Quartic curve - what is the genus?

Question

I am studying the following quartic curve:

$f(x,y) = c_1x^2 + c_2x^4 + c_3x^2y + c_4x^2y^2 + c_5y^2 + c_ 6y^3 + c_7y^4$

where $c_i$ are constant (in fact they are expressions in terms of other constants). Starting to learn a bit about curves, I found that a necessary condition for a point $(x_0, y_0)$ be singular (a double point) is that

$$F(x_0, y_0) = 0,\qquad F_x (x_0, y_0) = 0,\qquad F_y (x_0, y_0) = 0$$

and that the second derivatives calculated at that point are not all equal to zero. Solving these three equations (trial and error) I got two solutions:

$$(x_0, y_0) = (0,0),\qquad (x_0, y_0) = (0, -2 c_5/c_6)$$

The second solution is a solution due to the fact that the coefficients $c_i$ are interrelated. For both points the second derivatives are not equal to zero.

Therefore, this curve has apparently has two double points, both with multiplicity equal to 2. Thus, this curve would have genus = 1, if there are no more singular points.

My questions are:

1. Is what I said above accurate?

2. Is there any simple way to test if there is more singular points?

3. If there are no more singular points, how can I parameterize a quartic curve like that? (I tried to transform this curve in an elliptic one, making $x^2 = z$, but I'm not sure if this is correct.)

Answer 1

This is not a real answer, since the curve you are interested in is not the generic one of the type you describe (you say that there are relations between the coefficients). However, if you are starting to learn about curves maybe you will be interested in seeing how the generic such curve can be studied by hand.

The proper setting for the question, as pointed out in a comment, is the projective plane $P^2$, so I'm going to add a variable $z$ and make everything homogeneous. Also, since you say nothing about the coefficients, I will work over the complex numbers.

Consider the linear system of plane quartics spanned by $z^2x^2$, $x^4$, $zx^2y$, $x^2y^2$, $z^2y^2$, $zy^3$ and $y^4$. The only base point of this system is the point $P=[1,0,0]$. It is easy to see that every curve of the system is singular at $P$ (it is true for all the generators) and that there is at least one curve (e.g. $z^2(x^2+y^2)=0$) that has an ordinary double point at $P$. Hence by Bertini's theorem the general curve of the system has an ordinary double point at $P$ and is smooth elsewhere. It is easy to show directly that such a curve cannot be reducible, so by the genus formula it has geometric genus 2.