Resolving a node singularity on a plane curve So, I am trying to solve Exercise 1.5.6 b) on page 37 in Hartshorne's Algebraic Geometry. For completeness I will include the exercise in my post: 

If $P$ is a node on a plane curve $Y$, show that $\phi^{-1}(P)$
  consists of two distinct nonsingular points on the blown-up curve
  $\tilde{Y}$.

So, first we move $P$ to the origin $O$ by a linear change of coordinates. Now let $Y = V(f)$ where $f$ is an irreducible polynomial in $k[x,y]$ and write $f$ as the sum $f = f_r + ... + f_d$ where each $f_i$ is a homogeneous polynomial of degree $i$ in $x$ and $y$. Since $O$ is a node, $r = 2$ and $f_2$ factors into two distinct linear forms. 
Okay, so what do I understand here? Probably not too much, but we have that $\phi^{-1}(O) = E\cap\tilde{Y}$ where $E$ is the exceptional curve, whose points is in one-to-one correspondance with lines through $O$ in $\mathbb{A}^2$. My question then becomes, how do we prove (not by some hand-waving argument or just saying that it is so) that by intersecting $E$ with the blown-up curve we isolate exactly the points corresponding to the tangent lines of $Y$ at $O$, and how do we then conclude that these two points are smooth on $\tilde{Y}$? 
Hartshorne makes no general argument for this in his presentation of the blowing-up, instead he just states that the effect of blowing up (after considering a concrete example) is to "separate out the branches of curves passing through $O$ according to their slopes" and that "if the slopes are different, their strict transform no longer meet in $X$. Instead they meet $E$ at points corresponding to the different slopes". 
Any insight here would be very much appreciated and helpful.         
 A: I don't remember how Hartshorne talks about blowups in Chapter 1, so this might not be exactly what you're looking for; I'll write it anyway in case you find it useful.
One nice way to calculate with these things is to use affine charts on the blowup. In the case of blowing up the origin in $\mathbf A^2_{x,y}$, the blowup is covered by two affine charts $\mathbf A^2_{t,u}$ and $\mathbf A^2_{v,w}$ in which the blowup map $\pi$ is given by the formulae $$(t,u) \mapsto (tu,u) \\ (v,w) \mapsto (v,vw)$$
respectively. Note that in the first chart, say, the equation of the exceptional curve is just $u=0$. 
Now you have a curve $Y$ with a node at the origin; after a suitable linear change of coordinates you can assume it is given by the equation $$y^2-x^2 = h$$
where $h$ only contains terms of degree 3 or higher.
Now let's find the proper transform $\tilde{Y}$  of $Y$ in the $(t,u)$ chart. We make the substitutions $x=tu, \ y=u$ so that our formula for the preimage $\pi^{-1}(Y)$ becomes
 $$ u^2(1-t^2)= u^3 \tilde{h}$$
for some polynomial $\tilde{h}$ that we don't care about.
To find the proper transform $\tilde{Y}$, we throw out the exceptional curve from the preimage. So we divide the above equation by $u^2$ to get the equation
$$1-t^2 = u \tilde{h}.$$
(Note that this is no longer divisible by $u$, so we no longer have the exceptional curve as a component.)
Finally, to find the points where this intersects the exceptional divisor, we set $u=0$, to get $t= \pm 1$. Note that the partial derivatives with respect to $t$ of the equation defining $\tilde{Y}$ are nonzero at these points, so they really are nonsingular points of $\tilde{Y}$. 
(If you want to be very pernickety, you should then repeat this computation in the other chart, and check that you find the same points of intersection.)  
