Q:

How to classify a plane complex curve?

Let $p_1, p_2, t_1, t_2, a \in \mathbb{C}$ be constants. Consider the following plane complex curve in $\mathbb{C}^2$ ($c_1, c_2$ are indetermniates)
\begin{align}
& {p_1}^2 {p_2}^2 c_1 {t_1}^2 t_2 + {p_1}^2 p_2 {c_1}^2 c_2 {t_1}^2 t_2 + {p_1}^2 p_2 c_2 {t_1}^4 + {p_1}^2 c_1 {c_2}^2 {t_2}^3 \\
& + 2 p_1 p_2 {c_1}^3 c_2 {t_2}^3 + p_1 p_2 {c_1}^2 {c_2}^2 {t_1}^2 t_2 + p_2 {c_1}^3 {c_2}^2 {t_2}^3 = a {p_1}^2 p_2 c_1 c_2 {t_1}^2 t_2.
\end{align}
According to the article, the genus of the curve is the number of integer lattice points in the interior of the Newton polygon of the curve. 
The Newton polygon of the curve is the convex hull of the points $(1,0)$, $(2,1)$, $(0,1)$, $(1,2)$, $(3,1)$, $(2,2)$, $(3,2)$, $(1,1)$. There are two integer lattice points in the Newton polygon: $(1,1)$, $(2,1)$. Therefore the genus of this curve is $2$. Is this correct? How to check that whether or not this curve is smooth? Is this curve an elliptic curve? Thank you very much.

A:

A computation with magma reveals that this curve is isomorphic, as a genus two curve over the function field $\mathbb{C}(p_1,p_2,t_1,t_2,a)$, to the hyperelliptic curve given by the Weierstrass equation 
$$y^2=f(x):=4t_2^6x^6 + 4p_1t_1^2t_2^4x^5 + (p_1^2t_1^4t_2^2 - 4p_1t_1^2t_2^4a -
    4p_2t_1^2t_2^4)x^4 + (-2p_1^2t_1^4t_2^2a - 4p_1p_2t_1^4t_2^2 +
    4p_1t_1^4t_2^3)x^3 + (2p_1^2t_1^6t_2 + p_1^2t_1^4t_2^2a^2 -
    4p_1^2t_1^2t_2^4)x^2 - 2p_1^2t_1^6t_2ax + p_1^2t_1^8
$$
This defines an hyperelliptic curve for any value of the parameters such that the discriminant of the polynomial $f(x)$ is non-zero. 
The discriminant is zero when $t_1=0$ or $t_2=0$ or $p_1=0$ (some cases one should consider apart) or a very large of huge degree irreducible polynomial in the parameters is zero.  

