Reason behind standard names of coefficients in long Weierstrass equation A long Weierstrass equation is an equation of the form 
$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
Why are the coefficients named $a_1, a_2, a_3, a_4$ and $a_6$ in this manner, corresponding to $xy, x^2, y, x$ and $1$ respectively? Why is $a_5$ absent?
 A: This is because one thinks of each symbol having a weight. If $y$ has weight 3, $x$ has weight 2, and each $a_i$ has weight $i$, then the total weight of each term is exactly 6.
But where do these numbers come from? To answer that question, one has to look at the order of the pole at $\infty$. Consider the homogeneous form of the Weierstrass equation:
$$v^2 w + a_1 u v w + a_3 v w^2 = u^3 + a_2 u^2 w + a_4 u w^2 + a_6 w^3$$
The point at $\infty$ has homogeneous coordinates $(u : v : w) = (0 : 1 : 0)$. Taking $x = u / w$ and $y = v / w$ recovers the original affine coordinates. On the other hand, if we take $\xi = u / v$ and $\eta = w / v$, we get:
$$\eta + a_1 \xi \eta + a_3 \eta^2 = \xi^3 + a_2 \xi^2 \eta + a_4 \xi \eta^2 + a_6 \eta^3$$
One can check that $\xi$ is a local coordinate function near $(\xi, \eta) = (0, 0)$, i.e. near $\infty$. On the other hand,
$$\eta = \frac{\xi^3}{1 + a_1 \xi + a_3 \eta - a_2 \xi^2 - a_4 \xi \eta - a_6 \eta^2}$$
and the denominator does not vanish at $\infty$, so we deduce $\eta$ has a zero of order 3 at $\infty$. Hence, $y = 1 / \eta$ has a pole of order 3 at $\infty$, and $x = \xi / \eta$ has a pole of order 2. That is why $y$ is considered to have weight 3 and $x$ is considered to have weight 2.
A: If you make a standard change of variables $X=u^2x$ and $Y=u^3y$, the shape of the Weierstrass equation is preserved, and the coefficients of the new equation are
$$Y^2+a_1uXY+a_3u^3Y =X^3+a_2u^2X^2+a_4u^4X+a_6u^6.$$ 
In other words, the new coefficients replacing the $a_i$ are of the form $a_iu^i$, for $i=1,2,3,4,6$. Since no coefficient changes by a power $u^5$, we don't call any of them by $a_5$.
