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This question comes from exercise $1.3$ in Rational Points on Elliptic Curves (Silverman & Tate). I am self studying and trying to work some of the exercises. This one is giving me some trouble.

Let $C$ be the conic given by the equation $$x^2 - 3xy+ 2y^2 - x + 1 = 0.$$ Let $L$ be the line $y = \alpha x + \beta$. Suppose that the intersection $L \cap C$ contains the point $\left(x_0, y_0\right)$. Assuming that the intersection consists of two distinct points, find the second point of $L \cap C$ in terms of $\alpha, \beta, x_0, y_0$.

We know the line intersections the conic at point $P = (x_0, y_0)$, so by the group law it also intersects at point $P^2$. I think my understanding of this is algebraically correct, but I don't know how to translate it to my analytic understanding and write $P^2$ in terms of$\alpha, \beta, x_0,$ and $y_0$.

Substituting $y$ yields \begin{align*} x^2 -3x(\alpha x +\beta) + 2(\alpha x +\beta)^2 -x + 1 &= 0\\ x^2 -3\alpha x^2 -3x\beta + 2\alpha^2x^2 + 4\alpha x\beta +2\beta^2 -x + 1 &= 0\\ \end{align*} which doesn't seem to lead anywhere.

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  • $\begingroup$ What do you mean by $P^2?$ $\endgroup$
    – Thomas Andrews
    Feb 6 at 17:40
  • $\begingroup$ In the book, when there is a point of intersection, we know there must be another point $P^2$ (I think it's just how it's denoted, not 100% sure). What I don't know is how to write $P^2$ in terms of the line. Or maybe I'm not looking at the problem the right way at all. $\endgroup$
    – Clyde Kertzer
    Feb 6 at 17:43
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    $\begingroup$ Have you tried substituting $y = \alpha x + \beta$ into the equation for $C$? $\endgroup$
    – Viktor Vaughn
    Feb 6 at 18:33
  • $\begingroup$ I’m not very familiar with these problems, but if you write your last equation like a second degree polynomial in $x$, you know that $x_0$ is one root, the other can be obtained by using the first Vieta's formulas (en.wikipedia.org/wiki/Vieta%27s_formulas) because your $a_{n-1}$ is in terms of $\alpha$ and $\beta$. $\endgroup$
    – Bongo
    Feb 6 at 20:00
  • $\begingroup$ You could begin by writing cone equation as $(x-\frac{3}{2}y)^2-\dfrac{1}{4}y^2-x+1=0$ then $(x-2y)(x-y)=x+1$ $\endgroup$
    – EDX
    Feb 6 at 20:19

1 Answer 1

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Let's write your quadratic as $$ (1 -3\alpha+2\alpha^2) x^2 -(1+3\beta - 4\alpha \beta)x +2\beta^2 + 1 = 0 $$ Now, by Vieta's formulas, the sum of the roots of the quadratic $ax^2 + b x + c$ is $-b/a$. So for this quadratic we have $$ x_0 + x_1 = \frac{1+3\beta - 4\alpha \beta }{1 -3\alpha+2\alpha^2} $$ Next we use the line equation to get $y_0 + y_1$: $$ y_0 + y_1=\alpha (x_1 + x_0) +2\beta = \alpha \frac{1+3\beta - 4\alpha \beta}{1 -3\alpha+2\alpha^2}+2\beta = \frac{\alpha + 2\beta -3\alpha\beta}{1 -3\alpha+2\alpha^2} $$ Solving then gives $$ x_1 = \frac{1+3\beta - 4\alpha \beta }{1 -3\alpha+2\alpha^2} - x_0\\ y_1 = \frac{\alpha + 2\beta -3\alpha\beta}{1 -3\alpha+2\alpha^2} - y_0. $$ Note that the denominator here is the coefficients of the quadratic form in the conic. I'm not sure of an intuitive way to get the coefficients in the numerator, though.

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