The document defines asymptotes as straight lines that touch a curve at infinity. It then shows the steps to find the asymptotes of the curve x3+2x2y-xy2-2y3+xy-y2-1=0. The steps involve putting y=mx+c into the equation, setting coefficients of x3 and x2 to zero, and solving the resulting equations for m and c. The asymptotes obtained are y=x, y=-x-1, and x+2y=1.

1. MATHEMATICS
TOPIC: ASYMPTOTES
DEFINITION
WORKING PRINCIPLE
PARALLEL ASYMPTOTES

2. ASYMPTOTES
DEFINITION: Asymptote is a straight line which
touches the curve y=f(x) at infinity.

4. Question: find the asymptotes of the curve
x3+2x2y-xy2-2y3+xy-y2-1=0
• solution:
given curve: x3+2x2y-xy2-2y3+xy-y2-1=0 eq 1st
• put y=mx+c into eq 1st
x3+2x2 (mx+c) -x (mx+c )2-2 (mx+c )3+x (mx+c )- (mx+c )2-1=0
• x3(1+2m-m2-2m3)+x2(c-6m2c+m-m2) ……………………..=0
• put coefficient of x3 and x2 to zero
1+2m-m2-2m3=0 eq 2nd
c-6m2c+m-m2=0 eq 3rd
• solving eq 2nd for m:
1+ 2m-m2(1+2m)=0
(1+2m)(1-m2)=0
(1+2m)(1-m)(1+m)=0
m=-1/2,1,-1
• solving eq 3rd for c:
c(2-2m-6m2)=-m2-m
c=(m -m2)/(2-2m-6m2 )

5. When m=1
c=0/(2-2-6)
c=0
• when m=-1
c= (1=1)/(2+2-6)
c=2/-2
c=-1
• when m=1/2
c=-1/2
required asymptotes y=mx+c
when m=1, c=0
y=x
• when m=-1, c=-1
y=-x-1
• when m=-1/2, c=1/2
y=-x/2=1/2
x+2y=1

6. •Asymptotes of curve (method 2)
working rule:
Step 1st : simplify f(x,y)=0 and put x=1 and y=m into highest degree
terms and say it Øn (m)
Similarly find Øn-1(m)
Step 2nd : put Øn (m)=0 and solve it for m. (say
m=m1,m2,m3………..,mn
Step 3rd : find c=-Øn-1(m)/Ø’n(m) for each m=m1,m2,m3………..,mn
Step 4th : find asymptotes by y=mx+c.

7. •

8. C=-(-3am)/3m2
C=a/m
When m=-1
C=-a
Then asyototes is:
y=mx+c
y=-x+(-a)
X+y+a=0

9. Question:
Find the asyototes of the curve
x3+2x2y-xy2-2y3+3xy+3y2+x+1=0.

10. Question: find the asymptotes of the curve
x3+2x2y-xy2-2y3+xy-y2-1=0
solution:
given curve: x3+2x2y-xy2-2y3+xy-y2-1=0 eq 1st
put y=mx+c into eq 1st
x3+2x2 (mx+c) -x (mx+c )2-2 (mx+c )3+x (mx+c )- (mx+c )2-1=0
x3(1+2m-m2-2m3)+x2(c-6m2c+m-m2) ……………………..=0
put coefficient of x3 and x2 to zero
1+2m-m2-2m3=0 eq 2nd
c-6m2c+m-m2=0 eq 3rd
solving eq 2nd for m:
1+ 2m-m2(1+2m)=0
(1+2m)(1-m2)=0
(1+2m)(1-m)(1+m)=0
m=-1/2,1,-1
solving eq 3rd for c:
c(2-2m-6m2)=-m2-m
c=(m -m2)/(2-2m-6m2 )

11. Whem m=1
c=0/(2-2-6)
c=0
when m=-1
c= (1=1)/(2+2-6)
c=2/-2
c=-1
when m=1/2
c=-1/2
required asymptotes y=mx+c
when m=1, c=0
y=x
when m=-1, c=-1
y=-x-1
when m=-1/2, c=1/2
y=-x/2=1/2
x+2y=1

12. thank you