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Find the alll asymptotes of the equation $$x^3-x^2y-xy^2+y^3+2x^2-4y^2+2xy+x+y+1=0.$$

Here's what i tried:

Equation for oblique asymptotes- \begin{align} \varphi_3(m)&=1-m-m^2+m^3\\ \varphi_2(m)&=2-4m^2+2m\\ \varphi_1(m)&=1+m\\ \varphi_0(m)&=1 \end{align}

$$\varphi_3(m)= m=1,1,-1.$$

for $m=-1$ it's $c\varphi_3(m)+\varphi_2(m)=0$

on solving the above it's coming $4=0$.

I don't know where I did wrong.

Ayush Kumar
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    Hello, I have made an edit to fix the latex, please check if I made any mistake. – Arctic Char Jan 01 '21 at 20:19
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    Hey you changed the coefficient of x^3 from 1 to 4 so that actually changes the whole question!! – Ayush Kumar Jan 02 '21 at 02:12
  • I have not looked as asymptotics in quite some time. Does the double root at $1$ change the process? I do not recall. – John Wayland Bales Jan 02 '21 at 03:04
  • Suppose the coefficient of x^ 3 is 4. Then φ3(m)=4-m-m^2+m^3 but if it's 1 then φ3(m)=1-m-m^2+m^3 and as far as i know we should get different values of m. – Ayush Kumar Jan 02 '21 at 04:29
  • @JohnWaylandBales nvm i got the mistake , actually it is cφ3(m)'+φ2(m)=0 , i was not differentiating the φ3(m) before multiplying it with c – Ayush Kumar Jan 02 '21 at 04:43
  • $\varphi_3 = m$ should have been $\varphi_3 = 0$. The curve contains two points at infinity, $(1, 1, 0)$ and $(1, -1, 0)$. If we take $x = 1$, the leading terms in the Taylor expansion around $(y, z) = (-1, 0)$ are $4(y + 1) - 4 z$. Homogenizing back gives an asymptote $y + x = z$. The leading terms in the Taylor expansion around $(y, z) = (1, 0)$ are $$2 (y - 1)^2 - 6 (y - 1) z + 2 z^2 = 2 (y - 1 - k_1 z) (y - 1 - k_2 z),$$ which gives two more asymptotes $y - x = k_{1, 2} z$. – Maxim Jan 06 '21 at 13:29

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