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.

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.
