Semigroup law on points on the curve $f(x) = \frac{1}{x}$ Consider the positive half of the curve $f: \Bbb{R} \to \Bbb{R}, f(x) = \frac{1}{x}$.  Let $A = (a,1/a), B = (b, 1/b)$ be any two points on the curve.  Draw a line through them  Find where this point intersects the curve $f$ reflected across the $y$-axis.  Now reflect the point back across the $y$-axis to yield the result that we define to be $A + B$.  This is diagrammed below.

It appears to be an associative operation:

However, proving it to be associative seems way more involved than just drawing lines.  What's an easy route to prove that curve point addition is associative?
My attempt: There is positive $y$ such that $A + (B-A)t = (-y, 1/y)$ iff $a + (b-a)t = -y$ and $1/a + (1/b - 1/a)t = 1/y$ which leads to the quadratic equation: $\left( \frac{a-b}{a} - a(\frac{1}{b} - \frac{1}{a})\right )t + (\frac{1}{b} - \frac{1}{a})(a-b)t^2 -2 = 0$. This seems too complicated to work with on paper.
 A: Unless I have misunderstood OP, this seems not to be the case. I have drawn the following in GeoGebra, on which it is clearly false:

For clarity I didn't draw A+(B+C) nor (A+B)+C. These points would be reflections of A(BC) and (AB)C, respectively, in $y$ axis. From the picture it's clear that these will be distinct points.
A: modulo my unreliable arithmetic, the semi-group operation is:
$$
\theta_a \circ \theta_b = (\theta_a+\theta_b)\frac{\left(1+\sqrt{1+\frac{4\theta_a\theta_b}{(\theta_a+\theta_b)^2}} \right)}2
$$
which does not look too promising on the associativity front
A: If I hoped for a group law, I would begin by looking for the neutral element. In the case of elliptic curves, one works in projective coordinates and the neutral element is the point at infinity. In your case, the hyperbola is $xy=1$ so when you projectivize it you get $xy-z^2=0$. The trouble is that $z$ may now introduce a $-$ sign, which completely your idea (i.e. you cannot projectivize half of a hyperbola, you can only projectivize it all). So, the question would be: why work with only half a hyperbola?
Edit:
In order to find the coordinates of $(a,\frac 1 a) \oplus (b,\frac 1 b)$ it is conveninent to proceed as follows: intersect the line joining $A$ and $B$ (of equation $\frac {y-\frac 1 b} {x-b} = \frac {\frac 1 b - \frac 1 a} {b-a}$ or, equivalently, $y = \frac {a+b-x} {ab}$) with the hyperbola $xy=-1$; since you are looking for a negative solution, $x=\frac {a+b - \sqrt {(a+b)^2 + 4ab}} 2$. Then the $x$ coordinate of $A \oplus B$ will be the opposite of this, i.e. $x=\frac {\sqrt {(a+b)^2 + 4ab} \space -a -b} 2$.
Now, let us check whether $\big( (1,1) \oplus (1,1) \big) \oplus (2,\frac 1 2) = (1,1) \oplus \big( (1,1) \oplus (2,\frac 1 2) \big)$ (these points have been chosen to simplify computations). Rather than do it by hand, the following Mathematica code is faster:
plus[{a_,_}, {b_,_}] := List[ (Sqrt [(a+b)^2+4 a b] - a - b) / 2, 2 / (Sqrt [(a+b)^2+4 a b] - a - b)];
(plus [plus [{1,1}, {1,1}], {2, 1/2}] - plus[{1,1}, plus[{1,1}, {2,1/2}]]) == {0,0}
You will get "False", i.e. the two points do not coincide, so your operation is not associative.
A final edit:
There is, indeed, some deep similarity between conics and elliptic curves. You can find out more in this paper by Franz Lemmermeyer that I have found delicious to read (mind you, from a certain point onwards he gets into higher mathematics).
