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

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

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

Also, $y\to \infty$, when $x\to 2$, $y\to 0$ when $x\to \infty$.

So, there are three asymptotes.

$$\begin{array}{cc}x& =1\\ x& =2\\ y& =0\end{array}$$

2. $y=\frac{x^2}{x^2+1}$

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 $x$.

$$\begin{array}{cc}y-1& =0\\ y& =1\end{array}$$

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

$$\begin{array}{cc}{x}^2+1& =0\end{array}$$

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

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

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

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

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

So asymptotes are,

$$\begin{array}{cc}x& =1\\ x& =3\\ y& =4\end{array}$$

4. $y=\frac{x^2+2x-1}{x}$

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, $y\to \infty$, when $x\to 0$.

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

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

$$\begin{array}{cc}m& =\underset{x\to \infty }{lim}\frac{y}{x}\\ & =\underset{x\to \infty }{lim}\frac{x+2-\frac{1}{x}}{x}\\ & =\underset{x\to \infty }{lim}1+\frac{2}{x}-\frac{1}{x^2}\\ m& =1\end{array}$$

Also,

$$\begin{array}{cc}c& =\underset{x\to \infty }{lim}(y-mx)\\ & =\underset{x\to \infty }{lim}(y-x)\\ & =\underset{x\to \infty }{lim}(\frac{x^2+2x-1}{x}-x)\\ & =\underset{x\to \infty }{lim}\left(\frac{x^2+2x-1-x^2}{x}\right)\\ & =\underset{x\to \infty }{lim}(2-\frac{1}{x})\\ c& =2\end{array}$$

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

All symptotes of the given equation are thus,

$$\begin{array}{cc}x& =0\\ y& =x+2\end{array}$$

22. $y=\frac{2x-3}{x^2-3x+2}$

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

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

$$\begin{array}{cc}x& =0\\ x& =1\\ x& =2\end{array}$$

6. $y=\frac{(x+2{)}^2(x-3)}{x-1}$

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

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

To find oblique asymptote,

$$\begin{array}{cc}m& =\underset{x\to \infty }{lim}\frac{y}{x}\\ m& =\underset{x\to \infty }{lim}\frac{\frac{x^3+x^2-8x-12}{x-1}}{x}\\ & =\underset{x\to \infty }{lim}\left(\frac{x^2+x-8-\frac{12}{x}}{x-1}\right)\\ & =\underset{x\to \infty }{lim}\left(\frac{x+1-\frac{8}{x}-\frac{12}{x^2}}{1-\frac{1}{x}}\right)\\ & =\infty \end{array}$$

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

7. $y=\frac{x^2+1}{1+x}$

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

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

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

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

$$\begin{array}{cc}m& =\underset{x\to \infty }{lim}\frac{y}{x}\\ m& =\underset{x\to \infty }{lim}(\frac{1}{x}+1-\frac{2}{1+x})\\ m& =1\\ c& =\underset{x\to \infty }{lim}(y-mx)\\ & =\underset{x\to \infty }{lim}(y-x)\\ & =\underset{x\to \infty }{lim}(1-\frac{2x}{1+x})\\ & =-1\end{array}$$

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

So all asymptotes of the equation are,

$$\begin{array}{cc}x+1& =0\\ y& =x-1\end{array}$$

8. $y=\frac{x}{x^2-1}$

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

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

The asymptotes are thus,

$$\begin{array}{cc}x& =\pm 1\\ y& =0\end{array}$$

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

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.

$$\begin{array}{cc}x-a& =0\\ x& =a\end{array}$$

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

Put $x=1,y=m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =1+m^2\\ {{\varphi }^{\prime }}_3\left(m\right)& =2m\\ {\varphi }_2\left(m\right)& =-a{m}^2\end{array}$$

Putting,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =0\\ 1+m^2& =0\end{array}$$

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

24. $x^3-y^3=3a{x}^2$

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 $x=1,y=m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =1-m^3\\ {{\varphi }^{\prime }}_3\left(m\right)& =-3m^2\\ {\varphi }_2\left(m\right)& =-3a\end{array}$$

To find value of $m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =0\\ 1-m^3& =0\\ m& =1\end{array}$$

Now,

$$\begin{array}{cc}c& =-\frac{{\varphi }_2\left(m\right)}{{{\varphi }^{\prime }}_3\left(m\right)}\\ & =-\frac{-3a}{-3m^2}\\ c& =-a\end{array}$$

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

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

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

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

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

$$\begin{array}{cc}x& =0\end{array}$$

The asymptotes are thus,

$$\begin{array}{cc}x& =0\\ y& =\pm a\end{array}$$

12. $(y-a{)}^2(x^2-a^2)=x^4+a^4$

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

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

Also, $y\to \infty$, when $x\to \infty$. 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 $x=1,y=m$,

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =m^2-1\\ {{\varphi }^{\prime }}_4\left(m\right)& =2m\\ {\varphi }_3\left(m\right)& =-2am\end{array}$$

For finding value of $m$,

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =0\\ m^2-1& =0\\ m& =\pm 1\end{array}$$

For finding value of $c$,

$$\begin{array}{cc}c& =-\frac{{\varphi }_3\left(m\right)}{{{\varphi }^{\prime }}_4\left(m\right)}\\ & =\frac{2am}{2m}=a\end{array}$$

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

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

Degree of equation is $3$.

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

Putting $x=1,y=m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =m^3-m\\ {{\varphi }^{\prime }}_3\left(m\right)& =3m^2-1\\ {\varphi }_2\left(m\right)& =-2m^2\end{array}$$

Finding value of $m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =0\\ m^3-m& =0\\ m(m^2-1)& =0\\ m& =0,\pm 1\end{array}$$

For finding $c$,

$$\begin{array}{cc}c& =-\frac{{\varphi }_2\left(m\right)}{{{\varphi }^{\prime }}_3\left(m\right)}\\ c& =\frac{2m^2}{3m^2-1}\end{array}$$

Now,

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

The equation of the asymptotes are,

$$\begin{array}{cc}y& =0\\ y\pm x& =1\end{array}$$

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

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,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =1+m^3\\ {{\varphi }^{\prime }}_3\left(m\right)=3m^2& \\ {\varphi }_2\left(m\right)& =-3am\end{array}$$

To find the value of $m$,

$$\begin{array}{cc}{\varphi }_3\left(m\right)& =0\\ 1+m^3& =0\\ (1+m)(m^2-m+1)& =0\\ m& =-1\text{and two imaginary roots}\end{array}$$

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

For $c$,

$$\begin{array}{cc}c& =-\frac{{\varphi }_2\left(m\right)}{{{\varphi }^{\prime }}_3\left(m\right)}\\ & =\frac{3am}{3m^2}\\ & =\frac{a}{m}=-a\end{array}$$

The asymptote is thus,

$$\begin{array}{cc}y+x+a& =0\end{array}$$

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

$$\begin{array}{cc}lim(y-1)& =lim(\frac{x^2}{x^2+1}-1)\\ & =lim(-\frac{1}{x^2+1})\end{array}$$

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

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

See the figure below.

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

1 iv.

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

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

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,

$$\begin{array}{cc}lim(y-x-2)& =lim(\frac{x^2+2x-1}{x}-x-2)\\ & =lim(-\frac{1}{x})\end{array}$$

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

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

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

1 v.

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

$$\begin{array}{cc}x& =1\\ x& =2\\ y& =0\end{array}$$

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

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

$$\begin{array}{cc}x\to +\infty & y\to +0\\ x\to -\infty & y\to -0\\ x\to +1& y\to +\infty \\ x\to -1& y\to -\infty \\ x\to +2& y\to +\infty \\ x\to -2& y\to -\infty \end{array}$$

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

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

13.1.3 Question 3

Find the asymptotes of the following curves

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

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

$$\begin{array}{cc}\underset{F_2}{\underset{}{\frac{x^2}{a^2}-\frac{y^2}{b^2}}}-\underset{F_0}{\underset{}{1}}& =0\end{array}$$

So by the method of inspection,

$$\begin{array}{cc}{F}_2& =0\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}& =0\\ (\frac{x}{a}+\frac{y}{b})(\frac{x}{a}-\frac{y}{b})& =0\\ (\frac{x}{a}+\frac{y}{b})=0,(\frac{x}{a}-\frac{y}{b})& =0\end{array}$$

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

$$\begin{array}{cc}y\pm \frac{b}{a}x& =0\end{array}$$

---

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

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

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 $x$,

$$\begin{array}{cc}{b}^2-y^2& =0\\ y& =\pm b\end{array}$$

Similarly there is no $y^4$, so

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

Asymptotes are thus,

$$\begin{array}{cc}y& =\pm b\\ x& =\pm a\end{array}$$

---

3. $y^2=\frac{(a-x{)}^2}{a^2+x^2}{x}^2$

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

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

$$\begin{array}{cc}{x}^2+a^2& =0\end{array}$$

$x$ 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 $y=m$,

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =-1+m^2\\ {{\varphi }^{\prime }}_4\left(m\right)& =2m\\ {\varphi }_3\left(m\right)& =2a\end{array}$$

To find slope of the asymptotes,

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =0\\ -1+m^2& =0\\ m& =\pm 1\end{array}$$

For $c$,

$$\begin{array}{cc}c& =-\frac{{\varphi }_3\left(m\right)}{{{\varphi }^{\prime }}_4\left(m\right)}\\ & =-\frac{2a}{2m}\\ & =-\frac{a}{m}\end{array}$$

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

$$\begin{array}{cc}y& =x-a\\ y+x& =a\end{array}$$

---

4. $(a+x{)}^2(b^2+x^2)=x^2{y}^2$

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

$$\begin{array}{cc}{x}^2& =0\\ x& =0\end{array}$$

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

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =1-m^2\\ {{\varphi }^{\prime }}_4\left(m\right)& =-2m\\ {\varphi }_3\left(m\right)& =2a\end{array}$$

The slope $m$ of asymptotes are given by

$$\begin{array}{cc}{\varphi }_4\left(m\right)& =0\\ 1-m^2& =0\\ m& =\pm 1\end{array}$$

For finding values of $c$,

$$\begin{array}{cc}c& =-\frac{{\varphi }_3\left(m\right)}{{{\varphi }^{\prime }}_4\left(m\right)}\\ & =-\frac{2a}{-2m}\\ c& =\frac{a}{m}\end{array}$$

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

$$\begin{array}{cc}y& =x+a\\ y+x+a& =0\end{array}$$

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22. $x^{2}y+x{y}^2=a^3$

The equation can be written as,

$$\begin{array}{cc}xy(x+y)-a^3& =0\\ \underset{F_3}{\underset{}{xy(x+y)}}-\underset{F_0}{\underset{}{a^3}}& =0\end{array}$$

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

$$\begin{array}{cc}x& =0\\ y& =0\\ (x+y)& =0\end{array}$$