1 CONIC SECTIONS While cutting crrot ou might hve noticed different shpes shown b the edges of the cut. Anlticll ou m cut it in three different ws, nmel (i) (ii) (iii) Cut is prllel to the bse (see Fig.1.1) Cut is slnting but does not pss through the bse (see Fig.1.) Cut is slnting nd psses through the bse (see Fig.1.3) Fig. 1.1 Fig. 1. Fig. 1.3 The different ws of cutting, give us slices of different shpes. In the first cse, the slice cut represent circle which we hve studied in previous lesson. In the second nd third cses the slices cut represent different geometricl curves, which we shll stud in this lesson. MATHEMATICS 391 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com1

OBJECTIVES After studing this leson, ou will be ble to : recognise circle, prbol nd ellipse s sections of cone; recognise the prbol nd ellipse s certin loci; identif the concept of eccentricit, directrix, focus nd vertex of conic section; identif the stndrd equtions of prbol nd ellipse; nd find the eqution of prbol given its directrix nd focus. EXPECTED BACKGROUND KNOWLEDGE Bsic knowledge of codinte Vrious fms of eqution of stright line Eqution of circle in vrious fms 1.1 CONIC SECTION In the introduction we hve noticed the vrious shpes of the slice of the crrot. Since the crrot is conicl in shpe so the section fmed re sections of cone. The re therefe clled conic sections. Mthemticll, conic section is the locus of point P which moves so tht its distnce from fixed point is lws in constnt rtio to its perpendiculr distnce from fixed line. The fixed point is clled the focus nd is usull denoted bs. The fixed stright line is clled the Directrix. The stright line pssing through the focus nd perpendiculr to the directrix is clled thexis. The constnt rtio is clled the eccentricit nd is denoted b e. Wht hppens when (i) e < 1 (ii) e = 1 (iii) e > 1 In these cses the conic section obtined re known s ellipse, prbol nd hperbol respectivel. In this lesson we shll stud bout ellipse nd prbol onl. 1. ELLIPSE Recll the cutting of slices of crrot. When we cut it obliquel, slnting without letting the knife pss through the bse, wht do we observe? 39 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com

You might hve come cross such shpes when ou cut boiled egg verticll. The slice thus obtined represents n ellipse. Let us define the ellipse mthemticll s follows: An ellipse is the locus of point which moves in plne such tht its distnce from fixed point bers constnt rtio to its distnce from fixed line nd this rtio is less thn unit. 1..1 STANDARD EQUATION OFAN ELLIPSE Let S be the focus, ZK be the directrix nd P be moving point. Drw SK perpendiculr from S on the directrix. Let e be the eccentricit. Divide SK internll nd externll t A nd A' (on KS produced) repectivel in the rtio e : 1, s e<1. SA = e. AK (1) nd S A = e. A K () Since A nd A' re points such tht their distnces from the focus bers constnt rtio e (e < 1) to their respective distnces from the directrix nd so the lie on the ellipse. These points re clled vertices of the ellipse. x' K' Z 1 A' S' (-e,0) B B' C ' Fig. 1.4 L N S( e,0) Let AA' be equl to nd C be its mid point, i.e., CA = CA' = The point C is clled the centre of the ellipse. Adding (1) nd (), we hve SA + SA = e. AK + e. A K A A = e( CK CA + A C + CK ) = e. CK L' P Z M x A K CK = (3) e MATHEMATICS 393 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com3

Subtrcting (1) from (), we hve SA SA = e( A K AK ) ( SC + CA ) ( CA CS ) = e. A A CS = e. CS = e (4) Let us choose C s igin, CAX s x-xis nd CY, line perpendiculr to CX s -xis. s of S re then (e, 0) nd eqution of the directrix is x = e Let the codintes of the moving point P be (x, ). Join SP, drw PM ZK. B definition SP = e. PM SP = e. PM SN + NP = e.(nk ) ( CN CS) + NP = e.( CK CN) ( x e) + = e x e x (1 e ) + = (1 e ) x + = 1 (1 e ) [On dividing b (1 e ) ] Putting ( 1 e ) = b, we hve the stndrd fm of the ellipse s x + b = 1 Mj xis : The line joining the two vertices A' nd A, i.e., A'A is clled the mj xis nd its length is. Min xis : The line pssing through the centre perpendiculr to the mj xis, i.e.,bb' is clled the min xis nd its length is b. Principl xis : The two xes together (mj nd min) re clled the principl xes of the ellipse. 394 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com4

Ltus rectum : The length of the line segment LL' is clled the ltus rectum. Eqution of the directrix : x = ± e b Eccentricit : e is given b e = 1 Exmple 1.1 Find the eqution of the ellipse whose focus is (1, 1), eccentricite = 1 nd the directrix is x = 3. Solution : Let P (h,k) be n point on the ellipse then b the definition, its distnce from the focus = e. Its distnce from directrix SP = e.pm (M is the foot of the perpendiculr drwn from P to the directrix). 1 h k 3 ( h 1) + ( k + 1) = 4 1+ 1 7( h + k ) + hk 10h + 10k + 7 = 0 The locus of P is 7( x + ) + x 10x + 10 + 7 = 0 which is the required eqution of the ellipse. Exmple 1. Find the eccentricit, codintes of the foci nd the length of the xis of the ellipse 3x + 4 = 1 Solution : The eqution of the ellipse cn be written in the following fm x + 4 3 = 1 On compring this eqution with tht of the stndrd eqution of the ellipse, we hve = 4 nd b = 3, then (i) e b = 1= 3 = 1 e = 4 4 1 (ii) codintes of the foci re (1,0) nd ( 1,0) 1 [ The codinte re ( ± e, 0)] MATHEMATICS 395 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com5

(iii) Length of the mj xes = = 4 nd length of the min xis = b = 3 = 3. CHECK YOUR PROGRESS 1.1 1. Find the eqution of the ellipse referred to its centre () whose ltus rectum is 5 nd whose eccentricit is 3 (b) whose min xis is equl to the distnce between the foci nd whose ltus rectum is 10. (c) whose foci re the points (4,0) nd ( 4,0) nd whose eccentricit is 1 3.. Find the eccentricit of the ellipse, if its ltus rectum be equl to one hlf its min xis. 1.3 PARABOLA Recll the cutting of slice of crrot. When we cut obliquel nd letting the knife pss through the bse, wht do we observe? Also when btsmn hits the bll in ir, hve ou ever noticed the pth of the bll? Is there n propert common to the edge of the slice of the crrot nd the pth trced out b the bll in the exmple cited bove? Yes, the edge of such slice nd pth of the bll hve the sme shpe which is known s prbol. Let us define prbol mthemticll. "A prbol is the locus of point which moves in plne so tht its distnce from fixed point in the plne is equl to its distnce from fixed line in the plne." 1.3.1 STANDARD EQUATION OFA PARABOLA Let S be the fixed point nd ZZ' be the directrix of the prbol. Drw SK perpendiculr to ZZ'. Bisect SK t A. Since SA = AK, b the definition of the prbol A lies on the prbol. A is clled the vertex of the prbol. Tke A s igin, AX s the x-xis nd AY perpendiculr to AX through A s the -xis. 396 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com6

Z Y M P ( x, ) K A L S (,0) N X L' Let P' Z' Fig. 1.5 KS = AS = AK = The codintes of A nd S re (0,0) nd (,0) respectivel. Let P(x,) be n point on the prbol. Drw PN AS produced AN = x nd NP = Join SP nd drw PM ZZ B definition of the prbol SP = PM SP = PM ( x ) + ( 0) = ( x + ) [ PM = NK = NA + AK = x + ] ( x ) ( x + ) = = 4x which is the stndrd eqution of the prbol. Note : In this eqution of the prbol (i) Vertex is (0,0) (ii) Focus is (,0) (iii) Eqution of the xis is = 0 (iv) Eqution of the directrix is x + = 0 (v) Ltus rectum = 4 MATHEMATICS 397 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com7

1.3. OTHER FORMS OF THE PARABOLA Wht will be the eqution of the prbol when (i) focus is (,0) nd directrix is x = 0 (ii) focus is (0,) nd directrix is + = 0, (iii) focus is (0, ) nd directrix is = 0? It cn esil be shown tht the eqution of the prbol with bove conditions tkes the following fms: (i) (ii) = 4x x = 4 (iii) x = 4 The figures re given below f the bove equtions of the prbols. X' A z K ' z' = 4x X (i) z x' A K ' x = 4 Fig. 1.6 (ii) x z' z x' Cresponding results of bove fms of prbols re s follows: Fms A K ' x = 4 z' x = 4x = 4 x x = 4 x = 4 s of vertex (0,0) (0,0) (0,0) (0,0) s of focus (,0) (,0) (0,) (0, ) s of directrix x = x = = = s of the xis = 0 = 0 x = 0 x = 0 (iii) length of Ltus rectum 4 4 4 4 398 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com8

Exmple 1.3 Find the eqution of the prbol whose focus is the igin nd whose directrix is the line x + 1 = 0. Solution : Let S (0,0) be the focus nd ZZ' be the directrix whose eqution is x + 1 = 0 Let P(x, ) be n point on the prbol. Let PM be perpendiculr to the directrix (See Fig. 1.5) B definition SP= PM SP = PM x (x + 1) + = ( + 1) 5x + 5 = 4x + + 1 + 4x 4x x + 4 4 x + + 4x 1 = 0. Exmple 1.4 Find the eqution of the prbol, whose focus is the point (, 3) nd whose directrix is the line x 4 + 3 = 0. Solution : Given focus is S(,3); nd the eqution of the directrix is x 4 + 3 = 0. As in the bove exmple ( x ) + ( 3) x 4 + 3 = 1 + 4 16x + + 8x 74x 78 + 1 = 0 CHECK YOUR PROGRESS 1. 1. x Find the eqution of the prbol whose focus is (, b) nd whose directrix is + = 1. b. Find the eqution of the prbol whose focus is (,3) nd whose directrix is 3 x + 4 = 1. MATHEMATICS 399 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com9

LET US SUM UP Conic Section "A conic section is the locus of pointp which moves so tht its distnce from fixed point is lws in constnt rtio to its perpendiculr distnce from fixed stright line". (i) Focus : The fixed point is clled the focus. (ii) Directrix : The fixed stright line is clled the directrix. (iii) Axis : The stright line pssing through the focus nd pependiculr to the directrix is clled the xis. (iv) Eccentricit : The constnt rtio is clled the eccentricit. (v) Ltus Rectum : The double dinte pssing through the focus nd prllel to the directrix is known s ltus rectum. (In Fig.1.5 LSL' is the ltus rectum). x Stndrd Eqution of the Ellipse is : + = 1 b (i) Mj xis = (ii) Min xis = b (iii) Eqution of directrix is x = ± (iv) Foci : ( e ± e,0) b (v) Eccentricit, i.e., e is given b e = 1 Stndrd Eqution of the Prbol is : = 4x (i) Vertex is (0,0) (ii) Focus is (,0) (iii) Axis of the prbol is = 0 (iv) Directrix of the prbol is x + = 0 (v) Ltus rectum = 4. OTHER FORMS OF THE PARABOLAARE (i) (ii) = 4x (concve to the left). x = 4 (concve upwrds). (iii) x = 4 (concve downwrds). 400 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com10

SUPPORTIVE WEB SITES http://www.wikipedi.g http://mthwld.wolfrm.com TERMINAL EXERCISE 1. Find the eqution of the ellipse in ech of the following cses, when () focus is (0, 1), directrix is x + = 0 nd e = 1. (b) focus is ( 1,, 1), directrix is x + 3 = 0 nd e = 1.. Find the codintes of the foci nd the eccentricit of ech of the following ellipses: () 4x + 9 = 1 (b) 5x + 4 = 100 3. Find the eqution of the prbol whose focus is ( 8, ) nd directrix is x + 9 = 0. MATHEMATICS 401 Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com11

ANSWERS CHECK YOUR PROGRESS 1.1 1. () 0x + 36 = 405 (b) x + = 100 (c) 8x + 9 = 115. 3 CHECK YOUR PROGRESS 1. 3 3 4 4 1. ( x b) x b + + b + b = 0.. 16x 9+ 94x 14 4x + 34 = 0 TERMINAL EXERCISE 1. () 7x + 7 x 16 + 8 = 0. () (b) 7x + 7 + x + 10x 10 + 7 = 0 5,0 ; 6 ± (b) ( 0, ± 1) 5 3 3. x + 4 + 4x + 116x + + 59 = 0 ; 1 5 40 MATHEMATICS Get Discount Coupons f our Coching institute nd FREE Stud Mteril t www.pickmycoaching.com1