A curve which touches each member of a given family of curves is called envelope of that family.

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Neal Owens
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1 ENVELOPE A curve which touches ech memer of given fmil of curves is clle enveloe of tht fmil. Proceure to fin enveloe for the given fmil of curves: Cse : Enveloe of one rmeter fmil of curves Let us consier = f(,α) to e the given fmil of curves with α s the rmeter. Ste : Differentite w.r.t to the rmeter α rtill, n fin the vlue of the rmeter Ste : B Sustituting the vlue of rmeter α in the given fmil of curves, we get the require enveloe. Secil Cse: If the given eqution of curve is qurtic in terms of rmeter, Aα +Bα+c=0, then enveloe is given iscriminnt = 0 B - 4AC=0 Cse : Enveloe of two rmeter fmil of curves. Let us consier = f(,α, β) to e the given fmil of curves, n reltion connecting the two rmeters α n β, g(α, β) = 0 Ste : Consier α s ineenent vrile n β eens α. Differentite = f(,α, β) n g(α, β) = 0, w.r. to the rmeter α rtill. Ste : Eliminting the rmeters α, β from the equtions resulting from ste n g(α, β) = 0, we get the require enveloe. Prolems on enveloe of one rmeter fmil of curves :. Fin the enveloe of Solution : Differentite m m where m is the rmeter n, re constnts m m with resect to the rmeter m, we get, 0 m m

2 Using eliminte m from ( ) which is the require eqution of enveloe of. Determine the enveloe of sin cos, where θ eing the rmeter. Solution : Differentite, sin cos with resect to θ, we get, cos sin As θ cnnot e eliminte etween n,we solve n for n in terms of θ. For this, multil sinθ n cosθ n then sutrcting, we get, (sin cos ). Using similr simlifiction, we get, ( sin cos ).. (Leinitz s rolem) Clculte the enveloe of fmil of circles whose centres lie on the -is n rii re roortionl to the sciss of the centre. Solution : Let (,0) e the centre of n one of the memer of fmil of curves with s the rmeter. Then the eqution of fmil of circles with centres on -is n rius roortionl to the sciss of the centre is

3 ( ) k where k is the roortionlit constnt. Differentiting with resect to, we get, ( ) k. k From, k k ( k) k k k 0, k 4. Fin the enveloe of sec cosec, where θ is the rmeter. Solution : The given eqution is rewritten s, tn4 ( ) tn 0, which is qurtic eqution in iscriminnt eqution : B -4AC = 0 ( ) 4 0 tn cot t tn. Therefore the require enveloe is given the 0. Enveloe of Two rmeter fmil of curves :. Fin the enveloe of fmil of stright lines +=, where n re rmetersconnecte the reltion = Solution :

4 Differentiting with resect to ( consiering s ineenent vrile n eens on ). 0 () Differentiting with resect to 0 (4) From () n (4), we hve n (5) Using (5) in, we get the enveloe s 4 =. Fin the enveloe of fmil of stright lines connecte the reltion, where n re rmeters Solution :

5 Differentiting with resect to 0 () Differentiting with resect to 0 (4) From () n (4), we hve n (5) 4 4 Using (5) in, we get the enveloe s. Fin the enveloe of fmil of stright lines, where n re rmeters connecte the reltion = c 5

6 = c 5 Differentiting with resect to, 0 () Differentiting with resect to 0 (4) From () n (4), we hve n (5) Using (5) in, we get the enveloe s 7 c Fin the enveloe of the fmil of circles whose centres lie on the ellise which ss through its centre. n

7 Solution: Let (α,β) e the centre of ritrr memer of fmil of circles which lie on the ellise, whose centre is (0,0). Therefore, eqution of the circles ssing through origin n hving centret (α,β) is 0 with Differentiting with resect to α ( α s ineenent vrile n β eens on α ), 0 () Differentiting with resect to α 0 (4) From () n (4), we hve k, where k = α+β

8 n k From, we hve, k (5) k (6) Using (5) n (6) in, we get the enveloe s 4 5. Determine the eqution of the enveloe of fmil of ellises where the rmeters n re connecte the reltion m, l l n m re non-zero constnts. l m Differentiting with resect to, 0 () Differentiting with resect to l m 0 m (4) l From () n (4), we hve

9 4 4 m l l m l m 4 l n 4 m l n m (5) Using (5) in, we get the enveloe s l m Prolems on Evolute s enveloe of its normls :. Determine the evolute of herol consiering it s n enveloe of its norml Solution : Let P ( cosht, sinht) e n oint on the given herol. Then t t cosh t sinh t coth t Eqution of norml line to the herol is ( sinh t) cosh t cosh t sinh t cosh t Differentiting rtill with resect to t, we hve,

10 cosh t sinh t 0 (sinh t) (cosh t) tnh t sinh t n cosh t () h h Where h ( ) ( ) Using () in, we get, h h ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). B consiering the evolute of curve s the enveloe of its norml, fin the evolute of cos sin, sin cos Solution : sin tn cos. Eqution of norml line to the herol is ( ( sin cos )) cos sin tn

11 sin sin sin cos cos cos sin cos sin cos Differentiting with resect to the rmeter θ, we hve cos sin 0 Multiling cosθ n sinθ n then sutrcting, we hve, cos () Similrl we get, sin (4) Eliminting θ etween () n (4) we get the require evolute s

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Note: This question paper consists of three sections A,B and C. SECTION A

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. TIME : 3hrs M. Mrks.75 Note: This question pper consists of three sections A,B nd C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. X = ) Find the eqution