Chapter 14 Asymptotes-II

14.1 Exercise 7

14.1.1 Question 4

Find the asymptotes of the following curves

$x^{2} y^{2} - x^{2} y - x y^{2} + x + y + 1 = 0$

There is no $x^{3}$ , so asymptotes parallel to $x$ -axis is given by equating coefficients of highest degree term of $x$ .

$\begin{aligned} y^{2} - y & = 0 \\ y (y - 1) & = 0 \\ y & = 0, y = 1 \end{aligned}$

Similarly, no $y^{3}$ is present. Asymptotes parallel to $y$ -axis is given by,

$\begin{aligned} x^{2} - x & = 0 \\ x (x - 1) & = 0 \\ x & = 0, x = 1 \end{aligned}$

Thus four asymptotes of the equation are,

$\begin{aligned} x & = 0 \\ x & = 1 \\ y & = 0 \\ y & = 1 \end{aligned}$

$x^{2} (x - y)^{2} - a^{2} (x^{2} + y^{2}) = 0$

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.

$\begin{aligned} x^{2} - a^{2} & = 0 \\ x & = \pm a \end{aligned}$

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

$\begin{aligned} ϕ_{4} (m) & = 1 - 2 m + m^{2} \\ {ϕ^{'}}_{4} (m) & = 2 m - 2 \\ {ϕ^{″}}_{4} (m) & = 2 \\ ϕ_{3} (m) & = 0 \\ ϕ_{2} (m) & = - a^{2} - a^{2} m^{2} \end{aligned}$

The slope of the asymptotes can be found by,

$\begin{aligned} ϕ_{4} (m) & = 0 \\ 1 - 2 m + m^{2} & = 0 \\ (m - 1) (m - 1) & = 0 \\ m & = 1, 1 \end{aligned}$

This is a case of repeated factors, so

$\begin{aligned} \frac{c^{2}}{2!} {ϕ^{″}}_{4} (m) + c {ϕ^{'}}_{3} (m) + ϕ_{2} (m) & = 0 \\ \frac{c^{2}}{2} \times 2 + 0 - a^{2} - a^{2} m^{2} & = 0 \\ c^{2} - a^{2} - a^{2} & = 0 \\ c & = \pm \sqrt{2} a \end{aligned}$

The asymptotes are thus,

$\begin{aligned} x & = \pm a \\ y & = x \pm \sqrt{2} a \end{aligned}$

$y^{3} + x^{2} y + 2 x y^{2} - y + 1 = 0$

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 $0$ .

$\begin{aligned} y & = 0 \end{aligned}$

The equation is of form $F_{3} + F_{1} = 0$ .

$\begin{aligned} \underset{F_{3}}{\underset{⏟}{y^{3} + x^{2} y + 2 x y^{2}}} + \underset{F_{1}}{\underset{⏟}{(- y + 1)}} = 0 \end{aligned}$

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

$\begin{aligned} F_{3} & = y^{3} + x^{2} y + 2 x y^{2} \\ = y^{2} (x + y) + x y (x + y) \\ = y (x + y) (x + y) \end{aligned}$

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

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

$\begin{aligned} ϕ_{3} (m) & = m^{3} + m + 2 m^{2} \\ {ϕ^{'}}_{3} (m) & = 3 m^{2} + 1 + 4 m \\ {ϕ^{″}}_{3} (m) & = 6 m + 4 \\ ϕ_{2} (m) & = 0 \\ ϕ_{1} (m) & = - m \end{aligned}$

The slopes of the asymptotes are,

$\begin{aligned} ϕ_{3} (m) & = 0 \\ m^{3} + m + 2 m^{2} & = 0 \\ m (m^{2} + 2 m + 1) & = 0 \\ m (m + 1) (m + 1) & = 0 \\ m & = 0, - 1, - 1 \end{aligned}$

When $m = 0$ , $c = - \frac{ϕ_{2} (m)}{{ϕ^{'}}_{3} (m)} = 0$ . This asymptote is already found. See above.

$m = - 1$ is repeated. So

$\begin{aligned} \frac{c^{2}}{2!} {ϕ^{″}}_{3} (m) + c {ϕ^{'}}_{2} (m) + ϕ_{1} (m) = 0 \\ \frac{c^{2}}{2} (6 m + 4) + 0 - m & = 0 \\ c^{2} & = 1 \\ c & = \pm 1 \end{aligned}$

Equation of the asymptotes are thus,

$\begin{aligned} y & = 0 \\ y + x & = \pm 1 \end{aligned}$

$y^{3} - x y^{2} - x^{2} y + x^{3} + x^{2} - y^{2} = 1$

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 $y = m x + c$ , put $x = 1$ and $y = m$ ,

$\begin{aligned} ϕ_{3} (m) & = m^{3} - m^{2} - m + 1 \\ {ϕ^{'}}_{3} (m) & = 3 m^{2} - 2 m - 1 \\ {ϕ^{″}}_{3} (m) & = 6 m - 2 \\ ϕ_{2} (m) & = 1 - m^{2} \\ {ϕ^{'}}_{2} (m) & = - 2 m \\ ϕ_{1} (m) & = 0 \end{aligned}$

The slope of the asymptotes are,

$\begin{aligned} ϕ_{3} (m) & = 0 \\ m^{3} - m^{2} - m + 1 & = 0 \\ m^{2} (m - 1) - 1 (m - 1) & = 0 \\ (m + 1) (m - 1) (m - 1) & = 0 \\ m & = 1, 1, - 1 \end{aligned}$

For $m = - 1$ ,

$\begin{aligned} c & = - \frac{ϕ_{2} (m)}{{ϕ^{'}}_{3} (m)} \\ = - \frac{1 - 1}{4} \\ = 0 \end{aligned}$

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

$\begin{aligned} \frac{c^{2}}{2!} {ϕ^{″}}_{3} (m) + c {ϕ^{'}}_{2} (m) + ϕ_{1} (m) = 0 \\ \frac{c^{2}}{2} (6 m - 2) - 2 m c + 0 & = 0 \\ c^{2} \times 2 - 2 c & = 0 \\ c^{2} - c & = 0 \\ c (c - 1) & = 0 \\ c & = 0, 1 \end{aligned}$

The asymptotes are thus,

$\begin{aligned} y + x & = 0 \\ y & = x \\ y & = x + 1 \end{aligned}$

$x^{3} - 2 x^{2} y + x y^{2} + x^{2} - x y + 2 = 0$

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,

$\begin{aligned} x & = 0 \end{aligned}$

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

$\begin{aligned} ϕ_{3} (m) & = 1 - 2 m + m^{2} \\ {ϕ^{'}}_{3} (m) & = 2 m - 2 \\ {ϕ^{″}}_{3} (m) & = 2 \\ ϕ_{2} (m) & = 1 - m \\ {ϕ^{'}}_{2} (m) & = - 1 \\ ϕ_{1} (m) & = 0 \end{aligned}$

The slope of asymptotes are given by,

$\begin{aligned} ϕ_{3} (m) & = 0 \\ 1 - 2 m + m^{2} & = 0 \\ (m - 1) (m - 1) & = 0 \\ m & = 1, 1 \end{aligned}$

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

$\begin{aligned} \frac{c^{2}}{2!} {ϕ^{″}}_{3} (m) + c {ϕ^{'}}_{2} (m) + ϕ_{1} (m) = 0 \\ \frac{c^{2}}{2} \times 2 - c + 0 & = 0 \\ c^{2} - c & = 0 \\ c (c - 1) & = 0 \\ c & = 0, 1 \end{aligned}$

The three asymptotes are thus,

$\begin{aligned} x & = 0 \\ y & = x \\ y & = x + 1 \end{aligned}$

$x^{3} - 2 y^{3} + 2 x^{2} y - x y^{2} + x y - y^{2} + 1 = 0$ [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 $y = m$ ,

$\begin{aligned} ϕ_{3} (m) & = 1 - 2 m^{3} + 2 m - m^{2} \\ {ϕ^{'}}_{3} (m) & = - 6 m^{2} + 2 - 2 m \end{aligned}$

The slope of the asymptotes are given by,

$\begin{aligned} ϕ_{3} (m) & = 0 \\ 1 - 2 m^{3} + 2 m - m^{2} & = 0 \\ - 1 (m^{2} - 1) - 2 m (m^{2} - 1) & = 0 \\ (m^{2} - 1) (- 1 - 2 m) & = 0 \\ (m + 1) (m - 1) (2 m + 1) & = 0 \\ m & = 1, - 1, - \frac{1}{2} \end{aligned}$

For $c$ ,

$\begin{aligned} c & = - \frac{ϕ_{2} (m)}{{ϕ^{'}}_{3} (m)} \\ = - \frac{m (1 - m)}{- 6 m^{2} - 2 m + 2} \\ c & = \frac{m (1 - m)}{6 m^{2} + 2 m - 2} \end{aligned}$

So,

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

The asymptotes are thus,

$y = x$
$y + x + 1 = 0$
$x + 2 y = 1$

$(x^{2} - y^{2}) (x + 2 y + 1) + x + y + 1 = 0$

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

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 $F_{3} (x, y) + F_{1} (x, y) = 0$ .

$\begin{aligned} \underset{F_{3}}{\underset{⏟}{(x + y) (x - y) (x + 2 y + 1)}} + \underset{F_{1}}{\underset{⏟}{x + y + 1}} & = 0 \end{aligned}$

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

$\begin{aligned} x + y & = 0 \\ x - y & = 0 \\ x + 2 y + 1 & = 0 \end{aligned}$

$x (x - y)^{2} - 3 (x^{2} - y^{2}) + 8 y = 0$ [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.

$\begin{aligned} x + 3 & = 0 \end{aligned}$

The equation can be written as $x (x - y)^{2} - 3 (x - y) (x + y) + 8 y = 0$ 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, $x$ ,

$\begin{aligned} (x - y)^{2} - 3 (x - y) (x + y) \frac{1}{x} + 8 \frac{y}{x} & = 0 \end{aligned}$

The two asymptotes parallel to $x - y$ are,

$\begin{aligned} (x - y)^{2} - 3 (x - y) lim_{\begin{matrix} x \to \infty \\ \frac{y}{x} \to 1 \end{matrix}} (x + y) \frac{1}{x} + lim_{\begin{matrix} x \to \infty \\ \frac{y}{x} \to 1 \end{matrix}} 8 \frac{y}{x} & = 0 \\ (x - y)^{2} - 3 (x - y) lim_{\begin{matrix} x \to \infty \\ \frac{y}{x} \to 1 \end{matrix}} {1 + \frac{y}{x}} + 8 & = 0 \\ (x - y)^{2} - 3 (x - y) \times 2 + 8 & = 0 \\ (x - y)^{2} - 6 (x - y) + 8 & = 0 \end{aligned}$

Solving for $x - y$ ,

$\begin{aligned} x - y & = \frac{6 \pm \sqrt{36 - 32}}{2} \\ x - y & = 2, 4 \end{aligned}$

So the asymptotes parallel to $x - y$ are

$\begin{aligned} x - y & = 2 \\ x - y & = 4 \end{aligned}$

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

$x^{3} + 3 x^{2} y - 4 y^{3} - x + y + 3 = 0$ [TU 2054, 2055]

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

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

$\begin{aligned} ϕ_{3} (m) & = 1 + 3 m - 4 m^{3} \\ {ϕ^{'}}_{3} (m) & = - 12 m^{2} + 3 \\ {ϕ^{″}}_{3} (m) & = - 24 m \\ ϕ_{2} (m) & = 0 \\ ϕ_{1} (m) & = m - 1 \end{aligned}$

The slope of the asymptotes are given by

$\begin{aligned} ϕ_{3} (m) & = 0 \\ 1 + 3 m - 4 m^{3} & = 0 \\ (1 - m) (4 m^{2} + 4 m + 1) & = 0 \\ (1 - m) (2 m + 1)^{2} & = 0 \\ m & = 1, - \frac{1}{2}, - \frac{1}{2} \end{aligned}$

For $m = 1$ ,

$\begin{aligned} c & = - \frac{ϕ_{2} (m)}{{ϕ^{'}}_{3} (m)} \\ = 0 \end{aligned}$

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

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

$\begin{aligned} \frac{c^{2}}{2!} {ϕ^{″}}_{3} (m) + c {ϕ^{'}}_{2} (m) + ϕ_{1} (m) & = 0 \\ \frac{c^{2}}{2!} (- 24 m) + 0 + m - 1 & = 0 \end{aligned}$

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

$\begin{aligned} - 12 c^{2} \times (- \frac{1}{2}) - \frac{1}{2} - 1 & = 0 \\ 6 c^{2} - \frac{3}{2} & = 0 \\ c^{2} & = \frac{1}{4} \\ c & = \pm \frac{1}{2} \end{aligned}$

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,

$\begin{aligned} x - y & = 0 \\ x + 2 y + 1 & = 0 \\ x + 2 y - 1 & = 0 \end{aligned}$

$(x - 1) (x - 2) (x + y) + x^{2} + x + 1 = 0$ [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.

$\begin{aligned} (x - 1) (x - 2) & = 0 \\ x - 1 & = 0, x - 2 & = 0 \end{aligned}$

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

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

$\begin{aligned} (x + y) \underset{F_{3 - 1}}{\underset{⏟}{(x - 1) (x - 2)}} + \underset{P_{3 - 1}}{\underset{⏟}{x^{2} + x + 1}} & = 0 \end{aligned}$

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

$\begin{aligned} x + y + lim_{\begin{matrix} x \to \infty \\ \frac{y}{x} \to - 1 \end{matrix}} \frac{x^{2} + x + 1}{(x - 1) (x - 2)} & = 0 \\ x + y + lim_{\begin{matrix} x \to \infty \\ \frac{y}{x} \to - 1 \end{matrix}} \frac{1 + \frac{1}{x} + \frac{1}{x^{2}}}{(1 - \frac{1}{x}) (1 - \frac{2}{x})} & = 0 \\ x + y + 1 & = 0 \end{aligned}$

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 $x^{2} y^{2} = a^{2} (x^{2} + y^{2})$ form a square of side $2 a$ .

The equation can be written as,

$\begin{aligned} x^{2} y^{2} - a^{2} x^{2} - a^{2} y^{2} & = 0 \end{aligned}$

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 $x$ .

$\begin{aligned} y^{2} - a^{2} & = 0 \\ y & = \pm a \end{aligned}$

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

$\begin{aligned} x^{2} - a^{2} & = 0 \\ x & = \pm a \end{aligned}$

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

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

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

Trigonometry to remember

$\sin (n π + θ) = (- 1)^{n} \sin θ$
$\cos (n π + θ) = (- 1)^{n} \cos θ$

Function	Domain	Range
$\sin^{- 1} (x)$	$[- 1, 1]$	$[- π / 2, π / 2]$
$\cos^{- 1} (x)$	$[- 1, 1]$	$[0, π]$
$\tan^{- 1} (x)$	$(- \infty, \infty)$	$(- π / 2, π / 2)$
$\cot^{- 1} (x)$	$(- \infty, \infty)$	$(0, π)$
$\sec^{- 1} (x)$	$(- \infty, - 1] \cup [1, \infty)$	$[0, π / 2) \cup (π / 2, π]$
$\csc^{- 1} (x)$	$(- \infty, - 1] \cup [1, \infty)$	$[- π / 2, 0) \cup (0, π / 2]$

Rules for finding asymptotes of polar curves

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

14.1.3 Question 6

Find the asymptotes of the curves

$2 r^{2} = \tan 2 θ$

Putting $u = \frac{1}{r}$ , then

$\begin{aligned} u^{2} & = 2 \cot 2 θ \\ u & = \sqrt{2 \cot 2 θ} = F (θ) \end{aligned}$

When $r \to \infty$ , $u \to 0$ , or

$\begin{aligned} \sqrt{2 \cot 2 θ} & = 0 \\ \cot 2 θ & = 0 \\ 2 θ & = \cot^{- 1} 0 \\ 2 θ & = \frac{π}{2}, not - π / 2 because the domain \\ of arccot function is closed interval (0, π) \\ θ & = \frac{π}{4} \end{aligned}$

i.e. when $r \to \infty$ , $θ \to π / 4$ . So,

$\begin{aligned} θ_{1} & = \frac{π}{4} \end{aligned}$

Differentiating $F (θ)$ w.r.t $θ$ ,

$\begin{aligned} F^{'} (θ) & = \frac{d u}{d θ} = \frac{- 4 \csc^{2} 2 θ}{\sqrt{2 \cot 2 θ}} \\ F^{'} (θ_{1} = π / 4) & = \frac{- 4}{0} = \infty \end{aligned}$

The equation of the asymptote is then given by,

$\begin{aligned} r \sin (θ - θ_{1}) & = \frac{1}{F^{'} (θ_{1})} \\ r \sin (θ - π / 4) & = \frac{1}{\infty} \\ r \sin (θ - π / 4) & = 0 \\ θ - π / 4 & = \sin^{- 1} 0 \\ θ - π / 4 & = 0, domain of arcsin is [- π / 2, π / 2] \\ θ & = \frac{π}{4} \end{aligned}$

$r θ = a$

Put $u = \frac{1}{r}$ .

So $u = \frac{θ}{a} = F (θ)$ .

When $r \to \infty$ , $u \to 0$ or

$\begin{aligned} \frac{θ}{a} & = 0 \\ θ & = 0 \end{aligned}$

i.e. when $r \to \infty$ , $θ \to 0$ . So

$\begin{aligned} θ_{1} & = 0 \end{aligned}$

Also,

$\begin{aligned} F^{'} (θ) & = \frac{1}{a} \\ F^{'} (θ_{1} = 0) & = \frac{1}{a} \end{aligned}$

The equation of the asymptote is thus,

$\begin{aligned} r \sin (θ - θ_{1}) & = \frac{1}{F^{'} (θ_{1})} \\ r \sin (θ - 0) & = a \\ r \sin θ & = a \end{aligned}$

Second method

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

$\begin{aligned} r & = \frac{a}{θ}, where θ > 0 \end{aligned}$

We know,

$\begin{aligned} x & = r \cos θ \\ x & = a \frac{\cos θ}{θ}, from above \end{aligned}$

Lets see the behavior of $x$ when $θ \to 0^{+}$ .

$\begin{aligned} x & = a lim_{θ \to 0^{+}} \frac{\cos θ}{θ} \\ = + \infty \end{aligned}$

Similarly,

$\begin{aligned} y & = r \sin θ \\ y & = a \frac{\sin θ}{θ} \end{aligned}$

Lets see the behavior of $y$ when $θ \to 0^{+}$ .

$\begin{aligned} y & = a lim_{θ \to 0^{+}} \frac{\sin θ}{θ} \\ y & = a \times 1 \\ y & = a \\ r \sin θ & = a \end{aligned}$

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

$r \sin θ = a$

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

$\begin{aligned} y & = a \end{aligned}$

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

$r θ \cos θ = a \cos 2 θ$

Asymptotes are the lines which touch the curve at infinity.

Putting $u = \frac{1}{r}$ , then

$\begin{aligned} u & = \frac{θ \cos θ}{a \cos 2 θ} = F (θ) \end{aligned}$

When $r \to \infty$ , $u \to 0$ , or

$\begin{aligned} \frac{θ \cos θ}{a \cos 2 θ} & = 0 \\ θ \cos θ & = 0 \\ θ & = 0, \cos θ = 0 \\ θ & = 0, θ = \cos^{- 1} 0 \\ θ & = 0, θ = \frac{π}{2}, and not - \frac{π}{2} \\ because inverse cos function is defined only in the interval [0, π] \end{aligned}$

i.e. when $r \to \infty$ , $θ \to 0, \frac{π}{2}$ . So,

$\begin{aligned} θ_{1} & = 0, \frac{π}{2} \end{aligned}$

Differentiating $F (θ)$ w.r.t $θ$ ,

$\begin{aligned} F^{'} (θ) & = \frac{a \cos 2 θ \frac{d}{d θ} (θ \cos θ) - θ \cos θ \frac{d}{d θ} (a \cos 2 θ)}{a^{2} \cos^{2} 2 θ} \\ F^{'} (θ) & = \frac{a \cos 2 θ (\cos θ - θ \sin θ) + 2 a θ \cos θ \sin 2 θ}{a^{2} \cos^{2} 2 θ} \end{aligned}$

$θ_{1}$	$F^{'} (θ_{1})$
$0$	$\frac{1}{a}$
$\frac{π}{2}$	$\frac{π}{2 a}$

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

$\begin{aligned} r \sin (θ - θ_{1}) & = \frac{1}{F^{'} (θ_{1})} \end{aligned}$

So in our case, asymptotes are,

$θ_{1}$	Equation of asymptote
$0$	$r \sin (θ - 0) = \frac{1}{\frac{1}{a}} r \sin θ = a$
$π / 2$	$r \sin (θ - \frac{π}{2}) = \frac{2 a}{π} 2 a + π r \cos θ = 0$