m = y a = 2 2 ) 2 2 m a + y 1 2 + 4 = 8x at P & Q, then find the point of intersection k b 0 n chord of contact is x try" & "should" try" with "I& "should" Will". Ineffective People ases. 2 2 2 )4 } h drawn from point (x t t to the parabola y ular tangents to a curve is called the Director Cir Director the is called to a curve tangents ular from a point P(x a point from .....(ii) ...... (i) orials on www.MathsBySuhag.com om the point (xthe om point θ 2 ’s own directrix. own ’s = 4bx is the tangents of the parabola 1 and 1 , , y 1, y 1 ) ) to the parabola y 1 θ ) ) is 2 1, with the axis of the parabola, y 1 ) ) & is contact chord the of 2 = 4ax. of tangents don't. cle.

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 9 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com 2. 1. Self PracticeProblems Solution. :Example Solution. :Example 2. 1. Self PracticeProblems Get Solution of These Packages & Learn by Video Tut 5 motn Highlights: Important 15.15. point: middle given a with Chord 14.14. 15.15. 14.14. Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful so equation chordof is yk – + 2a(x h) = k so equation chordof is yk – + 2a(x h) = k but slopebut = Since it passes through (p, q) Find Find the locus of mid - point of chord of parabola Ans. Find the equation of chord of parabola y ∴ Let P(h, k) be the mid point of chord of parabola y Find the locus of middle point of the chord of the ∴ ∴ Let P(h, k) be the mid point of chord of parabola y Find the locus of middle point of the chord of the (x who y² = 4ax parabola the of chord the of Equation 1 + 2y Ans. – x line the on ‘P’ y point variable a from If Prove that locus of a point whose chord of contact (viii) (vii) (vi) ( v) (iv) (iii) (ii) (i) Ans. motn Highlights: Highlights: Important Important point: point: middle middle given given a a with with Chord Chord motn Highlights: Important point: middle given a with Chord 2 = 8x then prove that chord of contact passes throu 1 , y , 1 ) is y y is ) k = mh mh = k of length a length of x – y – 2 = 0 y –= 2 – x = y is locus y Required locus is qk – 2a + h)(p = k (1, 8) y If normal are drawn from a point P(h, k) to par to the k) P(h, a point from drawn are normal If points. these at tangents the by formed on points three by formed triangle the of area The parabola.the of chord focal is the 4ax, y² = parabola the of rectum latus Semi ⇒ ⇒ T,in then: meet Q P and at tangents the If vertex. on it perpendicular & the a to parabola tangent Any of a P (atpoint radii touch as diameter chord focal on any a circle hence in chord a focal of extremities the at tangents The theat angle a of to betw The a tangent portion cut off parabola that reflectio after conclude theparabola we of to axis the this parallel From directrix. the on P from between angle the of par bisectors the are the parabola the of ‘P’ point any at words In other ST = SG = where ‘S’ is SP focus. the normal & tangent the If y (2ax – y 2 2 – 2ax 2apqy2ax – –0. + = = − 2 k y 4 a a b 1 = m = − 2 ST S. focus the at angles equal subtend TQ and TP 2am 2am x 2 2 ) ) = 4a y 2 . a 1 = SP. SQ & SP.SQ =

2 focus. 1 m (x (x a + − 2 am b t − 2 2 x – 4ah on a normal at the point P.. point the at a on normal 2 3 3 1 , 2at) as diameter touches the tangent at the the verte touches tangent , as 2at) diameter ) ) ie am i.e. ≡ T = S = T 2 = 4x whose mid point is (4, 2). 2 2 1 ⇒ – 4ah. – 4ah. 3 + m(2a m(2a + try" & "should" try" with "I& "should" Will". Ineffective People parabola y parabola y y 2 2 The triangles SPT and STQ are similar. STQ and SPT The triangles = 4ax, = 4ax, 2 w.r.t. parabola passes through focus is directrix = 4ax which touches the parabola x gh a point, fixed also find that point. 0 ar f agn’ ae rw t te parabola the to drawn are tangent’s of pair 0 = se middle point is point se middle − h) + k = 0. = k + h) n. orials on www.MathsBySuhag.com 2 een the directrix & the curve & subtends a the curve right een the directrix 2 = 4ax which pass through a given point (p, q). abola yabola = 4ax whose slope is ‘m’. tersect at right angles on the directrix, and directrix, the on angles right at tersect harmonic mean between segments of any of segments between mean harmonic a parabola is twice the area of the triangle the of area the twice is a parabola the tangent and the normal at a point P at a on point and normal the tangent the from the focus meet on the tangent at theat tangent on the meet focus the from es the directrix. Also a circle on any focal a circle Also es directrix. the the focal radius SP & the perpendicular the & SP radius focal the all rays emanating from S will become will S from emanating rays all abola intersect the axis at T & G then G & T at axis the intersect abola 2 = 4ax then = 4ax x and intercepts a chord and x intercepts 2 = 4by. don't.

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 10 of 91 FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Get Solution of These Packages & Learn by Video Tut Successful People Replace the words like; "wish", " like; the "wish", words Replace People Successful Note : Students must try to proof all the above prop above the all proof to try must Students : Note (x) (ix) ⇒ ⇒ ⇒ ⇒ Where mWhere Length of subnormal is constant for all points on t on points all rectum.for constant is subnormal of Length bisecte is subtangent the P. that Note point the of pa on the y) P(x, point any at subtangent of Length to be normals distinct and real three for Condition the of Centroid conorma three the of ordinates the of sum algebraic concurrent three the of slopes the of sum algebraic m 1 + m+ 2 h > 2a & k & 2a > h + m+ 1, m 3 =0;m ; 0 = 2, & m ∆ formed by three co three by formed 2 3 < are the slopes of the three concurrent normals. N normals. concurrent three the of slopes the are 27 4 a (h – 2a)– (h 1 m 3. 2 − + + m normal points lies on the xon the lies points normal try" & "should" try" with "I& "should" Will". Ineffective People 2 m 3 + + m orials on www.MathsBySuhag.com 3 erties. m d at the vertex.the d at drawn froma point P (h, k) is k) P (h, point froma drawn rabola y² = 4ax equals twice the abscissa the twice equals 4ax = y² rabola 1 normals is zero.is normals he parabola & is equal to the semi latus semi the to equal is & parabola he = l points on the parabola is zerois parabola on the points l 2 a a − h ;m ; − axis. 1 m 2 m ote thatote 3 = − k a . don't.

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