Singularity of Morphism and Its Extension I really need help with this question. 

Assume you have a morphism $\varphi: \mathbb{A}^2 \rightarrow
> \mathbb{A}^1$ such that $\varphi (x,y)= x^2-y^4$.
1) Find all points in $\mathbb{C}$ such that $\varphi^{-1}(a)$ is
  singular. And its type of singularity.
2) If $Y_{a}$ is the closure of curve $\varphi^{-1}(a)$ in
  $\mathbb{P}^{2}$ and L is line at $\infty$ (z=0), find the point
  $Y_{a}\cap L$.
3) Let $\psi: \mathbb{P}^{2}\rightarrow \mathbb{P}^{1}$ be rational
  map that extends $\varphi$, find the domain of $\psi$ and the fiber
  $\psi^{-1}(\infty)$.

This is a reducible curve that I've never seen before. I do not know what can I do!
I found $(0,0)$ is a singular point by using Jacobian. For the second one, I took homogenization of curve but I couldn't how to find the intersection with the line at infinity. For the last one, I have no idea.
 A: 1) As you said, by the Jacobian criterion indeed $\varphi^{-1}(0)$ is the unique singular fiber  and the singularity type is two tangent parabolas $(x-y^2)(x+y^2) = 0$. 
2) The fiber has homogenous equation $x^2z^2 = y^4 + az^4 $. Its intersection with the line at infinity given by $z=0$ and the previous equation, i.e this is the point $(1,0,0)$ (with multiplicity $4$).
3)  We take $$\psi(x,y,z) = \frac{x^2z^2 - y^4}{z^4}$$ 
and the domain of $\psi$ is exactly when $z^4 \neq 0$ or $x^2z^2 - y^4 \neq 0$. This means that the domain is  $\Bbb P^2 \backslash \{(1,0,0)\}$. This is not surprising since by the previous question, $(1,0,0)$ was in the closure of each fibers !
Finally, the fiber over $\infty$ is simply given by the line $z=0$, minus the point $[1:0:0]$. For completness, the other projective fibers over $[a:1] \in \Bbb P^1 $  are given by $\{ (x,y,1) : x^2 = y^4 + a \}$. Notice that it does not include the point $(1,0,0)$ by what we said, even though this point is in the closure of all the fibers. 
A final remark : this is possible to find a surface $X$ and a morphism $f : X \to \Bbb P^2$ so that $\psi$ becomes defined everywhere, this morphism is called a blow-up and making your map defined everywhere is typically why algebraic geometers introduce blow-up.
A: About your final remark, If I make blow-up, can I get a such a morphism $f:X\rightarrow \mathbb{P}^{1}$ which an extension of $\varphi$ ? Does it have to be $\mathbb{P}^{2}$ 
