Question

find the equation of the asymptotes of the curve x^(3)-2y^(3)+xy(2x-y)+y(x-y)+1=0​

Answer 1

Answer:

I don't know

Step-by-step explanation:

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Answer 2

CONCEPT:

Asymptotes are lines which tends to  approach a definite curve  but will not meet the curve at finite distance.The value will be closer and closer but not meet the line yet.mathematically there are 3 types of asymptote they are horizontal ,vertical and oblique

we want to find the equation of asymptote with degrees of variables while resulting x=a and y=a

GIVEN:

x^(3)-2y^(3)+xy(2x-y)+y(x-y)+1=0​

FIND:

find asymptotes for x^(3)-2y^(3)+xy(2x-y)+y(x-y)+1=0​

SOLUTION:

Lets rewrite the equation after opening the bracket.

then it becomes :

$x^{3} -2x^{2} y-2y^{3} -y^{2} x+yx-y^{2} +1=0\\x^{3} -2x^{2} y-2y^{3} -y^{2} x+yx-y^{2} +1=0\\\\let "\alpha " be x^{3} -2x^{2}y -2y^{3} \\let "\beta " be y^{2} x+yx-y^{2} and\\q=1$

let y=mx+c be the asymptotes.

put x=1 and y=m and $\alpha$(1,m)

then $\alpha$=1-2m-2$m^{3}$=0

m=1/2

c=$\beta$/differential of alpha

=$m^{2}$+m-$m^{2}$/2-6$m^{2}$=m/2-6$m^{2}$=c=1/1=1

so the equation of asymptote is given by y=mx+c=y=1/2x+1=x/2+1

so asymptote equation is given by y=x/2+1

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