Chapter 13 Asymptotes-I

Total number of asymptotes cannot exceed the highest degree of the equation.
It is not necessary that all the curves have asymptotes.
If a curve of $n^\text{th}$ degree has $n$ asymptotes, then they cut the curve in $n(n-2)$ points.

Tips on finding vertical and horizontal asymptotes can be found here.

13.1 Exercise 7

13.1.1 Question 1

Find the asymptotes of the curves

$y= \dfrac{x}{(x-1)^2(x-2)}$

Here, $y \rightarrow \infty$, when $x \rightarrow 1$.

Also, $y \rightarrow \infty$, when $x \rightarrow 2$, $\quad y \rightarrow 0$ when $x \rightarrow \infty$.

So, there are three asymptotes.

\[\begin{equation*} \begin{split} x &=1 \\ x &= 2\\ y &= 0 \end{split} \end{equation*}\]

$y=\dfrac{x^2}{x^2 + 1}$

The equation is of degree $3$. Here $x^3$ is absent, so asymptotes parallel to $x$-axis is obtained by equating to zero the coefficients of highest degree term in $x$.

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

Also $y^3$ is absent, so asymptotes parallel to $y$-axis is obtained by equating to zero the coefficients of highest degree term in $y$.

\[\begin{equation*} \begin{split} x^2 + 1 &= 0 \end{split} \end{equation*}\]

This doesn’t have real roots. So $y=1$ is the only asymptote.

In other words, there is no value of $x$ such that $y \rightarrow \infty$. So, the equation doesn’t have vertical asymptote, horizontal asymptote only despite the degree of equation being $3$.

$y = \dfrac{4x^2 + 4x -3}{x^2 -4x +3}$

The equation can be written as $y = \dfrac{(2x + 3)(2x-1)}{(x-1)(x-3)}$.

Here, $y \rightarrow \infty$, when $x \rightarrow 1$, $\quad y \rightarrow \infty$, when $x \rightarrow 3$ and $y \rightarrow 4$, when $x \rightarrow \infty$.

So asymptotes are,

\[\begin{equation*} \begin{split} x &= 1\\ x &= 3\\ y &= 4 \end{split} \end{equation*}\]

$y = \dfrac{x^2 + 2x -1}{x}$

The function is a rational function and is improper. Trying to reduce fraction reveal difficult to manipulate roots of the numerator. So, no attempt is being made to reduce it further.

Here, $y \rightarrow \infty$, when $x \rightarrow 0$.

Also, $y \rightarrow \infty$, when $x \rightarrow \infty$. Hence, there is no asymptote parallel to $x$-axis.

To obtain oblique asymptote of form $y = mx +c$, we know,

\[\begin{equation*} \begin{split} m &= \lim_{x \to \infty} \dfrac{y}{x}\\ &= \lim_{x \to \infty} \dfrac{ x + 2 -\frac{1}{x}}{x}\\ &= \lim_{x \to \infty} 1 + \frac{2}{x} -\frac{1}{x^2}\\ m &= 1 \end{split} \end{equation*}\]

Also,

\[\begin{equation*} \begin{split} c &= \lim_{x \to \infty} (y - mx)\\ &= \lim_{x \to \infty} (y - x)\\ &= \lim_{x \to \infty} \left(\dfrac{x^2 + 2x -1}{x} - x\right)\\ &= \lim_{x \to \infty} \left(\dfrac{x^2 + 2x -1 -x^2}{x}\right)\\ &= \lim_{x \to \infty} \left(2 -\dfrac{1}{x}\right)\\ c &= 2 \end{split} \end{equation*}\]

The oblique asymptote is thus $y = x+2$.

All symptotes of the given equation are thus,

\[\begin{equation*} \begin{split} x &= 0\\ y &= x +2 \end{split} \end{equation*}\]

$y = \dfrac{2x -3}{x^2 - 3x + 2}$

The equation has degree $3$. So it can have maximum of three asymptotes. It can be written as $y = \dfrac{2x-3}{(x-1)(x-2)}$.

Here, $y \rightarrow \infty$, when $x \rightarrow 1$ and $x \rightarrow 2$. Also $y \rightarrow 0$, when $x \rightarrow \infty$. It cannot have more than $3$ asymptotes because the degree is $3$. No need to proceed further. Thus the asymptotes are,

\[\begin{equation*} \begin{split} x &= 0\\ x &= 1\\ x &= 2 \end{split} \end{equation*}\]

$y = \dfrac{(x+2)^2 (x-3)}{x-1}$

Here, $y \rightarrow \infty$, when $x \rightarrow 1$.

Also, $y \rightarrow \infty$, when $x \rightarrow \infty$. So no asymptotes parallel to $x$-axis.

To find oblique asymptote,

\[\begin{equation*} \begin{split} m &= \lim_{x \to \infty} \dfrac{y}{x}\\ m &= \lim_{x \to \infty} \dfrac{\dfrac{x^3 + x^2 - 8x -12}{x-1}}{x}\\ &= \lim_{x \to \infty} \left(\dfrac{x^2 + x-8 - \frac{12}{x}}{x -1}\right)\\ &= \lim_{x \to \infty} \left(\dfrac{x + 1 - \frac{8}{x}- \frac{12}{x^2}}{1 - \frac{1}{x}}\right)\\ &= \infty \end{split} \end{equation*}\]

Thus, $m$ is infinite. So, it does not have oblique asymptote. The only asymptote is $x=1$.

$y= \dfrac{x^2 + 1}{1 + x}$

The equation can be written as $y = 1 + x - \dfrac{2x}{1 + x}$.

Here, $y \rightarrow \infty$, when $x \rightarrow -1$.

Also, $y \rightarrow \infty$, when $x \rightarrow \infty$, So, no asymptotes parallel to $x$-axis.

For oblique asymptote of form $y = mx + c$,

\[\begin{equation*} \begin{split} m &= \lim_{x \to \infty} \dfrac{y}{x}\\ m &= \lim_{x \to \infty} \left(\frac{1}{x} + 1 - \frac{2}{1+x}\right)\\ m &= 1\\ c &= \lim_{x \to \infty} (y - mx)\\ &= \lim_{x \to \infty} (y - x)\\ &= \lim_{x \to \infty} \left(1 - \frac{2x}{1+x}\right)\\ &= -1 \end{split} \end{equation*}\]

Oblique asymptote is therefore $y = x -1$.

So all asymptotes of the equation are,

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

$y = \dfrac{x}{x^2 -1}$

It can be written as $y = \dfrac{x}{(x+1)(x-1)}$.

Here, $y \rightarrow \infty$, when $x \rightarrow 1$ and $x \rightarrow -1$. Also, $y \rightarrow 0$, when $x \rightarrow \infty$.

The asymptotes are thus,

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

$(x^2 + y^2)x - ay^2 = 0$

The equation is of degree $3$. Here $y^3$ is absent, so asymptotes parallel to $y$-axis is obtained by equating the coefficients of highest degree of $y$ to zero.

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

We expect two more asymptotes. Let $y = mx +c$ be the equation of asymptote.

Put $x= 1, y=m$,

\[\begin{equation*} \begin{split} \phi_3(m) &= 1 + m^2 \\ {\phi'}_3(m) &= 2m\\ \phi_2(m) &= -am^2 \end{split} \end{equation*}\]

Putting,

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

$m$ does not have real roots. So the equation does not have other asymptotes.

$x^3 -y^3 = 3ax^2$

The degree of equation is $3$.

The equation does not have asymptotes parallel to $x$-axis and $y$-axis because the coefficients of $x^3$ and $y^3$ are constants.

Let $y = mx +c$ be the oblique asymptotes to the curve. Putting $x= 1, y= m$,

\[\begin{equation*} \begin{split} \phi_3(m) &= 1 -m^3 \\ {\phi'}_3(m) &= -3m^2 \\ \phi_2(m) &= -3a \end{split} \end{equation*}\]

To find value of $m$,

\[\begin{equation*} \begin{split} \phi_3(m) &= 0 \\ 1 -m^3 &= 0\\ m &= 1 \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} c &= - \dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ &= -\dfrac{-3a}{-3m^2}\\ c &= -a \end{split} \end{equation*}\]

So the equation of asymptote is $y = x-a$.

$xy^2 - a^2(x-a) = 0$

Degree of equation is $3$. Here, $x^3$ is absent, so asymptotes parallel to $x$-axis are given by,

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

Here $y^3$ is also absent, so asymptote parallel to $y$-axis is,

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

The asymptotes are thus,

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

$(y-a)^2(x^2 - a^2) = x^4 + a^4$

The equation can be written as $(y-a)^2 = x^2 -a^2 + \dfrac{2x^2 a^2}{x^2 -a^2}$.

Here, $y \rightarrow \infty$, when $x \rightarrow \pm a$.

Also, $y \rightarrow \infty$, when $x \rightarrow \infty$. So, no asymptotes parallel to $x$-axis.

Degree of equation is $4$. We expect $4$ asymptotes. Let $y = mx +c$ be the equation of the remaining asymptotes.

Putting $x=1, y=m$,

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

For finding value of $m$,

\[\begin{equation*} \begin{split} \phi_4(m) &= 0\\ m^2 -1 &= 0\\ m &= \pm 1 \end{split} \end{equation*}\]

For finding value of $c$,

\[\begin{equation*} \begin{split} c &= - \dfrac{\phi_3(m)}{{\phi'}_4(m)} \\ &= \dfrac{2am}{2m} = a \end{split} \end{equation*}\]

Thus rest of the asymptotes are $y \pm x = a$. Thus the given equation has $4$ asymptotes.

$y(y-1)^2 - x^2y = 0$

Degree of equation is $3$.

Let $y = mx +c$ be the equation of the asymptotes.

Putting $x=1, y=m$,

\[\begin{equation*} \begin{split} \phi_3(m) &= m^3 -m \\ {\phi'}_3(m) &= 3m^2 -1\\ \phi_2(m) &= -2m^2 \end{split} \end{equation*}\]

Finding value of $m$,

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

For finding $c$,

\[\begin{equation*} \begin{split} c &= - \dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ c &= \dfrac{2m^2}{3m^2 -1} \end{split} \end{equation*}\]

Now,

$m$	$c$
$0$	$0$
$1$	$1$
$-1$	$1$

The equation of the asymptotes are,

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

$x^3 + y^3 = 3axy$

The degree of equation is $3$.

The coefficient of $x^3$ is $1$ which is constant, so there are no asymptotes parallel to $x$-axis. Also, the coefficient of $y^3$ is $1$ which is constant, so no asymptotes parallel to $y$-axis.

Let $y=mx + c$ be the equation of the asymptotes. Then,

\[\begin{equation*} \begin{split} \phi_3(m) &= 1 + m^3 \\ {\phi'}_3(m) = 3m^2 \\ \phi_2(m) &= -3am \end{split} \end{equation*}\]

To find the value of $m$,

\[\begin{equation*} \begin{split} \phi_3(m) &= 0\\ 1 + m^3 &= 0\\ (1+m)(m^2 - m+1) &= 0\\ m &= -1 \text{ and two imaginary roots}\\ \end{split} \end{equation*}\]

Asymptote does not exist for imaginary values, so we consider only $m=-1$.

For $c$,

\[\begin{equation*} \begin{split} c &= - \dfrac{\phi_2(m)}{{\phi'}_3(m)}\\ &= \dfrac{3am}{3m^2}\\ &= \dfrac{a}{m} = -a \end{split} \end{equation*}\]

The asymptote is thus,

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

13.1.2 Question 2

Determine the relative positions of the curves and the asymptotes in question number 1 [ii, iv, v] above.

1 ii.

For $y=\dfrac{x^2}{x^2 + 1}$, the only asymptote is $y=1$.

\[\begin{equation*} \begin{split} \lim (y-1) &= \lim \left(\dfrac{x^2}{x^2 + 1} - 1\right)\\ &= \lim \left(-\dfrac{1}{x^2 + 1}\right) \\ \end{split} \end{equation*}\]

When $x \rightarrow +\infty$, $y-1 \rightarrow 0$ through values less than $0$ i.e negative values and when $x \rightarrow -\infty$, $y-1 \rightarrow 0$ through values less than $0$ i.e negative values.

The curve thus approaches the asymptote $y=1$ from below both when $x \rightarrow +\infty$ and $x \rightarrow -\infty$.

See the figure below.

$Curve and asymptote of $\dfrac{x^2}{x^2 + 1}$$

Figure 13.1: Curve and asymptote of $\dfrac{x^2}{x^2 + 1}$

1 iv.

For $y = \dfrac{x^2 + 2x -1}{x}$, the asymptotes are $x=0$ and $y=x+2$.

When $x \rightarrow +0$, $y \rightarrow -\infty$ and when $x \rightarrow -0$, $y \rightarrow +\infty$.

When the curve approaches $x=0$ from positive side, the curve deflects toward $-\infty$ and when approaches $x=0$ from negative side, the curve deflects toward $+\infty$.

Also,

\[\begin{equation*} \begin{split} \lim (y-x-2) &= \lim \left(\dfrac{x^2 + 2x -1}{x} - x- 2\right)\\ &= \lim \left(-\dfrac{1}{x}\right) \end{split} \end{equation*}\]

When $x \rightarrow +\infty$, $y-x-2 \rightarrow -0$ and $x \rightarrow -\infty$, $y-x-2 \rightarrow +0$.

Therefore, the curve approaches $y-x-2=0$, from below when $x \rightarrow +\infty$ and from above when $x \rightarrow -\infty$. See 13.2.

$Curve and asymptotes of $\dfrac{x^2 + 2x -1}{x}$$

Figure 13.2: Curve and asymptotes of $\dfrac{x^2 + 2x -1}{x}$

1 v.

The asymptotes of $y = \dfrac{2x -3}{x^2 - 3x + 2}$ are,

\[\begin{equation*} \begin{split} x &= 1\\ x &= 2\\ y &= 0 \end{split} \end{equation*}\]

When $x=0$, $y =-\dfrac{3}{2}$, so $\left(0, -\dfrac{3}{2}\right)$ is the only crossing point here.

The equation can be written as $\dfrac{2x -3}{(x-1)(x-2)}$. So when,

\[\begin{equation*} \begin{split} x \rightarrow +\infty & \quad y \rightarrow +0 \\ x \rightarrow -\infty & \quad y \rightarrow -0 \\ x \rightarrow +1 & \quad y \rightarrow +\infty \\ x \rightarrow -1 & \quad y \rightarrow -\infty \\ x \rightarrow +2 & \quad y \rightarrow +\infty \\ x \rightarrow -2 & \quad y \rightarrow -\infty \\ \end{split} \end{equation*}\]

The relative position of the curve and the asymptotes is shown in 13.3.

$Curve and asymptotes of $\dfrac{2x -3}{x^2 - 3x + 2}$$

Figure 13.3: Curve and asymptotes of $\dfrac{2x -3}{x^2 - 3x + 2}$

13.1.3 Question 3

Find the asymptotes of the following curves

$\dfrac{x^2}{a^2}- \dfrac{y^2}{b^2} = 1$

The equation is of form $F_2 - F_0 = 0$.

\[\begin{equation*} \begin{split} \underbrace{\dfrac{x^2}{a^2}- \dfrac{y^2}{b^2}}_{F_2} - \underbrace{1}_{F_0} &= 0 \end{split} \end{equation*}\]

So by the method of inspection,

\[\begin{equation*} \begin{split} F_2 &= 0\\ \dfrac{x^2}{a^2}- \dfrac{y^2}{b^2} &= 0\\ \left(\frac{x}{a} + \frac{y}{b}\right)\left(\frac{x}{a} - \frac{y}{b}\right) &= 0\\ \left(\frac{x}{a} + \frac{y}{b}\right) = 0, \left(\frac{x}{a} - \frac{y}{b}\right) &= 0 \end{split} \end{equation*}\]

Two linear factors are found which are different. The asymptotes are thus,

\[\begin{equation*} \begin{split} y \pm \frac{b}{a}x &= 0 \end{split} \end{equation*}\]

$\dfrac{a^2}{x^2} + \dfrac{b^2}{y^2} = 1$

The equation can be written as $a^2y^2 + b^2 x^2 = x^2y^2$.

The degree of equation is $4$. There is no $x^4$. So asymptotes parallel to $x$-axis are obtained by equating the coefficients of highest degree term of $x$,

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

Similarly there is no $y^4$, so

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

Asymptotes are thus,

\[\begin{equation*} \begin{split} y &= \pm b \\ x &= \pm a \end{split} \end{equation*}\]

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

The equation can also be written as $x^2y^2 + a^2y^2 = (a-x)^2 x^2$.

For no values of $x$, $y \rightarrow \infty$, so no horizontal asymptotes in this case. Degree of equation is $4$, There is no $y^4$. Asymptotes parallel to $y$-axis are,

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

$x$ does not have real roots. So there are no asymptotes parallel to $y$-axis.

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

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

To find slope of the asymptotes,

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

For $c$,

\[\begin{equation*} \begin{split} c &= -\dfrac{\phi_3(m)}{{\phi'}_4(m)}\\ &= -\dfrac{2a}{2m}\\ &= - \dfrac{a}{m} \end{split} \end{equation*}\]

For $m=1, \quad c=-a$ and for $m=-1, \quad c =a$. The equation of asymptotes are thus,

\[\begin{equation*} \begin{split} y &= x-a\\ y + x &= a \end{split} \end{equation*}\]

$(a+x)^2(b^2 + x^2) = x^2y^2$

Degree of equation is $4$. No $y^4$, so

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

Let $y = mx +c$ be the equation of asymptotes, putting $x=1, y=m$,

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

The slope $m$ of asymptotes are given by

\[\begin{equation*} \begin{split} \phi_4(m) &= 0\\ 1 - m^2 &= 0\\ m &= \pm 1 \end{split} \end{equation*}\]

For finding values of $c$,

\[\begin{equation*} \begin{split} c &= - \dfrac{\phi_3(m)}{{\phi'}_4(m)}\\ &= - \dfrac{2a}{-2m}\\ c &= \dfrac{a}{m} \end{split} \end{equation*}\]

For $m =1, \quad c=a$ and for $m = -1, \quad c =-a$. Thus the asymptotes of the equation are

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

$x^2y + xy^2 = a^3$

The equation can be written as,

\[\begin{equation*} \begin{split} xy(x+y) - a^3 &= 0\\ \underbrace{xy(x+y)}_{F_3} - \underbrace{a^3}_{F_0} &= 0 \end{split} \end{equation*}\]

The $F_3$ has three non-repeating linear factors. The asymptotes thus can be obtained by method of inspection, equating $F_3=0$.

\[\begin{equation*} \begin{split} x &= 0\\ y &= 0\\ (x+y) &= 0 \end{split} \end{equation*}\]

\(m\)	\(c\)
\(0\)	\(0\)
\(1\)	\(1\)
\(-1\)	\(1\)