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1 UNIT-3-ASYMPTOTES Def. Asymptotes. A straight line is said to be an asymptote of an infinite branch of a curve , if ,as the point P recedes to ∞ along the branch , the ⊥ distance of P from the straight line tends to zero.(it is a line that is tangent at ∞ to the curve ) To obtain asymptotes parallel to the coordinate axes : Rule 1. Asymptotes parallel to y-axis are obtained by equating to zero the coefficients of highest degree terms of y . Rule 2. Asymptotes parallel to x-axis are obtained by equating to zero the coefficients of highest degree terms of x . Exception: If these coefficients are constants , then there are no asymptotes parallel to coordinate axes. Ex. Find asymptotes parallel to coordinate axes of the curves: a2 b2 a2 b2 (i) + 2 =1 (ii) − 2 =1 (iii) x2 y 2 − y 2 − 2 =0 x2 y x2 y Sol. (i) Eq. is x 2 y 2 − a 2 y 2 − b 2 x 2 = 0 . Highest power of y in it is 2 & coefficient of y 2 is x 2 −a 2 . ∴asymptotes parallel to y − axis are x = ±a . Highest power of x in it is 2 & coefficient of x2 is y 2 −b 2 ∴ asymptotes parallell to x − axis are y = ±b . (ii) The equation is x y − a y + b x = 0 2 2 2 2 2 2 Highest power of y in it is 2 & coefficient of y 2 is x 2 −a 2 . ∴asymptotes parallel to y − axis are x = ±a . Highest power of x in it is 2 & coefficient of x2 is y 2 +b 2 ∴ There is no asymptotes parallell to x − axis because 2 2 ( y + b ) does not have any linear factors . (iii) The equation is x y − y − 2 = 0 2 2 2 Asymptotes parallel to y-axis are x 2 −1 =0 i.e. x =±1 Asymptote parallel to x-axis is y =0 i.e. y =0 . 2 To Obtain Oblique Asymptotes : Let y = mx + c be an oblique asymptotes . Then m= xLt∞( y / x) → and c= xLt∞( y − mx ) . → Ex. Find all the asymptotes of the following curves : (i) x 3 − x 2 y + ay 2 = 0 , (ii) x 3 − y 3 + ax 2 = 0 Sol.(i) No asymptotes parallel to coordinate axes . For oblique asymptotes divide the equation by x 3 & we have 2 y y2 1 − + a 3 = 0 ; now taking limits as x → ∞ , we get x x 1 − m + a ⋅ m 2 ⋅ 0 = 0 ⇒ m =1 . Hence c= xLt∞( y − x) → y2 And from the given equation we have ( y − x) = a . x2 y2 y2 ∴ c = Lt ( y − x) = Lt a = a Lt 2 = a ⋅ m 2 , but m = 1 x →∞ x →∞ x2 x →∞ x ∴c = a . Hence the asymptote is y = x + a . (ii) Dividing the equation by x 3 and taking limit as x→ ∞ we have 1 − m + a ⋅ 0 = 0 ⇒ m =1 ⇒ m =1 . 3 3 ∴ c = Lt ( y − x ) = Lt ax 2 a a = = ( m =1) x→∞ x→∞ x 2 + y 2 + xy 2 1+ m + m 3 a Hence the asymptote is y = x + . 3 Working Procedure for Oblique Asymptotes. 1.Write general algebraic equation of the curve of nth degree as x nφn ( y / x) + x n−1φn−1 ( y / x ) + ⋅ ⋅ ⋅ ⋅ +xφ1 ( y / x ) + φ0 ( y / x ) = 0 Dividing by x n and then taking limit as x → ∞ ,we have φn (m) = 0 ,which will give n values of m , say , m1 , m2 , m3 ,..., mn . We will have n asymptotes y = mi x + ci , for i=1,2,3,4,…,n . 2. To determine ci , we replace (y/x) in the equation by mi + ci / x and get ci c c x nφn (mi + ) + x n −1φn −1 (mi + i ) + x n −2φn −2 (mi + i ) + ........ = 0 . x x x Expanding φ s by Taylor’s expansion about the point mi , we ' have x nφn ( mi ) + x n−1{ciφn ( mi ) + φn−1 ( mi )} + ′ 2 n− 2 ci x { φ n′ (mi ) + ciφ n′ − 1 (m)i + φ n− 2 (mi ) }+……=0 2 Since φn (m) = 0 that gives values of m , by dividing by x n−1 we have 1c 2 ciφn ( mi ) + φn−1 (mi ) + i φn′(mi ) + ciφn−1 (mi ) + φn−2 ( mi ) + ... = 0 ′ ′ ′ x 2 Taking limit as x → ∞ we get φn −1 (mi ) ci = − ′ , if φn (mi ) ≠ 0 . (1) ′ φ n (mi ) 3 Hence corresponding to every mi ; ci is known & also the asymptotes are determined . ′ 3. (a) If φn (mi ) = 0 but φn−1 (mi ) ≠ 0 , then ci is not det er min able . And hence no asymptotes corresponding to mi . ′ (b) If φn (mi ) = 0 = φn−1 (mi ) , then ci is det er min ed from 2 ci ′ ′ φn′(mi ) + ciφn−1 (mi ) + φ n−2 ( mi ) = 0 (2) 2 which will give us two values of ci for the same value of mi .Hence we will get two parallel asymptotes . Imp. Note: φn (m) is obtainable from the nth degree terms by putting y = m & x =1 . Similarly φn −1 , φn−2 ... and so on Ex. Find asymptotes of the following curves : ( a ) x 3 + y 3 − 3axy = 0 (b) ( x − y ) 2 ( x 2 + y 2 ) −10 x 2 ( x − y ) +12 y 2 + 2 x + y = 0 Sol. (a ) No asymptotes parallel to coordinate axes . Let the oblique asymptotes be y = mx + c . φ3 ( m) = 1 + m 3 , φ3 (m) = 3m 2 , φ2 ( m) = −3am . ′ φ2 (−1) 3a φ3 (m) = 0 ⇒m = −1 ∴c = − = − = −a ′ φ3 ( −1) 3 Hence asymptote is y = −x − a ⇒ y + x + a = 0 . (b) No asymptotes parallel to x-axis and y-axis . φ4 (m) = (1 − m) 2 (1 + m 2 ) = 0 ⇒m =1 , 1 ; φ3 (m) = −10 (1 − m) ′ ′ φ4 ( m) = 4m 3 − 6m 2 + 4m − 2 ; φ4 (1) = 0 & φ3 (1) = 0 Hence c can’t be determined using eq. (1) ′ ′ For using eq.(2) ,we find φ4′(m) and φ3 (m) ′ ′ φ4′(m) = 12 m 2 − 12 m + 4 , φ3 (m) = 10 & φ2 ( m) = 12 m 2 c2 2 ∴ φ4′(1) + cφ3 (1) + φ2 (1) = 0 ⇒ ′ ′ 4 ⋅ c + 10 c + 12 = 0 2 2 giving 2c 2 +10 c +12 = 0 . Hence (c + 2)( 2c + 6) = 0 ∴c = −2 & c = −3 . Aymptotes are y = x − 2 & y = x −3 . Ex. Find the asymptotes of (y-2x)2(y-x)+(y+3x)(y-2x)+2x+2y-1=0 {Ans. y=2x-2 , y=2x-3 } Asymptotes of Polar Curves: 4 1 An asymptote of a polar curve = f (θ ) is given by r 1 r sin( θ − α ) = , where α is root of f (θ ) = 0 f ′(α ) Ex. Find asymptotes of the curve rθ = a . 1 θ θ Sol. Eq . can be written as = , f (θ ) = = 0 ⇒θ = 0 r a a f ′(0) =1 / a . Hence asymptote is r sin θ = a Circular Asymptotes Let the equation of the curve be r = f (θ) . If θLt∞ f (θ ) = a , → then the equation of circular asymptote is r =a . 6θ 2 + 5θ − 1 Ex. Find the circular asymptote of the curve r= . 2θ 2 − 3θ + 7 6θ 2 + 5θ − 1 Sol θLt∞ f (θ ) = θLt∞ = 3 . Therefore asymptote is r = 3 . → → 2θ 2 − 3θ + 7 Parametric Curves Let the equation of the curve be x =φ(t ) & y =ψ (t ) . Find the value of t =t0 which makes φ(t ) or ψ(t ) or both as ∞ . Then y = mx + c will be the asymptotes where m and c are given as ψ (t ) m = Lt & c = Lt (ψ (t ) − mφ (t )) t →t0 φ (t ) t →t0 3at 3at 2 Ex. Find asymptotes of x = & y= 1+ t3 1 + t3 Sol. t = −1 makes x & y both as ∞. ∴ m = t →−1( y / x) = −1 Lt c = Lt ( y − mx ) = −a . Hence the asymptote is x + y + a = 0 t →−1

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