Question

find the asymptotes of the following curve
$x3 - x2y - xy2 + y3 + 2x2 - 4y2 + 2xy + x + y - 1 = 0$
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Answer 1

Answer:

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Step-by-step explanation:

please follow

Answer 2

Answer:

The asymptote of the curve $x^{3}-x^{2}y-xy^{2}+y^3+2x^{2}-4y^{2}+2xy+x+y-1=0$ is $y=1-x$.

Step-by-step explanation:

Consider the curve as follows:

$x^{3}-x^{2}y-xy^{2}+y^3+2x^{2}-4y^{2}+2xy+x+y-1=0$

⇒ $h_{3}+h_{2}+x+y-1=0$,

where  $h_{3}=x^{3}-x^{2}y-xy^{2}+y^3$ (combined power of $x$ and $y$ is $3$) and  $h_2=2x^{2}-4y^{2}+2xy$ (combined power of $x$ and $y$ is $2$)

Let suppose $y=mx+c$ is an asymptote where $m$ and $c$ are constants.

Consider $h_3$ as follows:

$h_{3}=x^{3}-x^{2}y-xy^{2}+y^3$  ...... (1)

Substitute $x=1$ and $y=m$ in the equation (1) as follows:

$h_{3}(m)=1-m-m^{2}+m^3$  ...... (2)

⇒ $m^3-m^{2}-m+1=0$

⇒ $m^2(m-1)-1(m-1)=0$

⇒ $(m^2-1)(m-1)=0$

⇒ $(m-1)(m+1)(m-1)=0$

Thus, $m=-1,1,1$.

Also, $h_2(m)=2-4m^{2}+2m$ (Since $x=1$ and $y=m$)

Formula to find value of $c$ is given as

$c=-\frac{h_2(m)}{h'_3(m)}$

From (2),

$h'_{3}(m)=-1-2m+3m^2$

Case1. When $m=-1$. Then,

$c=-\frac{h_2(m)}{h'_3(m)}$

$c=-\frac{2-4m^{2}+2m}{-1-2m+3m^2}$

$c=-\frac{2-4-2}{-1+2+3}$

$c=1$

Case2. When $m=1$. Then,

$c=-\frac{h_2(m)}{h'_3(m)}$

$c=-\frac{2-4m^{2}+2m}{-1-2m+3m^2}$

$c=-\frac{2-4+2}{-1-2+3}$

$c=\frac{0}{0}$ , undefined form

$y=1-x$ (Since $m=-1$ and $c=1$ in $y=mx+c$)

Therefore, the asymptote of the curve $x^{3}-x^{2}y-xy^{2}+y^3+2x^{2}-4y^{2}+2xy+x+y-1=0$ is $y=1-x$.

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