Chapter 14 Asymptotes-II

14.1 Exercise 7

14.1.1 Question 4

Find the asymptotes of the following curves

  1. x2y2−x2y−xy2+x+y+1=0

There is no x3, so asymptotes parallel to x-axis is given by equating coefficients of highest degree term of x.

y2−y=0y(y−1)=0y=0,y=1

Similarly, no y3 is present. Asymptotes parallel to y-axis is given by,

x2−x=0x(x−1)=0x=0,x=1

Thus four asymptotes of the equation are,

x=0x=1y=0y=1

  1. x2(x−y)2−a2(x2+y2)=0

Degree of equation is 4. The equation does not have asymptotes parallel to x-axis. The asymptotes parallel to y-axis is given by equating coefficients of y2 to zero.

x2−a2=0x=±a

We expect 4 asymptotes, let y=mx+c be the equation of the rest, Putting x=1 and y=m,

ϕ4(m)=1−2m+m2ϕ′4(m)=2m−2ϕ″

The slope of the asymptotes can be found by,

\begin{equation*} \begin{split} \phi_4(m) &= 0\\ 1-2m + m^2 &= 0\\ (m-1)(m-1) &= 0\\ m &= 1, 1 \end{split} \end{equation*}

This is a case of repeated factors, so

\begin{equation*} \begin{split} \dfrac{c^2}{2!}{\phi''}_4(m) + c {\phi'}_3(m) + \phi_2(m) &= 0 \\ \dfrac{c^2}{2} \times 2 + 0 -a^2 - a^2m^2 &= 0\\ c^2 - a^2 - a^2 &= 0\\ c &= \pm \sqrt{2}a \end{split} \end{equation*}

The asymptotes are thus,

\begin{equation*} \begin{split} x &= \pm a\\ y &= x \pm \sqrt{2}a \end{split} \end{equation*}

  1. y^3 + x^2y + 2xy^2 -y +1 = 0

There are no asymptotes parallel to y-axis. Degree of equation is 3, so asymptote parallel to x-axis is obtained by equating the coefficients of highest degree term of x to 0.

\begin{equation*} \begin{split} y &= 0 \end{split} \end{equation*}

The equation is of form F_3 + F_1 = 0.

\begin{equation*} \begin{split} \underbrace{y^3 + x^2y + 2xy^2}_{F_3} + \underbrace{(-y +1)}_{F_1} = 0 \end{split} \end{equation*}

By inspection method we can obtain asymptotes by equating F_3 = 0. But we have to make sure that no two linear factors of F_3 are coincident or differ by constant.

The linear factors of F_3 are

\begin{equation*} \begin{split} F_3 &= y^3 + x^2y + 2xy^2 \\ &= y^2(x+y) + xy(x +y) \\ &= y(x+y)(x+y) \end{split} \end{equation*}

Two linear factors are repeated which violates the method of inspection. So we cannot take this approach.

Let y=mx + c be the equation of rest of the asymptotes. Putting x=1 and y=m,

\begin{equation*} \begin{split} \phi_3(m) &= m^3 + m + 2m^2 \\ {\phi'}_3(m) &= 3m^2 + 1 + 4m \\ {\phi''}_3(m) &= 6m + 4 \\ \phi_2(m) &= 0\\ \phi_1(m) &= -m \end{split} \end{equation*}

The slopes of the asymptotes are,

\begin{equation*} \begin{split} \phi_3(m) &= 0\\ m^3 + m + 2m^2 &= 0\\ m(m^2 + 2m + 1) &= 0\\ m(m+1)(m+1) &= 0\\ m &= 0, -1, -1 \end{split} \end{equation*}

When m=0, c=-\dfrac{\phi_2(m)}{{\phi'}_3(m)} = 0. This asymptote is already found. See above.

m=-1 is repeated. So

\begin{equation*} \begin{split} \dfrac{c^2}{2!}{\phi''}_3(m) + c {\phi'}_2(m) + \phi_1(m) = 0 \\ \dfrac{c^2}{2}(6m +4) + 0 -m &= 0\\ c^2 &= 1\\ c &= \pm 1 \end{split} \end{equation*}

Equation of the asymptotes are thus,

\begin{equation*} \begin{split} y &= 0\\ y + x &= \pm 1 \end{split} \end{equation*}

  1. y^3 - xy^2 - x^2y + x^3 + x^2 -y^2 =1

There are no asymptotes parallel to x-axis and y-axis because coefficients of x^3 and y^3 are constants and the degree of equation is 3.

For finding oblique asymptotes in the form y = mx + c, put x=1 and y=m,

\begin{equation*} \begin{split} \phi_3(m) &= m^3 -m^2 -m + 1\\ {\phi'}_3(m) &= 3m^2 -2m -1\\ {\phi''}_3(m) &= 6m - 2 \\ \phi_2(m) &= 1-m^2 \\ {\phi'}_2(m) &= -2m \\ \phi_1(m) &= 0 \end{split} \end{equation*}

The slope of the asymptotes are,

\begin{equation*} \begin{split} \phi_3(m) &= 0\\ m^3 -m^2 -m + 1 &= 0\\ m^2(m-1) -1(m-1) &= 0\\ (m+1)(m-1)(m-1) &= 0\\ m &= 1, 1, -1 \end{split} \end{equation*}

For m=-1,

\begin{equation*} \begin{split} c &= -\dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ &= - \dfrac{1-1}{4} \\ &= 0\\ \end{split} \end{equation*}

For m=1, which is the repeated value of m, for finding c, we have,

\begin{equation*} \begin{split} \dfrac{c^2}{2!}{\phi''}_3(m) + c {\phi'}_2(m) + \phi_1(m) = 0 \\ \dfrac{c^2}{2}(6m-2) - 2mc + 0 &= 0\\ c^2 \times 2 - 2c &= 0\\ c^2 - c &= 0\\ c(c-1) &= 0\\ c &= 0, 1 \end{split} \end{equation*}

The asymptotes are thus,

\begin{equation*} \begin{split} y + x &= 0\\ y &= x\\ y &= x + 1 \end{split} \end{equation*}

  1. x^3 - 2x^2y + xy^2 + x^2 - xy +2 =0

The degree of equation is 3. The equation does not have asymptote parallel to x-axis. The equation does not have y^3, so the asymptote parallel to y-axis is,

\begin{equation*} \begin{split} x &= 0 \end{split} \end{equation*}

Let y = mx +c be the equation of asymptotes. Putting x=1 and y=m, we get

\begin{equation*} \begin{split} \phi_3(m) &= 1 -2m +m^2 \\ {\phi'}_3(m) &= 2m -2 \\ {\phi''}_3(m) &= 2\\ \phi_2(m) &= 1-m \\ {\phi'}_2(m) &= -1 \\ \phi_1(m) &= 0 \end{split} \end{equation*}

The slope of asymptotes are given by,

\begin{equation*} \begin{split} \phi_3(m) &= 0\\ 1 -2m +m^2 &= 0\\ (m-1)(m-1) &= 0\\ m &= 1, 1 \end{split} \end{equation*}

This is a case of two repeated roots i.e. two values of m are same, so

\begin{equation*} \begin{split} \dfrac{c^2}{2!}{\phi''}_3(m) + c {\phi'}_2(m) + \phi_1(m) = 0 \\ \dfrac{c^2}{2} \times 2 - c + 0 &= 0\\ c^2 -c &= 0\\ c(c-1) &= 0\\ c &= 0, 1 \end{split} \end{equation*}

The three asymptotes are thus,

\begin{equation*} \begin{split} x &= 0\\ y &= x\\ y &= x + 1 \end{split} \end{equation*}

  1. x^3 - 2y^3 + 2x^2y -xy^2 + xy -y^2 + 1=0 [TU 2062]

There are no asymptotes parallel to x-axis and y-axis because the degress is 3 and coefficients of x^3 and y^3 are constants.

Let x=1 and y=m,

\begin{equation*} \begin{split} \phi_3(m) &= 1- 2m^3 + 2m -m^2 \\ {\phi'}_3(m) &= -6m^2 + 2 -2m \\ \end{split} \end{equation*}

The slope of the asymptotes are given by,

\begin{equation*} \begin{split} \phi_3(m) &= 0\\ 1- 2m^3 + 2m -m^2 &= 0\\ -1(m^2 -1) - 2m (m^2 -1) &= 0\\ (m^2 - 1)(-1-2m) &= 0\\ (m+1)(m-1)(2m + 1) &= 0\\ m &= 1, -1, -\frac{1}{2} \end{split} \end{equation*}

For c,

\begin{equation*} \begin{split} c &= - \dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ &= -\dfrac{m(1-m)}{-6m^2 -2m +2}\\ c &= \dfrac{m(1-m)}{6m^2 + 2m - 2}\\ \end{split} \end{equation*}

So,

m c
1 0
-1 -1
-\frac{1}{2} \frac{1}{2}

The asymptotes are thus,

  • y= x
  • y + x + 1 = 0
  • x + 2y = 1
  1. (x^2 -y^2)(x+2y+1) + x + y + 1 =0

The equation can be written as (x+y)(x-y)(x+2y+1) + x + y + 1 =0.

The equation has no asymptotes parallel to x-axis. The coefficient of y^3 is constant, so no asymptotes parallel to x-axis.

The equation is of form F_3(x,y) + F_1(x,y) = 0.

\begin{equation*} \begin{split} \underbrace{(x+y)(x-y)(x+2y+1)}_{F_3} + \underbrace{x + y + 1}_{F_1} &= 0 \end{split} \end{equation*}

F_3(x,y) has degree 3 and is product of three different non-repeating linear factors. By method of inspection, thus the asymptotes are obtained by equating F_3(x,y) = 0,

\begin{equation*} \begin{split} x + y &= 0\\ x - y &= 0\\ x + 2y + 1 &= 0 \end{split} \end{equation*}

  1. x(x-y)^2 -3(x^2 - y^2) + 8y = 0 [TU 2060]

Coefficient of x^3 is constant, no asymptotes parallel to x-axis.

The degree of equation is 3. There is no y^3, so asymptotes parallel to y-axis is obtained by equating the coefficients of highest degree terms to zero.

\begin{equation*} \begin{split} x + 3 &= 0 \end{split} \end{equation*}

The equation can be written as x(x-y)^2 -3(x-y)(x+y) + 8y = 0 which is of form

(y-m_1 x)^2 F_{n-2} + (y-m_1 x) G_{n-2} + P_{n-2} = 0.

Dividing the equation both sides by, x,

\begin{equation*} \begin{split} (x-y)^2 -3(x-y)(x+y)\dfrac{1}{x} + 8\dfrac{y}{x} &= 0\\ \end{split} \end{equation*}

The two asymptotes parallel to x-y are,

\begin{equation*} \begin{split} (x-y)^2 -3(x-y)\lim_{\substack{x \to \infty \\ \frac{y}{x} \to 1}} (x+y)\dfrac{1}{x} + \lim_{\substack{x \to \infty \\ \frac{y}{x} \to 1}} 8\dfrac{y}{x} &= 0\\ (x-y)^2 -3(x-y)\lim_{\substack{x \to \infty \\ \frac{y}{x} \to 1}} \{1 + \frac{y}{x}\} + 8 &= 0\\ (x-y)^2 -3(x-y)\times 2 + 8 &= 0\\ (x-y)^2 -6(x-y) + 8 &= 0\\ \end{split} \end{equation*}

Solving for x-y,

\begin{equation*} \begin{split} x -y &= \dfrac{6 \pm \sqrt{36-32}}{2}\\ x -y &= 2, 4 \end{split} \end{equation*}

So the asymptotes parallel to x-y are

\begin{equation*} \begin{split} x - y &= 2 \\ x - y &= 4 \end{split} \end{equation*}

Thus, the given equation has maximum of three asymptotes, all has been found.

  1. x^3 + 3x^2 y - 4y^3 -x + y + 3= 0 [TU 2054, 2055]

There are no asymptotes parallel to x-axis and y-axis.

Lets put x=1 and y=m. Then,

\begin{equation*} \begin{split} \phi_3(m) &= 1 + 3m - 4m^3 \\ {\phi'}_3(m) &= -12 m^2 + 3\\ {\phi''}_3(m) &= -24m \\ \phi_2(m) &= 0\\ \phi_1(m) &= m - 1\\ \end{split} \end{equation*}

The slope of the asymptotes are given by

\begin{equation*} \begin{split} \phi_3(m) &= 0 \\ 1 + 3m - 4m^3 &= 0\\ (1-m)(4m^2 + 4m +1) &= 0\\ (1-m)(2m + 1)^2 &= 0\\ m &= 1, -\frac{1}{2}, -\frac{1}{2} \end{split} \end{equation*}

For m=1,

\begin{equation*} \begin{split} c &= -\dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ &= 0 \end{split} \end{equation*}

So, for m=1, y=x is an asymptote.

Now two m values are same i.e -\frac{1}{2}. So to find c,

\begin{equation*} \begin{split} \dfrac{c^2}{2!}{\phi''}_3(m) + c {\phi'}_2(m) + \phi_1(m) &= 0\\ \dfrac{c^2}{2!}(-24m) + 0 + m - 1 &= 0\\ \end{split} \end{equation*}

Putting c = -\frac{1}{2},

\begin{equation*} \begin{split} -12c^2 \times \left(-\frac{1}{2}\right) -\frac{1}{2} -1 &= 0\\ 6c^2 - \frac{3}{2} &= 0\\ c^2 &= \frac{1}{4}\\ c &= \pm \frac{1}{2} \end{split} \end{equation*}

So y = -\frac{1}{2}x \pm \frac{1}{2} are the asymptotes corresponding to m = -\frac{1}{2}. Thus all three asymptotes of the equation are,

\begin{equation*} \begin{split} x - y &= 0 \\ x + 2y + 1 &= 0\\ x + 2y - 1 &= 0 \end{split} \end{equation*}

  1. (x-1)(x-2)(x+y) + x^2 + x+ 1=0 [TU 2059]

The degree of equation is 3. There is no y^3, so asymptotes parallel to y-axis is obtained by equating the coefficients of highest degree term of y, i.e.

\begin{equation*} \begin{split} (x-1)(x-2) &= 0\\ x - 1 &= 0, \quad x-2 &= 0 \end{split} \end{equation*}

The coefficient of x^3 is constant. So no asymptotes parallel to x-axis.

The equation is of form (y-m_1 x)F_{n-1} + P_{n-1} = 0.

\begin{equation*} \begin{split} (x+y)\underbrace{(x-1)(x-2)}_{F_{3-1}} + \underbrace{x^2 + x+ 1}_{P_{3-1}} &= 0 \end{split} \end{equation*}

So the asymptote parallel to x+y is obtained by,

\begin{equation*} \begin{split} x+y + \lim_{\substack{x \to \infty \\ \frac{y}{x} \to -1}} \dfrac{x^2 + x + 1}{(x-1)(x-2)} &= 0\\ x+y + \lim_{\substack{x \to \infty \\ \frac{y}{x} \to -1}} \dfrac{1 + \frac{1}{x} + \frac{1}{x^2}}{(1-\frac{1}{x})(1-\frac{2}{x})} &= 0\\ x + y + 1 &= 0 \end{split} \end{equation*}

Thus three asymptotes are,

  • x-1 = 0
  • x-2 = 0
  • x + y + 1 = 0

14.1.2 Question 5 [TU 2059]

Show that the asymptotes of the curve x^2y^2 = a^2(x^2 + y^2) form a square of side 2a.

The equation can be written as,

\begin{equation*} \begin{split} x^2 y^2 - a^2 x^2 - a^2 y^2 &= 0\\ \end{split} \end{equation*}

Degree of equation is 4. No x^4 and y^4 terms. So asymptotes parallel to x-axis is obtained by equating the coefficients of highest degree term of x.

\begin{equation*} \begin{split} y^2 - a^2 &= 0\\ y &= \pm a \end{split} \end{equation*}

So asymptotes parallel to y-axis is obtained by equating the coefficients of highest degree term of y.

\begin{equation*} \begin{split} x^2 - a^2 &= 0\\ x &= \pm a \end{split} \end{equation*}

The distance between the asymptotes x = a and x = -a is 2a.

Similarly, the distance between the asymptotes y = a and y = -a is 2a. Thus, two distances are equal and hence it is a square. See the graph below,

Curve and asymptotes of $x^2y^2 = a^2(x^2 + y^2)$

Figure 14.1: Curve and asymptotes of x^2y^2 = a^2(x^2 + y^2)

Trigonometry to remember

  • \sin (n\pi + \theta) = (-1)^n \sin \theta
  • \cos (n\pi + \theta) = (-1)^n \cos \theta
Function Domain Range
\sin^{-1}(x) [-1, 1] [-\pi/2, \pi/2]
\cos^{-1}(x) [-1, 1] [0, \pi]
\tan^{-1}(x) (-\infty, \infty) (-\pi/2, \pi/2)
\cot^{-1}(x) (-\infty, \infty) (0, \pi)
\sec^{-1}(x) (-\infty, -1]\cup [1, \infty) [0, \pi/2) \cup (\pi/2, \pi]
\csc^{-1}(x) (-\infty, -1]\cup [1, \infty) [-\pi/2, 0) \cup (0, \pi/2]

Rules for finding asymptotes of polar curves

  • Put u =\frac{1}{r} and write the equation in form u = F(\theta).
  • Find \theta for which F(\theta) = 0. This value will be \theta_1.
  • Find F'(\theta_1).
  • The equation of the asymptotes are then obtained by plugging by \theta_1 and F'(\theta_1) pairs in the equation r \sin (\theta - \theta_1) = \dfrac{1}{F'(\theta_1)}.

14.1.3 Question 6

Find the asymptotes of the curves

  1. 2r^2 = \tan 2\theta

Putting u=\dfrac{1}{r}, then

\begin{equation*} \begin{split} u^2 &= 2 \cot 2\theta \\ u &= \sqrt{2 \cot 2\theta} = F(\theta) \end{split} \end{equation*}

When r \rightarrow \infty, u \rightarrow 0, or

\begin{equation*} \begin{split} \sqrt{2 \cot 2\theta} &= 0\\ \cot 2 \theta &= 0\\ 2 \theta &= \cot^{-1} 0\\ 2 \theta &= \dfrac{\pi}{2}, \text{ not } -\pi/2 \text{ because the domain} \\ & \text{of arccot function is closed interval } (0, \pi)\\ \theta &= \dfrac{\pi}{4} \end{split} \end{equation*}

i.e. when r \rightarrow \infty, \theta \rightarrow \pi/4. So,

\begin{equation*} \begin{split} \theta_1 &= \dfrac{\pi}{4} \end{split} \end{equation*}

Differentiating F(\theta) w.r.t \theta,

\begin{equation*} \begin{split} F'(\theta) &= \dfrac{du}{d\theta} = \dfrac{-4\csc^2 2 \theta}{\sqrt{2\cot 2\theta}}\\ F'(\theta_1 = \pi/4) &= \dfrac{-4}{0} = \infty \end{split} \end{equation*}

The equation of the asymptote is then given by,

\begin{equation*} \begin{split} r\sin (\theta - \theta_1) &= \dfrac{1}{F'(\theta_1)}\\ r\sin (\theta - \pi/4) &= \dfrac{1}{\infty}\\ r\sin (\theta - \pi/4) &= 0\\ \theta - \pi/4 &= \sin^{-1}0\\ \theta - \pi/4 &= 0 , \text{ domain of arcsin is } [-\pi/2, \pi/2]\\ \theta &= \dfrac{\pi}{4} \end{split} \end{equation*}

  1. r\theta = a

Put u = \dfrac{1}{r}.

So u = \dfrac{\theta}{a} = F(\theta).

When r \rightarrow \infty, u \rightarrow 0 or

\begin{equation*} \begin{split} \frac{\theta}{a} &= 0\\ \theta &= 0 \end{split} \end{equation*}

i.e. when r \rightarrow \infty, \theta \rightarrow 0. So

\begin{equation*} \begin{split} \theta_1 &= 0 \end{split} \end{equation*}

Also,

\begin{equation*} \begin{split} F'(\theta) &= \dfrac{1}{a}\\ F'(\theta_1 = 0) &= \dfrac{1}{a} \end{split} \end{equation*}

The equation of the asymptote is thus,

\begin{equation*} \begin{split} r \sin (\theta - \theta_1) &= \dfrac{1}{F'(\theta_1)}\\ r \sin (\theta - 0) &= a\\ r \sin \theta &= a \end{split} \end{equation*}

Second method

Here, the given equation is a hyperbolic spiral. The equation can be written as,

\begin{equation*} \begin{split} r &= \dfrac{a}{\theta}, \text{ where }\theta > 0\\ \end{split} \end{equation*}

We know,

\begin{equation*} \begin{split} x &= r \cos \theta\\ x &= a \dfrac{\cos \theta}{\theta}, \text{ from above}\\\\ \end{split} \end{equation*}

Lets see the behavior of x when \theta \rightarrow 0^{+}.

\begin{equation*} \begin{split} x &= a \lim_{\theta \to 0^{+}} \dfrac{\cos \theta}{\theta}\\ &= +\infty \end{split} \end{equation*}

Similarly,

\begin{equation*} \begin{split} y & = r \sin \theta \\ y &= a \dfrac{\sin \theta}{\theta} \end{split} \end{equation*}

Lets see the behavior of y when \theta \rightarrow 0^{+}.

\begin{equation*} \begin{split} y &= a\lim_{\theta \to 0^{+}} \dfrac{\sin \theta}{\theta}\\ y &= a \times 1\\ y &= a \\ r \sin \theta &= a \end{split} \end{equation*}

Thus, y =a or r \sin \theta = a is the horizontal asymptote of the given equation.

  1. r \sin \theta = a

Transforming the equation into cartesian form, the equation can be written as,

\begin{equation*} \begin{split} y &= a \end{split} \end{equation*}

This is equation of a straight line. Asymptote in case of straight line does not make sense. No asymptote !

  1. r \theta \cos \theta = a \cos 2\theta

Asymptotes are the lines which touch the curve at infinity.

Putting u=\dfrac{1}{r}, then

\begin{equation*} \begin{split} u &= \dfrac{\theta \cos \theta}{a \cos 2\theta} = F(\theta)\\ \end{split} \end{equation*}

When r \rightarrow \infty, u \rightarrow 0, or

\begin{equation*} \begin{split} \dfrac{\theta \cos \theta}{a\cos 2 \theta} &= 0\\ \theta \cos \theta &= 0\\ \theta &= 0, \quad \cos \theta = 0\\ \theta &= 0, \quad \theta = \cos^{-1}0\\ \theta &= 0, \quad \theta = \dfrac{\pi}{2},\text{ and not} -\dfrac{\pi}{2} \\ & \text{ because inverse cos function is defined only in the interval }[0, \pi] \end{split} \end{equation*}

i.e. when r \rightarrow \infty, \theta \rightarrow 0, \dfrac{\pi}{2}. So,

\begin{equation*} \begin{split} \theta_1 &= 0, \dfrac{\pi}{2} \end{split} \end{equation*}

Differentiating F(\theta) w.r.t \theta,

\begin{equation*} \begin{split} F'(\theta) &= \dfrac{a\cos 2 \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (\theta \cos \theta) - \theta \cos \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (a\cos 2\theta)}{a^2 \cos^2 2\theta}\\ F'(\theta) &= \dfrac{a\cos 2\theta (\cos \theta - \theta \sin \theta) + 2a\theta \cos \theta \sin 2 \theta}{a^2\cos^2 2 \theta}\\ \end{split} \end{equation*}

\theta_1 F'(\theta_1)
0 \frac{1}{a}
\dfrac{\pi}{2} \dfrac{\pi}{2a}

The equation of the asymptote in case of polar curves is given by,

\begin{equation*} \begin{split} r \sin (\theta - \theta_1) &= \dfrac{1}{F'(\theta_1)} \end{split} \end{equation*}

So in our case, asymptotes are,

\theta_1 Equation of asymptote
0 r \sin (\theta - 0)= \dfrac{1}{\frac{1}{a}}\\ r\sin \theta= a
\pi/2 r \sin \left(\theta - \frac{\pi}{2}\right)=\frac{2a}{\pi}\\ 2a + \pi r \cos \theta = 0