Why there are two point at infinity on certain elliptic curve In article Adams, W. W., & Razar, M. J. (1980). Multiples of points on elliptic curves and continued fractions. Proc. London Math. Soc, 41, 481-498. is said on page 482 that elliptic curve 
$$E_a: w^2=v^3+Av+B$$
has one point $O$ at infinity.
I understand it in such way: there is projective curve 
$$E_p: uw^2=v^3+Au^2v+Bu^3$$
and points that in $E_p - E_a$ (as sets) are points at infinity.
In our case there is only one point, which third coordinate equal to $0$: $(0:1:0)$ (we set $u=0$ in $E_p$ polynomial, it automatically forces that $v=0$).
Then authors let $P = (a,b) \ne O$ and 

construct a plane quatric model for E, with respect to which $O$ and
  $P$ are the two points (valuations) at infinity. Let
$x = \frac{w+b}{v-a}, y = 2v+a - (\frac{w+b}{v-a})^2$

after "some algebra" they have
$$y^2=x^4-6ax^2-8bx+c$$
where $c=-4A-3a^2$.
Later they assert that there are two points at infinity $P$ and $O$.
If we look only at last equation we can get only one infinity point -- again (0:1:0). The second should arrive somehow from the map, but I don't understand how.
So the question is: how to explain existence of two points on infinity after change of variables? And how to calculate points on infinity in general case? And how to interpret map between projective varieties in terms of map of affine varieties (like in this case)?
 A: If you just take the projective closure in ${\mathbb P}^2$, then your curve will have only one point at infinity, but this point is singular. What you really want to consider is the smooth projective model of the curve, which you obtain by resolving the singularity at infinity. This leads to two distinct points.
One concrete way of doing this is to consider the weighted projective plane ${\mathbb P}^2_{1,2,1}$ and coordinates $x,y,z$; then the homogeneous equation is
$$y^2 = x^4 - 6 a x^2 z^2 - 8 b x z^3 + c z^4$$
and the points at infinity on this (smooth) model are $(1 : \pm 1 ; 0)$.
A: Possibly it's easier to take two affine curves and glue them together, rather than taking weighted projective space (which admittedly is slicker). So take
$$
  C_1 : y^2 = x^4 + ax^3 + bx^2 + cx + d
$$
and 
$$
  C_2 : v^2 = 1 + au + bu^2 + cu^3 + du^4
$$
and glue them together (along appropriate subsets) via
$$
  x = 1/u\quad\text{and}\quad y=v/u^2.
$$
The points at infinity on the smooth projective closure of $C_1$ are the points
on $C_2$ with $u=0$, which are the two points $(0,1)$ and $(0,-1)$. 
A: Elliptic curves double cover the Riemann sphere with 4 branch points.  When you look at an affine model of the elliptic curve, naturally the points at infinity are missing.  If the point at infinity is not one of the branch points, it will correspond to two points on the elliptic curve that are missing.  That's a way of seeing it without writing down any formulas.
