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The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u,t), y -> f2(u,t) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v +a6

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u,t), f1(u, v), y -> f2(u,t)f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v +a6

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v +a6
+ a6

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

UPDATE:

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

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

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

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 transformation):transformations):

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

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

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):

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

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

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):

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

UPDATE2:

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

The following code outputs rational parametrization of genus 0 curve of degree 5:

x, y = QQ['x,y'].gens()
C=Curve(2*x^5 + x^2*y - 4*x^3*y + 2*x*y^2 + 2*x*y^3 + 1*y^5)
print(C.is_smooth())
print(C.genus())
C.rational_parameterization()

False
0
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by 2*x^5 + y^5 - 4*x^3*y + 2*x*y^3 + x^2*y + 2*x*y^2
  Defn: Defined on coordinates by sending (t) to
        ((-2*t^2 - t)/(4*t^5 + 1), (4*t^4 + 2*t^3)/(4*t^5 + 1))

Is there a function that instead of parametrization outputs transformation x -> f1(u, v), y -> f2(u, v) so that the curve is transformed into a conic section curve (any curve of degree 2)?

The curve would look like this with given coefficients a1..a6.

a1*u^2 + a2*v^2 + a3*u*v + a4*u + a5*v + a6

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):

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

UPDATE2:

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