I need a birational transformation of singular quartics into weierstrass form. the quartics are of genus one and have the following form: $$(x^2 + c) (y^2 + i) + k = 0$$ where $c,i,k$ are given rational numbers, and $k$ is of big height. The only known rational solutions: the singular points in homogenous coordinates: $(x,y,z)=(0,1,0),(1,0,0)$.
Methods described in the literature (e.g. Lawrence Washington, Elliptic curves, p.37) require a non-singular rational solution. but after this text a singular quartic can be transformed birationally into a nonsingular curve, where the singular solutions are mapped into two nonsingular solutions. Unfortunately, I have not found in this work and in google, how to do this transformation. I ask for information, how to transform singular quartics as above into Weierstrass form or how to transform into a nonsingular curve.
This is just an extened comment. I would conjecture that it is unlikely that there is a nice closed from for a Weierstrass equation for the curve $$C: (X^2 + rZ^2)(Y^2 + sZ^2) + tZ^4 = 0$$
over the function field $\mathbb{Q}(r,s,t)$ - maybe some expert on elliptic surfaces will come along and say that this is codswhallop.
Why do I say this? Let's ask Magma about the case where $r = -u^2$ (we will see why this is helpful). Take the affine piece where $y = 1$ and compute the blowup (as in e.g., Hartshorne I Example 4.9.1) see that the curve $$C' : sx^2z^2 + x^2 + (rs + t)z^2 + r = 0$$ is birational to $C$. Notice that $C'$ has the nonsingular point $(u, 0)$.
Now let's ask Magma what the Weierstrass equation is, shall we?
Which outputs the lovely equation