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I have an elliptic curve $ y^2=x^3+109x^2+224x$ and a point $P(-100;260)$ on it. And I need to find point $2P$. I took a formulas $$x_2=\left(\frac{ax_1-b}{y_1}\right)^2 -a+x_1$$ and $$y_2=-y_1+\frac{ax_1-b}{y_1}(x_1-x_2)$$ put into this formulas $a=109$, $b=224$, $x_1=-100$, $y_1=260$ and have $2P=(\frac{6850936}{4225}; \frac{37736919137}{514164}) $ but this point is not lie on curve. In the article the result is $2P=(\frac{8836}{25}; -\frac{950716}{125})$ but I have no idea how can I find them.

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    $\begingroup$ The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula. $\endgroup$ – Dietrich Burde Dec 10 '18 at 13:30
  • $\begingroup$ formulas I have taken are not from this article. Is it right formulas for this curve? $\endgroup$ – aid78 Dec 10 '18 at 14:00
  • $\begingroup$ No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$. $\endgroup$ – Dietrich Burde Dec 10 '18 at 15:49
  • $\begingroup$ @DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-\frac{109}{3}$ got equation $ y^2=x^3-\frac{11209}{3}x+\frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point. $\endgroup$ – aid78 Dec 10 '18 at 16:39
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For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.

If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form $$y = y_1 + s(x - x_1)$$

where $$2y_1 s = f'(x_1) \iff s = \frac{f'(x_1)}{2y_1} = \frac{3x_1^2 + 2a x_1 + b}{2y_1}$$

If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root of the cubic equation

$$\big(y_1 + s(x- x_1)\big)^2 = f(x) = x^3 +ax^2 + bx + c$$

Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to the coefficient of $x^2$, we obtain

$$2 x_1 + x_3 = s^2 - a \implies x_3 = s^2 - 2x_1 - a$$

The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis. This means $$\begin{cases} x_2 &= x_3 &= s^2 - 2s_1 - a\\ y_2 &= -y_3 &= -y_1 + s(x_1 - x_2) \end{cases} \quad\text{ where }\quad s = \frac{3x_1^2 + 2ax_1 + b}{2y_1}$$

For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get $s = \frac{81}{5}$ and

$$ \begin{cases} x_2 &= \left(\frac{81}{5}\right)^2 - 2(-100) - 109 = \frac{8836}{25}\\ y_2 &= -260 + \frac{81}{5}\left(-100 - \frac{8836}{25}\right) = -\frac{950716}{125} \end{cases}$$

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