Question

find the asymptotes of the curve x^3 - 5x^2 y + 8xy^2 - 4y^3 + x^2 - 3xy + 2y^2 - 1 = 0.

Answer 1

Answer:

Step-by-step explanation:

The given curve is

$\rm{{x}^{3}-5{x}^{2}y+8{x}{y}^{2}-4{y}^{3}+{x}^{2}-3{x}{y}+2{y}^{2}-1=0}$

$\rm{\implies\,{x}^{3}-{x}^{2}y-4{x}^{2}y+4{x}{y}^{2}+4x{y}^{2}-4{y}^{3}+{x}^{2}-2{x}{y}-{x}{y}+2{y}^{2}-1=0}$

$\rm{\implies\,{x}^{2}(x-y)-4xy(x-y)+4{y}^{2}(x-y)+x(x-2y)-y(x-2y)-1=0}$

$\rm{\implies\,(x-y)\left({x}^{2}-4xy+4{y}^{2}\right)+(x-y)(x-2y)-1=0}$

$\rm{\implies\,(x-y)\left(x-2y\right)^{2}+(x-y)(x-2y)-1=0}$

$\rm{\implies\,(x-y)\left(x-2y\right)\left(x-2y+1\right)-1=0}$

$\rm{\implies\,(x-y)\left(x-2y\right)\left(x-2y+1\right)=1}$

Since this is a cubic curve, so the asymptotes are given by,

$\rm{x-y=0\,\,\,\,,\,\,\,\,x-2y=0\,\,\,\,,\,\,\,\,x-2y+1=0}$

Answer 2

Answer:

Step-by-step explanation: