# Find intersection points of $x^2 - 3xy+ 2y^2 - x + 1 = 0$ and $y = \alpha x + \beta$

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.

• What do you mean by $P^2?$ Feb 6 at 17:40
• 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. Feb 6 at 17:43
• Have you tried substituting $y = \alpha x + \beta$ into the equation for $C$? Feb 6 at 18:33
• 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 $\beta$. Feb 6 at 20:00
• You could begin by writing cone equation as $(x-\frac{3}{2}y)^2-\dfrac{1}{4}y^2-x+1=0$ then $(x-2y)(x-y)=x+1$
– EDX
Feb 6 at 20:19

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.