Exercise 5.7.18 (Family of cubics $axy = (x+1)(y+1)(x+y+b)$)

Proof.

(a)

Consider the cubic curve ${𝒞}_f(ℝ)$, where $$\begin{array}{llll}\hfill f(x,y)& =\mathit{axy}-(x+1)(y+1)(x+y+b).& \hfill & \end{array}$$

If $a=0$, then ${𝒞}_f(ℝ)$ is the union of three lines $x=-1$, $y=-1$ and $y=-x-b$. We suppose now that $a\ne 0$. Let $L:y=\mathit{mx}+r$ be a non vertical line. By Theorem 5.15, $L\subseteq {𝒞}_f(ℝ)$, if and only if there is a polynomial $k(x,y)$ such that

$$f(x,y)=(y-\mathit{mx}-r)k(x,y),$$

or equivalently $f(x,\mathit{mx}+r)$ is identically zero. So, for all $x\in ℝ$,

$$\begin{array}{llll}\hfill 0=& f(x,\mathit{mx}+r)& \hfill & \\ \hfill =& -\left(\mathit{mx}+b+r+x\right)\left(\mathit{mx}+r+1\right)\left(x+1\right)+\left(\mathit{mx}+r\right)\mathit{ax}& \hfill & \\ \hfill =& -\left(m^2+m\right)x^3+\left(\mathit{am}-\mathit{bm}-m^2-2\mathit{mr}-2m-r-1\right)x^2& \hfill & \\ \hfill & -\left(\mathit{bm}-\mathit{ar}+\mathit{br}+2\mathit{mr}+r^2+b+m+2r+1\right)x-b-r-\mathit{br}-r^2.& \hfill & \end{array}$$

All the coefficients of this polynomial in $x$ are null. In particular $m^2+m=0$, so $m=0$ or $m=-1$.

• If $m=0$, then the coefficient of $x^2$ gives $r=-1$, thus $f(x,y)=-\mathit{ax}$, which is not identically zero for $a\ne 0$.
• Since $f(x,y)=f(y,x)$, there is no solution for a vertical line ( $m=\infty$).

• It remains only the case $m=-1$. Then

  $$f(x,-x+r)=-\left(a-b-r\right)x^2+r\left(a-b-r\right)x-b-r-\mathit{br}-r^2,$$

  so

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}r\hfill & =a-b,\hfill \\ 0\hfill & =r^2+(b+1)r+b.\hfill \end{array}& & \hfill \text{(1)}\end{array}$$

  The system (1) implies $0={(a-b)}^2+(b+1)(a-b)+b=a^2-\mathit{ab}+a=a(a-b+1)$, where $a\ne 0$, so

  • if $a-b\ne -1$, the system (1) has no solution, and the curve ${𝒞}_f(ℝ)$ contains no line.
  • If $a-b=-1$, then $f(x,-x-1)=0$ for all $x\in ℝ$, so the line $x+y+1=0$ is contained in ${𝒞}_f(ℝ)$.

In conclusion, ${𝒞}_f(ℝ)$ contains a line if an only if $a=0$ or $b=a+1$.

Verification: If $a=b-1$,

$$f(x,y)=(b-1)\mathit{xy}-(x+1)(y+1)(x+y+b)=-\left(\mathit{xy}+b+x+y\right)\left(x+y+1\right).$$

(b)

We obtain the points at infinity of the curve with the homogeneous equation $$\mathit{axyt}=(x+t)(y+t)(x+y+\mathit{bt}).$$

For $t=0$, this gives

$$0=\mathit{xy}(x+y),$$

so $x=0$ or $y=0$ or $y=-x$. There are three points at infinity:

$$A(0:1:0),B(1:0:0),C(1:-1:0).$$