How to construct a curve of degree $6$ having exactly two ordinary singular points of order $3$ and $4$? How to construct a curve of degree $6$ having exactly two ordinary singular points of order $3$ and $4$?
This is related to genus-degree formula of algebraic curves. I am just playing around to construct the curves of given degree and given genus.
EDIT: I set, $d=6,r_1=3,r_2=4$ in the following genus-degree formula trying to create a curve of genus $1$ and degree $6$:
$$\frac{(d-1)(d-2)}{2}-\frac{r_1(r_1-1)}{2}-\frac{r_2(r_2-1)}{2}=1$$
 A: As pointed out by Mohan, such a curve must be reducible, with one of the factors the line through the singular points. Conversely, if we have a quintic $f(x,y,z)$ with ordinary singular points of orders $2$ and $3$, we can multiply $f(x,y,z)$ by the defining equation of the line through the singular points and, as long as that line isn't tangent to the quintic, we get what you want.
Let's put the singular points at $(1:0:0)$ and $(0:1:0)$ and let $f = \sum_{i+j+k=5} f_{ijk} x^i y^j z^k$. Having a singular point of order $2$ at $(1:0:0)$ is equivalent to saying that $f_{500} = f_{410} = f_{401}=0$ and having a singular point of order $3$ at $(0:1:0)$ is equivalent to saying that $f_{050} = f_{140} = f_{041} = f_{230} = f_{131} = f_{032} = 0$. The line through these singular points is $z=0$ and saying that this is not tangent to $f(x,y,z)=0$ is equivalent to saying that $f_{320} \neq 0$. So our polynomial is of the form $\sum_{i=0}^3 \sum_{j=0}^2 f_{ij(5-i-j)} x^i y^j z^{5-i-j}$ with $f_{320} \neq 0$.  The condition that our singular points be ordinary says in addition that the cubic $\sum_{i=0}^3 f_{i2(3-i)} t^i$ does not have multiple roots and neither does the quadratic $\sum_{j=0}^2 f_{3j(2-j)} t^j$, since the roots of these polynomials are the slopes of the tangent lines to the branches of our curve through the singularities.
In summary, just write down some random polymomial $z \sum_{i=0}^3 \sum_{j=0}^2 f_{ij(5-i-j)} x^i y^j z^{5-i-j}$. (The factor of $z$ is the line through the singularities.) With high probability, $f_{320}$ will be nonzero and the relevant quadratic and cubic will be square free, so you will win.
