How to transform genus zero curve to conic section?

Question

I have a curve of genus zero:

$$C: 2 x^5-4 x^3 y+x^2 y+2 x y^3+2 x y^2+y^5$$

or in homogenized form:

$$C': 2 x^5-4 x^3 y z+x^2 y z^2+2 x y^3 z+2 x y^2 z^2+y^5$$

How to transform it to some conic section curve with rational substitutions $x\to f_1(u, v), y\to f_2(u, v)$?

(or do you know Sagemath code that would do that?)

I computed singular points of the curve and also rational parametrization of the curve if that can be useful to find the transformation.

Singular points: $$(-1:1:1), (0:0:1), (\frac{1}{2}+\frac{i \sqrt{3}}{2}:-\frac{1}{2}+\frac{i \sqrt{3}}{2}:1), (\frac{1}{2}-\frac{i \sqrt{3}}{2}:-\frac{1}{2}-\frac{i \sqrt{3}}{2}:1)$$

Rational parametrization (in one parameter or in two parameters): $$x\to -\frac{p (2 p+1)}{4 p^5+1}, y\to \frac{2 p^3 (2 p+1)}{4 p^5+1}$$ $$x\to -\frac{p q^3 (2 p+q)}{4 p^5+q^5}, y\to \frac{2 p^3 q (2 p+q)}{4 p^5+q^5}$$

Also some small rational points on the curve:

$$(0,0), (-1,1), (\frac{1}{3},-\frac{2}{3}), (-\frac{3}{5},\frac{6}{5}), (-\frac{8}{9},\frac{4}{9}), (-\frac{54}{53},\frac{48}{53}), (-\frac{96}{13},\frac{108}{13})$$

UPDATE:

Here is an example of a curve of degree 3 and genus 0 that I was able to transform to conic - specifically parabola (with forward and backward transformations):

$$y^2=x^3-x^2,\left(x\to \frac{v}{u^2},y\to \frac{v}{u^3}\right)\\ v=u^2+1,\left(u\to \frac{x}{y},v\to \frac{x^3}{y^2}\right)$$

Answer 1

The accepted answer does not give any details of how the transformation was found, so let me give an alternative answer. First, you have a parametrisation so you have a morphism $C \to \mathbb{P}^1$. Embedding $\mathbb{P}^1$ as a conic can be done in many ways (choose three generic homogeneous degree 2 polynomials in $k[x,y]$).

Let me then deal with the more interesting case: We have a genus $0$ curve $C \subset \mathbb{P}^n$ and we don't know whether or not it has a smooth rational point. The key point to note is that if $C/k$ is a genus $0$ curve, then (by Riemann-Roch) the complete linear system of the anti-canonical divisor $|-K_C|$ defines a degree $1$ rational map $C \dashrightarrow \mathbb{P}^2$.

I don't know how to make this algorithmic in SageMath, but Magma has very good canonical divisor functionality for curves:

P2<x,y,z> := ProjectiveSpace(Rationals(), 2);
f := 2*x^5 - 4*x^3*y*z + x^2*y*z^2 + 2*x*y^3*z + 2*x*y^2*z^2+ y^5;
C := Curve(P2, f);
K := CanonicalDivisor(C);
lin_sys := Basis(-K);
phi := map<C -> P2 | lin_sys>;
X := Image(phi);
phi := map<C -> X | lin_sys>;
assert IsInvertible(phi);

// print Inverse(phi); // access the inverse of phi

// In response to the comment below, here's how to make the 
// map nicer because we know a point
P1 := ProjectiveSpace(Rationals(), 1);
lin_sys_P := Basis(Divisor(C![1,-2,3]));
phi := map<C -> P1 | lin_sys_P>;
assert IsInvertible(phi);

// print Inverse(phi); // access the inverse of phi
```

Answer 2

$$2 x^5-4 x^3 y+x^2 y+2 x y^3+2 x y^2+y^5=0,\left(x\to -\frac{8 u v^3-u+4 v^3-2 v}{16 v^5-1},y\to \frac{2 v \left(8 u v^3-u+4 v^3-2 v\right)}{16 v^5-1}\right)\\v-u^2=0,\left(u\to \frac{-2 x^5+2 x^3 y-x y^3-y^5}{2 x \left(x^3+y^3\right)},v\to -\frac{y}{2 x}\right)$$