Small resolution of threefold with a node I heard that a one-parameter family of surfaces acquiring a node (explicitly below) can be made into a smooth family through a small resolution of the ambient threefold. I want to know why.
Explicity, consider a one parameter family of (analytic) surfaces 
$$X=\{x^2+y^2+z^2+t^2=0\}\xrightarrow{f} \Delta, \ \ \ \ (x,y,z,t)\mapsto t$$
in a small neighborhood of $0\in \mathbb C^4$ over a disk $\Delta$. Then both the fiber $X_0=f^{-1}(0)$ and the total space $X$ have a node at $0$. It is claimed that a small resolution $\hat{X}$ of the total space $X$ produces a smooth family of surfaces $\hat{X}\to \Delta$. Here "small" means exceptional locus has dimension one, or (in this case) is simply a copy of $\mathbb P^1$. 
Assuming such resolution exists, I'd like to ask:

Question 1: How to show $\hat{X}\to \Delta$ is smooth? 

Remark: This is intuitively true to me because topologically $X_0$ can be obtained from nearby $X_t$ by contracting the vanishing cycle $\cong S^2$. On the other hand, the small resolution replace the node by a $\mathbb P^1\cong S^2$, so it seems to reverse the process and make $\hat{X}_0$  topologically the same to $X_t$. However, I want to see how this is worked out in local coordinates.
I also heard/read that such a small resolution is obtained by big blowup $Bl_0X$ with an exceptional divisor $E\cong \mathbb P^1\times \mathbb P^1$ a smooth quadric surface, then blowdown one of the ruling. By my computation, the normal bundle of a ruling to $Bl_0X$ is $\mathcal{O}_{\mathbb P^1}(-1)+\mathcal{O}_{\mathbb P^1}$, but I'd like to ask

Question 2: What is the criterion to blowdown a ruling of quadric surface in a threefold?

Thanks in advance if anyone has a solution or reference!
 A: For question 1, recall that a smooth map is characterized by a finiteness condition, flatness, and a regular fiber over each geometric point. The finiteness condition is clearly satisfied, and the fact that the family of surfaces is smooth gives you the third condition. It remains to show that this map is flat: this is accomplished by miracle flatness which states that a map from Cohen-Macaulay to regular with equidimensional fibers is in fact flat. 
For question 2, there is a classical result due to Castelnuovo which tells you when you can blow down a curve in a smooth projective surface and get a smooth projective surface back. Here's the treatment from Hartshorne:

Theorem V.5.7 (Castelnuovo): If $Y$ is a curve on a nonsingular projective surface $X$, with $Y\cong \Bbb P^1$ and $Y^2=-1$, then there exists a morphism $f:X\to X_0$ to a (nonsingular projective) surface $X_0$ and a point $P\in X_0$ so that $X$ is isomorphic via $f$ to the blowup of $X_0$ at the point $p$, and $Y$ is the exceptional curve.

In general, you can contract any $\Bbb P^1$ on a surface if that $\Bbb P^1$ has negative self-intersection, but the result will be singular unless that self-intersection is $-1$. In general, you can contract any curve with negative self intersection and get a complex-analytic space out of it, but it need not be a variety.
