About birational map between curves of the form $y^{2}=g(x)$ I'm trying to solve the following exercise but I'm stuck after trying for a long time. 
Suppose that $g(x)=ax^{4}+bx^{3}+cx^{2}+dx+e\in{k[x]}$ and similarly $g'(x)=a'x^{4}+b'x^{3}+c'x^{2}+d'x+e'\in{k[x]}$. Assume that $C:y^{2}=g(x)$ and $C':y^{2}=g'(x)$ are nonsingular curves (of genus 1) and that there is an $k$-isomorphism $\phi:C\rightarrow{C'}$. The problem is to find explicitly what is $\phi$.
In the proof there are three steps that I don't understand. First, one should take a root $\xi$ of $g(x)$. Then the claim is that $\phi$ should map the point $(\xi,0)$ to a point of the form $(\xi',0)$ where $\xi'$ is a root of $g'(x)$. My quetion is why this happens? Why the image is not a point of the form $(a,b)$ with $a$ not a root of $g'(x)$ and $b\neq{0}$?. 
Now once we know that the image of $(\xi,0)$ is of the form $(\xi',0)$, then it follows that $\phi$ maps the roots of $g(x)$ bijectively onto the roots of $g'(x)$. This is ok. But then it follows that $g(x)=\lambda^{2}(\gamma{x}+\delta)^{4}g'(\frac{\alpha{x}+\beta}{\gamma{x}+\delta})$ for some $\alpha,\beta,\gamma,\delta,\lambda\in{k}$. Why this happens?? There must be some bilinearity involved as a consequence of the bijection between roots of $g(x)$ and $g'(x)$ but I can't clearly see what is happening here.
Finally, we should have that $\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$ is an invertible matrix and then from this I am supposed to find $\phi$. My guess is that $\phi$ is of the form $(x,y)\mapsto(\frac{\alpha{x}+\beta}{\gamma{x}+\delta},\frac{\lambda^{-1}y}{(\gamma{x}+\delta)^{2}})$ or something like this but I'm not even sure if this makes sense (due to $(\gamma{x}+\delta)^{2}$).
I will deeply appreciate any help about this.
 A: The model of the affine curves $E: y^{2} = g(x)$ is reminiscent of the fact that they are ramified coverings of the projective line $\mathbb{P}^{1}$ ramified at four points (this is well known: look at the image of the rational function $y/x$) and the ramification points are the roots of $g(x)$  and it is unramified at infinity.
So if we had an isomorphism $\phi$  then it would take the ramification points to ramification points (this is because the map $\phi$ is an isomorphism of varieties not just abstract curves). This answers your first question why roots should go to roots. 
The map $\phi$ induces a map on the underlying $\mathbb{P}^{1}$ as follows. for any point in $p \in \mathbb{P}^{1}$ choose a point $q$ lying in $E : y^{2} = g(x)$ (there will be two choices) and then look at $\phi(q)\in E^{\prime} : y^{2}= g^{\prime}(x)$ and then map it down to the $\mathbb{P}^{1}$ that is covered by $E^{'}$; because $\phi$ is an isomorphism you can revert this process so the induced map is an isomorphism of $\mathbb{P}^{1}$ so it is, in  explicit co-ordinates, a standard transformation of $PGL_{2}$. Writing this in the coordinates $[x:y]$ will give you the relations between $g(x)$ and $g^{\prime}(x)$.   
