Higher Engineering Mathematics

206 GRAPHS

Problem 9. Determine the asymptotes par-

allel to the x- 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 of

xto zero givesy^2 − 9 =0 from which,y^2 =9 and

y=± 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 ofy

to zero givesx^2 − 9 =0 from which,x^2 =9 and

x=± 3.

Hence asymptotes occur aty=±3 andx=± 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 pow-
ers ofxto zero and determine the values ofm
andc.y=mx+cgives the asymptote.

Problem 10. Determine the asymptotes for the

function:y(x+1)=(x−3)(x+2) and sketch

the 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.

Equating the coefficient of the next highest

power ofxto zero givesm+c+ 1 =0.

and sincem=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 )

is 1. Equating this to zero gives 1=0 which is not

an equation of a line. Hence there is no asymptote

parallel to thex-axis.

To determine any asymptotes parallel to they-axis:

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

the highest power ofyisx+1. Equating this to

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

x=− 1 is an asymptote.

Whenx= 0 ,y(1)=(−3)(2), i.e.y=− 6.

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

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

is shown in Fig. 19.34.

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 the coefficient of the highest power

of x (i.e. x^3 in this case) to zero gives

1 −m^2 =0, from which,m=±1.

Equating the coefficient of the next highest

power ofx(i.e.x^2 in this case) to zero gives

− 2 mc=0, from which,c=0.