How $a^2(x^2-y^2)-b=0$ is a degenerate conic consisting of twice the line at infinity for $a=0$ and $b=1$?

Question

According to Wikipedia:

The conic section with equation $x^2-y^2 = 0$ is degenerate as its equation can be written as $(x-y)(x+y)= 0$, and corresponds to two intersecting lines forming an "X". This degenerate conic occurs as the limit case $a=1, b=0$ in the pencil of hyperbolas of equations $a(x^2-y^2) - b=0$. The limiting case $a=0, b=1$ is an example of a degenerate conic consisting of twice the line at infinity.

If $a=0$, and $b=1$, the equation becomes $1=0$, how is that possible?

Answer 1

If we rewrite the equation we find $y^{2}=x^{2}-\frac{1}{a}$. Now think about what happens if $a$ tends to $0$, clearly $x^{2}$ has to tend to infinity, while every value of $y$ is still possible. So for "$a=0$", i.e. the limit case of these conics we are left with the lines $(\infty,y)$ and $(-\infty,y)$ with $y\in\mathbb{R}$.

I agree it is not written down with a lot of mathematical rigour on the wikipedia page.

Answer 2

When talking about points and lines at infinity, it can be helpful to work in the extended Euclidean plane (projective plane with a distinguished line at infinity) and homogenize the equations. Substituting $x\to x/w$ and $y\to y/w$, the equation becomes $$a(x^2-y^2)-bw^2=0$$ and with $a=0$ and $b=1$ it’s simply $w^2=0$. This equation is satisfied by every point that has homogeneous coordinates with $w=0$, i.e., by every point at infinity. The solution set is therefore the line at infinity and, just as the solution set of $(x-y)^2=0$ is considered to be a double line, so, too, is this line doubled.

Answer 3

As an alternative to amd's excellent answer: Your observation that the (affine!) equation becomes $1=0$ shows that the limiting conic has no affine points. Hence as a conic in the projective plane, it is contained in the line at infinity. Because of its degree, it must be twice the line at infinity.