Discrete valuation on an algebraic function field 
Let $f=y^2-(x^3+2x^2+1)\in \mathbb{Q}[x,y]$ and $L=\mathbb{Q}(\alpha,\beta)$ is an algebraic function field given by $f(\alpha,\beta)=0$, where $\alpha=x+(f),\beta=y+(f)\in \mathbb{Q}[\alpha,\beta]\subset L$. Suppose $\nu$ is a normalised discrete valuation on $L$, such that $\nu(\mathbb{Q}\setminus\{0\})=0$, $\nu(\alpha-1)>0$, $\nu(\beta-2)>0$. Find all $(l_0,l_1,l_2)\in \mathbb{Q}^3$, such that 
  $$\nu(l_0+l_1\alpha+l_2\beta)=1,\ 2,\ 3,$$
  respectively. 

I have no idea how to solve the problem. But I notice that $(1,2)\in V_f(\mathbb{Q})$ is a zero of $f$, and $f$ is smooth at $(1,2)$, since 
$$\nabla(f)=(-3x^2-4x,2y)$$
does not vanish there. By definition, if $\nu$ is a normalised discrete valuation, then $\exists M\in \mathbb{P}_{L/\mathbb{Q}}$ a place of the algebraic function field $L$, such that $M=(t)$ and $\nu=\nu_t$, where
$$\mathbb{P}_{L/\mathbb{Q}}:=\{M\subset L\mid \exists R \ \text{valuation ring of}\ L:\mathbb{Q}\subset R\subset L, M\ \text{maximal ideal of}\ R\},$$
and $\nu_t(a):=\max\{j\ge 0\mid p^j\vert a\}$ is the normalised valuation of $t$. 
Thank you very much in advance for your help!
 A: Let $E: y^2 = x^3 + 2x^2 + 1$ be the elliptic curve associated to $f$, let $P = (1,2)$ and let $\ell = \ell_0 + \ell_1\alpha + \ell_2\beta$. As mentioned in the comments, the geometric meaning of the values for the valuation of $\ell$ is:
\begin{align*}
\nu(\ell) \geq
\begin{Bmatrix}
1\\ 2\\ 3
\end{Bmatrix}
\iff
\begin{Bmatrix}
\ell \text{ passes through $P$}\\
\ell \text{ is tangent to $E$ at $P$}\\
\ell \text{ is tangent to $E$ at $P$ and $P$ is a flex}
\end{Bmatrix} \, .
\end{align*}
To find the explicit values $\ell_0, \ell_1, \ell_2$, we examine each case. Note that the Taylor expansion of $f$ about $(1,2)$ is
$$
f = -7(x-1) + 4(y-2) - 5(x-1)^2 + (y-1)^2 - (x-1)^3 \, .
$$
$\nu(\ell) = 3$: To show there are no such lines, it suffices to show that $P$ is not a flex of $E$. A point $Q$ on an elliptic curve is a flex iff it is a $3$-torsion point iff $[3]Q = \infty$ iff $[-2]Q = Q$. But $[-2]P = (-15/16, -89/64) \neq P$, so $P$ is not $3$-torsion. One could also show directly that $P$ is not a flex by showing that the Hessian matrix of the homogenization of $f$ is not $0$ at $P$. (See Fulton's curve book, exercise 5.23 (p. 59) for more on flexes and the Hessian.) The Hessian is
$$
\left(\begin{array}{rrr}
-6 \, x - 4 \, z & 0 & -4 \, x \\
0 & 2 \, z & 2 \, y \\
-4 \, x & 2 \, y & -6 \, z
\end{array}\right)
$$
which is nonzero for $(x,y,z) = (1,2,1)$.
$\nu(\ell) = 2$: As you noted, $E$ is smooth so there is a unique tangent line at $P$, which we can read off the Taylor series: $\ell = -7(x-1) + 4(y-2) = - 1 - 7x + 4y$. Thus
$$
\begin{pmatrix}
\ell_0\\ \ell_1\\ \ell_2
\end{pmatrix}
=
c
\begin{pmatrix}
-1\\ -7\\ 4
\end{pmatrix}
$$
for any $c \neq 0$.
$\nu(\ell) = 1$: Then $0 = \ell(P) = \ell_0 + \ell_1 + 2 \ell_2$, so $\ell_0 = -\ell_1 - 2\ell_2$. So the set of solutions are
$$
\begin{pmatrix}
\ell_0\\ \ell_1\\ \ell_2
\end{pmatrix}
=
\begin{pmatrix}
-\ell_1 - 2\ell_2\\
\ell_1\\
\ell_2
\end{pmatrix}
$$
except that we must also posit that
$$
\begin{pmatrix}
\ell_0\\ \ell_1\\ \ell_2
\end{pmatrix}
\neq
c
\begin{pmatrix}
-1\\ -7\\ 4
\end{pmatrix}
$$
for any $c \neq 0$, or else we fall back into the $\nu(\ell) = 2$ case.
(Note that it's somewhat unwieldy to think about the triples $(l_0,l_1,l_2)$ as elements of $\mathbb{Q}^3$, as scaling a triple by a nonzero scalar results in the same line. It's more natural to consider them as elements $(\ell_0 : \ell_1 : \ell_2)$ of $\mathbb{P}^2$.)
Just as a note, there are elliptic curves where the $\nu(\ell) = 3$ case is possible: it just happened in this case that $E$ had no $3$-torsion points defined over $\mathbb{Q}$. For instance, consider the elliptic curve $E': y^2 = x^3 + 1$ and the point $P' = (0,1)$. Letting $g = y^2 - (x^3 + 1)$, the Taylor expansion of $g$ at $P'$ is
$$
g = 2(y-1) + (y-1)^2 - x^3
$$
so in the local ring at $P'$ we have $y-1 = \frac{x^3}{y+1}$. This shows that the tangent line $y=1$ at $P'$ vanishes to order $3$.
