Higher Engineering Mathematics, Sixth Edition

Functions and their curves 193

Problem 9. Determine the asymptotes parallel to

thex-andy-axes for the function

x^2 y^2 = 9 (x^2 +y^2 ).

Asymptotes parallel to thex-axis:

Rearrangingx^2 y^2 = 9 (x^2 +y^2 )gives

x^2 y^2 − 9 x^2 − 9 y^2 = 0

hence x^2 (y^2 − 9 )− 9 y^2 = 0

Equating the coefficient of the highest power ofxto zero
givesy^2 − 9 =0 from which,y^2 =9andy=± 3.

Asymptotes parallel to they-axis:

Since x^2 y^2 − 9 x^2 − 9 y^2 = 0

then y^2 (x^2 − 9 )− 9 x^2 = 0

Equating the coefficient of the highest power ofyto zero
givesx^2 − 9 =0 from which,x^2 =9andx=± 3.
Hence asymptotes occur aty=±3andx=± 3.

Other asymptotes

To determine asymptotes other than those parallel to
x-andy-axes a simple procedure is:

(i) substitutey=mx+cin the given equation

(ii) simplify the expression

(iii) equate the coefficients of the two highest powers

ofxto zero and determine the values ofmandc.

y=mx+cgives the asymptote.

Problem 10. Determine the asymptotes for the

function:y(x+ 1 )=(x− 3 )(x+ 2 )andsketchthe

curve.

Following the above procedure:

(i) Substitutingy=mx+cinto

y(x+ 1 )=(x− 3 )(x+ 2 )gives:

(mx+c)(x+ 1 )=(x− 3 )(x+ 2 )

(ii) Simplifying gives

mx^2 +mx+cx+c=x^2 −x− 6

and (m− 1 )x^2 +(m+c+ 1 )x+c+ 6 = 0

(iii) Equating the coefficient of the highest power

of x to zero gives m− 1 =0 from which,

m= 1.

Equatingthe coefficient of the next highest power

ofxto zero givesm+c+ 1 =0.

and since m=1, 1+c+ 1 =0 from which,

c=− 2.

Hencey=mx+c= 1 x−2.

i.e.y=x−2 is an asymptote.

To determine any asymptotes parallel to thex-axis:

Rearranging y(x+ 1 )=(x− 3 )(x+ 2 )

gives yx+y=x^2 −x− 6

The coefficient of the highest power ofx(i.e.x^2 )is1.

Equatingthistozerogives1=0 whichisnot anequation

of a line. Hence there is no asymptote parallel to the

x-axis.

To determine any asymptotes parallel to they-axis:

Since y(x+ 1 )=(x− 3 )(x+ 2 ) the coefficient of

the highest power of y is x+1. Equating this to

zero givesx+ 1 =0, from which,x=−1. Hencex=− 1

is an asymptote.

Whenx= 0 ,y( 1 )=(− 3 )( 2 ),i.e.y=− 6.

Wheny= 0 , 0 =(x− 3 )(x+ 2 ),i.e.x=3andx=− 2.

A sketch of the functiony(x+ 1 )=(x− 3 )(x+ 2 )is

shown in Fig. 18.34.

Problem 11. Determine the asymptotes for the

functionx^3 −xy^2 + 2 x− 9 =0.

Following the procedure:

(i) Substitutingy=mx+cgives

x^3 −x(mx+c)^2 + 2 x− 9 =0.

(ii) Simplifying gives

x^3 −x[m^2 x^2 + 2 mcx+c^2 ]+ 2 x− 9 = 0

i.e. x^3 −m^2 x^3 − 2 mcx^2 −c^2 x+ 2 x− 9 = 0

and x^3 ( 1 −m^2 )− 2 mcx^2 −c^2 x+ 2 x− 9 = 0

(iii) Equating thecoefficient of the highest power ofx

(i.e.x^3 in this case) to zero gives 1−m^2 =0, from

which,m=±1.

Equatingthe coefficient of the next highest power

ofx(i.e.x^2 in this case) to zero gives− 2 mc=0,

from which,c=0.