# Elliptic curve- Component of point

If $E/ \mathbb{Q}$ elliptic curve in the general Form of Weierstrass and $P=(x,y)$ a rational point of it, show that the first coordinate of the point $2P$ is

$$x(2P)=\frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6}$$ where $$b_2=a_1^2+4a_2$$ $$b_4=a_1a_3+2a_4$$ $$b_6=a_3^2+4a_6$$ $$b_8=a_1^2a_6-a_1a_3a_4+4a_2a_6+a_2a_3^2-a_4^2$$



I have done the following:

$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$, then

$$\lambda=\frac{3x_1^2+2a_2x_1+a_4-a_1y_1}{2y_1+a_1x_1+a_3}, v=\frac{-x_1^3+a_4x_1+2a_6-a_3y_1}{2y_1+a_1x_1+a_3}$$ and then $$2P=(\lambda^2+a_1 \lambda-a_2-x_1-x_2, -(\lambda+a_1)x_3-v-a_3)$$ $P_1=P_2 \Rightarrow x_1=x_2=x, y_1=y_2=y$ $$\lambda^2+a_1 \lambda-a_2-x-x= \\ \left ( \frac{3x^2+2a_2x+a_4-a_1y}{2y+a_1x+a_3} \right )^2+a_1 \frac{3x^2+2a_2x+a_4-a_1y}{2y+a_1x+a_3}-a_2-2x \\ =\frac{(3x^2+2a_2x+a_4-a_1y)^2}{(2y+a_1x+a_3)^2} +a_1 \frac{(3x^2+2a_2x+a_4-a_1y)(2y+a_1x+a_3)}{(2y+a_1x+a_3)^2}-\frac{(a_2+2x)(2y+a_1x+a_3)^2}{(2y+a_1x+a_3)^2} \\ = \frac{12a_2x^3-6a_1x^2y+4a_2^2x^2+6a_4x^2-4a_1a_2xy+4a_2a_4x+a_1^2y^2-2a_1a_4y+a_4^2+9x^4}{(2y+a_1x+a_3)^2}+\frac{3a_1^2x^3+6a_1x^2y+2a_1^2a_2x^2+3a_1a_3x^2-a_1^3xy+4a_1a_2xy+2a_1a_2a_3x+a_1^2a_4x-2a_1^2y^2-a_1^2a_3y+2a_1a_4y+a_1a_2a_3}{(2y+a_1x+a_3)^2}-\frac{2a_1^2x^3+8a_1x^2y+a_1^2a_2x^2+4a_1a_3x^2+4a_1a_2xy+8a_3xy+2a_3^2x+2a_1a_2a_3x+4a_2y^2+4a_2a_3y+a_2a_3^2+8xy^2}{(2y+a_1x+a_3)^2}$$

Is it right so far? How can we continue in order to show the desired result?

EDIT:

EDIT 2

How can we find the tangent of the curve at the point $P$ ?

• Do you know what $2P$ is, geometrically? With this and your other question, it seems you are trying to apply formulas without really knowing what you're doing. – fkraiem Dec 28 '14 at 0:57
• This is not the place for a lecture, you should really read some background material from a textbook or something. I recommend the book of Hoffstein, Pipher and Silverman for an introduction to elliptic curves requiring not too much background. – fkraiem Dec 28 '14 at 1:36
• @fkraiem I tried to use Wolfram but it failed... Which other program could I use? – evinda Dec 30 '14 at 15:14
• Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer Dec 30 '14 at 15:55
• No problem, @evinda, it's just to remove some clutter from under the question after the fact. Having such clarifying comment threads is no problem while they are still used (I didn't notice that the last interactions here were so recent, I'd have waited longer if I had noticed). – Daniel Fischer Dec 30 '14 at 16:04

In the first line you already have given the general Weierstrass form for the affine curve. The other one is just the associated projective curve, where you obtain the affine part by setting $z=1$. So the difference is only whether you want to consider the affine or projective Weierstrass form for your elliptic curve. Usually, the affine form is used, with the formula for $2P$, as you have written (with $P=(x_1,y_1)$ ?). Note that there are typos in your formula, e.g. $2_3y_1$, and what are $x_2,x_3$ ?
• Yes, we have $x_3=λ^2+λa_1−a_2−2x$, with $\lambda$ as above. Then you are done. There is no need to expand this further. – Dietrich Burde Dec 23 '14 at 9:52
• Dietrich The answer should be $$x(2P)=\frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6}$$ where $b_2=a_1^2+4a_2, \\ b_4=a_1 a_3+2a_4, \\ b_6=a_3^2+4a_6, \\ b_8=a_1^2a_6-a_1a_3a_4+4a_2a_6+a _2a_3^2-a_4^2$ – evinda Dec 23 '14 at 11:18
• In the formula that I found there is also the variable $y$. What have I done wrong? – evinda Dec 23 '14 at 11:19
• You have not done something wrong. The results are equivalent, because $P=(x,y)$ satisfies the elliptic curve equation. – Dietrich Burde Dec 23 '14 at 11:59