# Asymptotes

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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
1c                                        
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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