Chapter 14 Asymptotes-II

14.1 Exercise 7

14.1.1 Question 4

Find the asymptotes of the following curves

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

There is no \(x^3\), so asymptotes parallel to \(x\)-axis is given by equating coefficients of highest degree term of \(x\).

\[\begin{equation*} \begin{split} y^2 - y &= 0\\ y(y-1) &= 0\\ y &= 0, \quad y= 1 \end{split} \end{equation*}\]

Similarly, no \(y^3\) is present. Asymptotes parallel to \(y\)-axis is given by,

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

Thus four asymptotes of the equation are,

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

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

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 \(y^2\) to zero.

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

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

\[\begin{equation*} \begin{split} \phi_4(m) &= 1-2m + m^2\\ {\phi'}_4(m) &= 2m -2 \\ {\phi''}_4(m) &= 2 \\ \phi_3(m) &= 0\\ \phi_2(m) &= -a^2 -a^2m^2 \end{split} \end{equation*}\]

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
\(0\) \[r \sin (\theta - 0)= \dfrac{1}{\frac{1}{a}}\\ r\sin \theta= a \]
\(\pi/2\) \[r \sin \left(\theta - \frac{\pi}{2}\right)=\frac{2a}{\pi}\\ 2a + \pi r \cos \theta = 0\]