# Group formed on Parabola similarly to how an Elliptic curve forms a group (by drawing lines, circles, intersecting, or taking tangent lines)

There's probably other ways of doing this, but I've found this to be the simplest way (group law) that does indeed work: To add points $$A, B \in \{(x, f(x)) : x \in \Bbb{C}\} = G$$ where $$f$$ is any parabola with vertex $$E \in G$$, we treat $$E$$ as zero. Now draw a line between $$A, B$$ and then draw a parallel line to this line that passes through $$E$$. The unique intersection point (other than $$E$$, unless of course $$A = B = E$$ or $$A = -B$$) is then the value of the group law.

I've checked all the axioms of a group using Geogebra. This also works on a circle if I recall correctly.

I'm wondering:

How do we express $$AB$$ the abelian group law algebraically?

Nice observation! You can just crank it out algebraically. Starting with two points $$(a, a^2), (b, b^2)$$ on the parabola $$y = x^2$$ (to keep things simple; we can reduce to this case WLOG by a suitable change of variables), the line between them has slope $$\frac{a^2 - b^2}{a - b} = a + b$$, and the line with this slope passing through the origin intersects the parabola at the origin and at $$(a + b, (a + b)^2)$$.
So this does give us an algebraic group although it is just the familiar additive group $$\mathbb{G}_a$$. The interesting thing is that this construction actually makes sense on any nondegenerate conic over any field. For example on the circle $$x^2 + y^2 = 1$$ we recover $$SO_2$$ and on the hyperbola $$xy = 1$$ we recover the multiplicative group $$\mathbb{G}_m$$. The analogy to elliptic curves is discussed in detail in Lemmermeyer's Conics - a poor man's elliptic curves.