Asymptotes

 Course Outline

&        Definition of asymptotes with illustration in figure.

&        Asymptotes parallel and non-parallel to the axes.

&        Asymptotes of algebraic curves.

&        Asymptotes of polar curves.

Asymptote

        A straight line is called asymptote to a curve if the perpendicular distance of the straight from a point on the curve tends to zero as the point moves to infinity along the curve.

Some important facts about asymptotes of a curve

(i)         The number of asymptotes of an algebraic curve does not exceed the degree of equation.

(ii)        If in an algebraic curve of degree n, yⁿ is absent then the asymptote parallel to y- axis can be obtained by equating the coefficient of highest degree terms of y to zero.

(iii)       If in an algebraic curve of degree n, xⁿ is absent then equation of asymptotes parallel to x- axis can be obtained by equating the coefficient of highest degree term of x to zero.

(iv)       If the term involving both xⁿ and yⁿ are absent, then equate the coefficient of highest available powers of z, y to zero, and we get asymptote parallel to
x- and y- axis.

Solved Problems

Q.1.      What do you mean by asymptote to a curve ? Show that the asympote of the curve x2y2 = a(x2 + y2) forms a square of side 2a.

Soln:    A straight line is said to be an asymptote to a curve if the perpendicular distance of the straight line from a point on the curve tends to zero as the point moves to infinity along the curve.

Now, the given curve is x²y² = a²(x² + y²).

             The curve is of degree four, so there are at the most four asymptotes.

             Equating the coefficients of the highest power of x with zero, we get

             y² – a² = 0                                                    [Q (y² – a²) x² – a²y² = 0]

             y2 = a2

\         y = ± a which are the asymptotes parallel to x- axis.

Again, equating the coefficient of highest power of y with zero, we have,

             x2 – a2 =  0                                                 [Q (x2 – y2) y2 – a2x2 = 0]

or,        x² = a²

\         x = ± a which are the asymptotes Parallel to y-axis.

There are no oblique asymptotes.

Hence the asymptotes are x = ±a, y = ±a

             Now, these are four asymptotes x = a, x = – a, y = a and y = – a. Solving there in pair, they intersect at points (a, a), (a, – a), (–a, a) and (– a, – a), which are the vertices of square of side 2a.

Q.2.      Find all asymptotes of the curve x(x² + y²) = a(x² – y²).               [T.U. 2073]

Soln:    The given curve is

                         x(x² + y²) = a(x² – y²)

             or,        x³ + xy² – ax² + ay² = 0

             The given curve is an algebraic curve of degree 3. So there are at the most three asymptotes real or imaginary. Due to the presence of x³ in the curve shows that there are no asymptotes parallel to x-axis. But the absence of y³ in the curve shows that there are asymptotes parallel to y-axis. The asymptotes parallel to y-axis are obtained by equating the coefficient of highest available power of y to zero. The coefficient of highest power of y is x + a and equating it to zero, we get x + a = 0. Which is the asymptote parallel to y-axis.

             Again, to find the oblique asymptotes of the form y = mx + c put x = 1 and
y = m in third and second degree term, we get

                         f₃ (m) = 1 + m², f₂(m) = – a + am²

             Now,    f₃ (m) = 0 Þ 1 + m² = 0 Þ m = ± Ö(- 1).

             Hence, there is no real oblique asymptote.

             \ Equation of asymptote of the curve is x + a = 0.

Q.3.      Define asymptotes of a curve. And fine the asymptotes of the curve x²/a² + y²/b² = 1 
+ = 1                                                                                                      

Soln:  Asymptote of a curve: A straight line is said to be an asymptote to a curve if the perpendicular distance of the straight line from a point on the curve tends to zero as the point moves to infinity along the curve. The finitely bounded curve does not have an asymptote.

             The given curve x²/a² + y²/b² = 1  can be written as

                         a²y² + b²x² = x²y² or, x²y² – a²y² – b²x² = 0

            The given curve is an algebraic curve of degree four. So, there are at the must four asymptotes real or imaginary. Absence of x⁴ and y⁴ in the curve shows that there are asymptotes parallel to x–axis and y–axis. Asymptotes parallel to x–axis and y–axis are obtained by equating the highest available powers of x and y respectively to zero.

             i.e.        y² – b² = 0

             \          y = ± b                [Q x²y² –a²y² – b²x² = 0 or, x²(y² – b²) – a²y² = 0]

             and       x² – a² = 0     

             \          x = ± a.                  [Q x²y²– a²y²– b²x² = 0 or, y²(x² – a²) –b²x² = 0]

             Hence, the required asymptotes of the given curve are,

             x = ± a and y = ± b Ans.

Q.4.      Find the asymptotes of the curve (x – 1) (x + 2) (x + y) + x² + x + 1 = 0.  

Soln:    The given curve is (x – 1) (x + 2) (x + y) + x² + x + 1 = 0

             or,        (x² + x – 2) (x + y) + x² + x + 1 = 0

             or,        x³ + x² – 2x + x²y + xy – 2y + x² + x + 1 = 0

             or,        x³ + x²y + 2x² + xy – x – 2y + 1 = 0 ..... (i)

             This is an algebraic curve of degree three, so there are at the most three asymptotes real or imaginary. Due to the presence of x³ in the curve this shows that there are no asymptotes parallel to x–axis. But absence of y³ in the curve shows that there are asymptotes parallel to y–axis. The asymptote parallel to y–axis are obtained by equating to zero the coefficient of highest degree term in y. The highest degree term is y in the given curve is (x²+x–2)y. Equating its coefficient to zero, i.e., x² + x – 2 = 0

             or,        (x + 2) (x – 1) = 0

             Thus x – 1 = 0 and x + 2 = 0 are asymptotes parallel to y–axis.

             To obtain the asymptote of the form y = mx + c ...(ii)

             Put x = 1 and y = m in 3rd, 2nd and 1st degree terms to get f₃ (m), f₂ (m) and f1 (m) respectively.

                         i.e.        f3 (m) = 1 + m

                                      f2 (m) = 2 + m

                                      f1 (m) = –1–2m

                         f3 (m) = 0 Þ  1 + m = 0. Also, f'3 (m) = 1

                         or,        m = – 1

             When, m = –1, c = – 1

             Substituting values of m and c in equation (ii) the oblique asymptote is,

                         y = – x – 1                   

            or,         y + x + 1 = 0

             Hence the required asymptotes are x – 1 = 0, x + 2 = 0 and x + y + 1 = 0. Ans.

Q.5.      Find all asymptotes of the curve y2(x2 – a2) = x2(x2 – 4a2).

Soln:    The given equation of curve is

                         y2(x2 – a2) = x2(x2 – 4a2)

             or,        y2(x2 – a2) – x2(x2 – 4a2) = 0

             This is an algebraic curve of degree 4. So there are at the most four asymptotes of real or imaginary. Due to the presence of x4 in the curve, this shows that there are no asymptotes parallel to x- axis. But absence of y4 in the curve shows that there are asymptotes parallel to y- axis. The asymptotes parallel to y- axis are obtained by equating to zero the coefficient of the highest degree term in y. The highest degree term in y is y2(x2 – a2).

             Equating to zero the coefficient of highest degree term.

             We have,

                         x² – a² = 0

             or,        x² = a²

                   or,        x = ±a.

             To find the oblique asymptote, put x = 1 and y = m in the fourth and third degree terms we get,

                         f4(m) = 1 – m2, f3(m) = 0

             Now,   f4 (m) = 0 Þ 1 – m2 = 0 \ m = ±1

           

  Putting the value of C and m in y = mx + C, we get

                         y = ±1x + 0

             or,        y = ± x

             Hence, required asymptotes  are y = ±a and y = ±x.

Q.6.      Find all the asymptotes of the curve x³ – 2x²y + xy² + x² –xy + 2 = 0.

[T.U. 2068]

Soln: The given curve is an algebraic curve of degree 3. So, there are at most three asymptotes real or imaginary. Due to the presence of x³ in the curve shows that there are no asymptotes parallel to x – axis. But absence of yin the curve shows that there are asymptotes parallel to y–axis. The asymptotes parallel to y–axis are obtained by equating the coefficient of highest available power of y to zero.  i.e. x = 0        

                   Now, we have to find oblique asymptotes of the form y = mx + c. For this, putting x = 1 and y = m in third, second and first degree terms of the given curve, we get

                          f₃(m) = 1 – 2m + m², f₂(m) = 1–m , f₁(m) = 0

             Now f₃(m) = 0 Þ 1 – 2m + m= 0 i.e. (1 – m)= 0

\         m = 1, 1 (twice repeated)

\         c²/2! f₃" (m) + cf₂'  (m) + f₁(m) = 0               [Q f'₃ (m) = – 2 + 2m

                                                                                     f₃" (m) = 2, f₂' (m) = –1]

             or,         c²/2! . 2 + c( – 1) = 0

             or,         c² – c = 0                       \             c = 0, 1

             Substituting m = 1 and c = 0, 1 in y = mx + c we get, y = x+0 and y = x+1
i.e. y = x and y = x + 1

             Hence the required asymptotes are x = 0, x – y = 0 and x – y + 1= 0 Ans.

Q.7.      Find the asymptotes of the curve x (x – y)2 – 3(x2 – y2) + 8y = 0.

Soln:    The given curve is x (x – y)2 – 3 (x2 – y2) + 8y = 0

             or,        x (x2 – 2xy + y2) – 3x2 + 3y2 + 8y = 0

             or,        x3 – 2x2y + xy2 – 3x2 + 3y2 + 8y = 0 ...... (i)

             This is an algebraic curve of degree three, so there are at the most three asymptotes real or imaginary. Due to presence of x3 in the curve, this shows that there are no asymptotes parallel to x–axis. But absence of y3 in the curve shows that there are asymptotes parallel to y–axis. The asymptotes parallel to y–axis are obtained by equating to zero the coefficient of highest degree term in y.

             The highest degree term in y in the given curve is y² (x + 3).

             Equating the coefficient of highest degree term to zero, we have

                         x + 3 = 0 which is the asymptote parallel to y–axis.

             Again to obtain the oblique asymptote of the form

                         y = mx + c ...... (ii)

             Put x = 1 and y = m in 3rd, 2nd and 1st degree terms of equation (i) to get
f3 (m), f2(m) and f1(m) respectively.

             i.e.,       f3 (m) = 1 – 2m + m2

                         f2 (m) = 3 – 3m2

             and       f1 (m) = 8m

                         f3 (m) = 0 Þ 1 – 2m + m2 = 0

             or,        (1 – m)2 = 0

             \         m = 1, 1 (twice repeated)

Also,    f'3 (m) = –2 + 2m

            f''3 (m) = 2

            f'2 (m) = – 6m

Now using c²/2! f''3 (m) + c f'2 (m) + f1 (m) = 0, we have

or,        c²/2!. 2 + c (–6m) + 8m = 0

or,        c2 – 6 c m + 8m = 0

When m = 1, we have

            c² – 6c + 8 = 0

or,        (c – 2) (c – 4) = 0 \   c = 2 or 4

Substituting the value of c and m in (i) , y = x + 2 and y = x + 4

Therefore, the required asymptotes are x+3 = 0, y–x = 2 and y–x = 4 Ans.

Q.9.      a) Find the asymptotes of the curve 2r² = tan 2q.                                       

Soln:    The given curve 2r² = tan 2q  can be written as

                          2r² =

             or,        2r² cos2q = sin 2q

             or,        2r² (cos² q – sin² q) = 2 sin q . cos q

             or,        2r² . (r² cos² q – r² sin² q) = 2 r sinq . r cos q

             or,        2(x² + y²) (x² – y²) = 2xy       

             or,        2(x⁴ – y⁴) = 2xy.

             or,        x⁴ – y⁴ – xy = 0 ........ (i)

             The curve (i) is the algebraic curve of degree four. So, there are at the most four asymptotes real or imaginary. Due to the presence of x⁴ and y⁴ in the curve shows that there are no asymptotes parallel to x–axis and y–axis respectively.

             Now, we find the asymptotes of the form

             y = mx + c .... ... ...(ii)

             For this putting x = 1 and y = m in four and third degree terms in (i) to obtain f₄ (m) and f₃ m), we have

                         f₄ (m) = 1 – m⁴ and f₃ (m) = 0

                         f'₄ (m) = – 4m³

             Now, f₄ (m) = 0 Þ 1 – m⁴ = 0  Þ (1– m²) ( 1+ m²) = 0

             i.e. m = ± 1 and other values of m are imaginary.

             We know that

                         c = – f₃(m) / f ¢₄(m) = - 0/4m³= 0.

             Substituting the value of c and m in (ii), we get

                         y = ± 1 x + 0                              or,        y = ± x

             Hence, the required asymptotes of the curve are

                y = ± x.

             or,  rsinq = ± rcosq                                 or,  = ± 1

             or,  tanq = ± 1                                         or, tanq = tan (± p/4)

             \                q = ± p/4 Ans.

 

 

b)         Find all asymptotes to the curve y² (x – 2) – x² (y – 1) = 0.

                                                                                                           [T.U. 2075]    

Soln:    The given curve is

            y² (x – 2) – x² (y – 1) = 0

This equation is of third degree and the term x³ and y³ are both absent. Hence there are parallel asymptotes of the form x = a and y = b to the coordinate axes.

To find the parallel asymptotes:

The asymptote parallel to y axis is given by

Coefficient of y² = 0

⟹ x – 2 = 0

and the asymptote parallel to x – axis is given by

coefficient of x² = 0.

or,     y – 1 = 0

To find the asymptote of the form y = mx + c:

Put x = 1 and y = m in 3^rd and 2^nd degree term, we get ϕ₃ (m) = m² – m,
ϕ₂ (m) = –2m² + 1

Now, ϕ₃ (m) = 0

⟹ m² – m = 0

or, m (m – 1) = 0

∴ m = 0, 1

ϕ (m) = 2m – 1

 

∴ Oblique asymptote is given by y = mx + c or, y = x + 1

Required asymptotes are x – 2 = 0, y – 1 = 0 and y = x + 1 Ans.