Resolution of Singularities: Base Point Consider the curve $y^2=4x^3-ax-b$, where $a$ is a fixed constant and $b$ is a free constant. For each value of $b$ we get a family of curves.

Part 1: Show that the family of curves intersect at infinity (the base-point).
My Solution: Using projective geometry (homogeneous coordinates), we have $x=u/w$, $y=v/w$, where $[x,y,1]=[u,v,w]\in\mathbb{CP}^2$. The equation of the level curves is $$\frac{v^2}{w^2}=4\frac{u^3}{w^3}-a\frac{u}{w}-b\qquad\Rightarrow\qquad v^2w=4u^3-auw^2-bw^3$$
For line at infinity, we let $w=0$ such that $0=4u^3\quad\Rightarrow\quad u=0$
Thus the base-point is $[0,1,0]$.
Is this correct?


Part 2: Resolve the base point through blow-ups.
I do not know how to go about answering this question.

Any help with Part 2 will be highly appreciated. 
I am relatively new to concepts such as homogeneous coordinates and blow-ups. Are there any recommended books for beginners on these concepts? Thanks.
 A: I will sketch the argument, since writing all the details would be too tedious, and would certainly include typos...
An important idea about blowups of points in the plane is the following: given two curves passing through a point $P$, the blowup $\operatorname{Bl}_P(\mathbb P^2)$ separates the curves (i.e. their strict transforms do not intersect) iff they have different tangent directions at $P$. The same idea works for germs of curves, which is why repeatedly blowing up a singularity (e.g. a node) desingularizes the curve. Moreover, blowing up reduces the tangency even if the curves hit each other at $P$ with higher order. So the idea is that by doing enough blowups, we can separate all the curves in the linear series for different $b$'s.
Here is how this works. In every blowup that follows there are two charts that glue to give the blowup. Only one chart will really by relevant at each stage, so I will simplify matters by working in that chart. You can check the other chart by mimicking my computations, if you wish. You correctly identified the basepoint $[0:1:0]$, which we should blow up. In the chart $v\neq 0$ the equation becomes 
$$0 = 4u^3 - auw^2 - bw^3 - w$$
by substituting $v = 1$. I want to work with variables $x,y$, so let's replace this by $0 = 4x^3 - axy^2 - by^3 - y$, and compute the first blowup at $(x,y) = (0,0)$ in the way explained here. There are two charts. In the first we do
$$0 = 4(x/y\cdot y)^3 - a(x/y\cdot y)y^2 - by^3 - y$$
obtaining
$$0 = y\left(4(x/y)^3y^2 - a(x/y)y^2 - by^2 - 1\right)$$
where $y = 0$ is the local equation of the exceptional divisor of the blowup, and the equation of the strict transform is in brackets. Note that the two equations do not intersect, since $y=0$ implies the equation in brackets equals $1$. This means that the strict transform in this chart doesn't touch the preimage of the basepoint, so we can ignore this chart.
In the second chart, we do
$$0 = 4x^3 - ax(y/x\cdot x)^2 - b(y/x\cdot x)^3 - (y/x\cdot x)$$
which gives
$$0 = x\left(4x^2 - ax^2(y/x)^2 - bx^2(y/x)^3 - y/x\right)$$
where $x=0$ now gives the exceptional divisor. The strict transform has the solution $(x,y/x) = (0,0)$. This does not depend on $b$, so this is a solution of every strict transform. Thus, this point is a basepoint of the blowup linear series. This means we have to repeat the procedure to eliminate the basepoint. Let's work with the equation $$0 = 4x^2 - ax^2y^2 - bx^2y^3 - y$$ instead of using the awkward coordinate $y/x$. 
Again, there will be two charts to deal with, but now I'll only work with one. We do
$$0 = 4x^2 - ax^2(y/x\cdot x)^2 - bx^2(y/x\cdot x)^3 - (y/x\cdot x)$$
which gives
$$0 = x\left(4x - ax^3(y/x)^2 - bx^4(y/x)^3 - y/x\right)$$
and again we see that $(x,y/x) = (0,0)$ is a solution of the exceptional and the strict transform. Thus we have a basepoint. We repeat the procedure, using the equation
$$0 = 4x - ax^3y^2 - bx^4y^3 - y$$
as our starting point.
There will be two charts, and I ignore one of them. There is an added twist at this stage. We do
$$0 = 4x - ax^3(y/x\cdot x)^2 - bx^4(y/x\cdot x)^3 - y/x\cdot x$$
which gives
$$0 = x\left(4 - ax^4(y/x)^2 - bx^6(y/x)^3 - y/x\right)$$
and we find that $(x,y/x) = (0,4)$ is a point on the intersection of the strict transform and the exceptional divisor, i.e. is a basepoint. To blow this point up we have to change coordinates so that the basepoint is the origin. E.g. we can take $(x,y) = (x, y/x-4)$, in other words $x = x, y/x = y+4$. We must rewrite the equation of the strict transform before proceeding. We obtain
$$0 = 4 - ax^4(y+4)^2 - bx^6(y+4)^3 - (y+4)$$
in other words
$$0 = - ax^4(y+4)^2 - bx^6(y+4)^3 - y$$
or 
$$0 = ax^4(y+4)^2 + bx^6(y+4)^3 + y$$
which we see passes through $(0,0)$. Thus, we can apply the previous techniques and continue on.
I'll let you take it from here. At some point the intersection point of the exceptional divisor and the strict transform will be uniquely determined by $b$, which means that all the curves intersect the exceptional in a different point, i.e. you have separated all the curves of the linear series.
