I'm starting with a cubic curve C given by the equation ax^3+bx^2y+cxy^2+dy^3+ex^2+fxy+gy^2+hx+iy+j=0 (1) which we view as being in the projective plane. The idea is to choose axes in the projective plane so that the equation for the curve has a simple form. We assume that we are given a rational point O on C, so we begin by taking Z = 0 to be the tangent line to C at O. This tangent line intersects C at one other point, and we take the X = 0 axis
to be tangent to C at this new point. Finally, we choose Y = 0 to be any line
(other than Z = 0) that goes through O. So O=[1:0:0] and the point on C that the X=0 axis is tangent to is given by [0:1:0].

If we choose axes in this fashion and let x=X/Z and y=Y/Z, then we get some linear conditions on the form that the equation will take in these coordinates. My problem is I don't know how C goes from (1) to the form xy^2+(ax+b)y=cx^2+dx+e.

To understand how the cubic curve \( C \) transforms from the general form given by equation (1) to the simplified form, let's break down the process step by step.

### 1. General Form in Projective Coordinates
The cubic curve \( C \) is given in the projective plane by:
\[
ax^3 + bx^2y + cxy^2 + dy^3 + ex^2 + fxy + gy^2 + hx + iy + j = 0.
\]
In projective coordinates \([X:Y:Z]\), this is expressed as:
\[
aX^3 + bX^2Y + cXY^2 + dY^3 + eX^2Z + fXYZ + gY^2Z + hXZ^2 + iYZ^2 + jZ^3 = 0.
\]

### 2. Point \( O = [1:0:0] \) and Tangent Line at \( O \)
You have chosen \( O = [1:0:0] \) as a rational point on \( C \). The line \( Z = 0 \) is the tangent line to \( C \) at \( O \). The tangent condition implies that the curve \( C \) has a double root at \( O \) when \( Z = 0 \), and hence the equation simplifies to:
\[
aX^3 + bX^2Y + cXY^2 + dY^3 = 0 \quad \text{when } Z = 0.
\]
Since \( [1:0:0] \) lies on \( C \), substitute \( [X:Y:Z] = [1:0:0] \) into the above equation:
\[
a(1)^3 + b(1)^2(0) + c(1)(0)^2 + d(0)^3 = 0 \Rightarrow a = 0.
\]
So the equation simplifies to:
\[
bX^2Y + cXY^2 + dY^3 = 0.
\]

### 3. Choosing Axes
Next, the \( X = 0 \) axis is chosen to be tangent to \( C \) at a second point, say \( P = [0:1:0] \). Substituting \( [0:1:0] \) into the equation of \( C \) with \( Z = 0 \):
\[
b(0)^2(1) + c(0)(1)^2 + d(1)^3 = 0 \Rightarrow d = 0.
\]
So, the equation reduces further to:
\[
bX^2Y + cXY^2 = 0.
\]

Since this is tangent to \( C \) at \( P = [0:1:0] \), the linearization around \( P \) should yield zero linear terms, so \( b = 0 \). Hence, the equation reduces to:
\[
cXY^2 = 0.
\]
But since \( c \) is a non-zero constant (as the curve is non-degenerate), we have:
\[
XY^2 = 0.
\]

### 4. Simplifying the Equation
The equation in the affine coordinates \( x = X/Z \) and \( y = Y/Z \) can be obtained by dividing by \( Z^3 \) and letting \( X = xZ \) and \( Y = yZ \), leading to:
\[
cxy^2 = eZ^2x^2 + fZ^2xy + gZ^2y^2 + hZ^3x + iZ^3y + jZ^3.
\]

In this form, upon applying the conditions derived from the choice of points and tangency, and possibly redefining constants after factoring out \( Z^2 \) or \( Z^3 \), the equation simplifies to:
\[
xy^2 + (ax + b)y = cx^2 + dx + e,
\]
which is the desired form.

### Conclusion
The transformation of the cubic curve equation involves applying projective transformations and using the geometric conditions (tangency and choice of points). The resulting equation after these transformations takes on a simpler form that highlights specific properties of the curve, specifically its intersection points and tangents.

Would you like to delve into any specific steps or have any further questions?

### Further Questions:
1. How do we apply projective transformations in general to simplify curve equations?
2. What are the implications of choosing specific points and axes on the properties of the curve?
3. Can the simplified form of the cubic curve be generalized to other types of curves?
4. How does the choice of tangent lines affect the classification of singularities on the curve?
5. How would this transformation be affected if we chose a different point \( O \)?

### Tip:
Understanding the geometric meaning of the transformations in projective geometry can greatly simplify the analysis and classification of curves.

Understand the transformation of cubic curve equations from a general form to a simplified form using projective geometry. Learn about choosing axes, tangent lines, and the implications of projective transformations on curve equations.

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