Silverman Arithmetic of Elliptic Curves Exercise 3.5(a)(ii) As the title says, this question is about the final part of Exercise 3.5(a)(ii). 
Let $E/K$ be given by a singular Weierstrass equation $f(x,y)$. Suppose that $E$ has a node, and let the tangent lines at the node be $y=\alpha_1+\beta_1$ and $y=\alpha_2+\beta_2$. First we make the assumption that the singular point lies at $(0,0)$, hence we can assume that $\beta_1=\beta_2=0$ and $f(x,y)=y^2+a_1xy-a_2x^2-x^3$. Moreover we are assuming the singular point to be a node, and $\alpha_1\notin K$.
By considering the Taylor expansion of $f(x,y)$ at $(0,0)$, we find that $y^2+a_1xy-a_2x^2=(y-\alpha_1 x)(y-\alpha_2 x)$. Setting $t=y/x$, it follows that $\alpha_1$ and $\alpha_2$ are roots of $t^2+a_1t-a_2\in K[t]$. Moreover $\alpha_1+\alpha_2=-a_1$ and $\alpha_1\alpha_2=-a_2$. Therefore we have that $K(\alpha_1,\alpha_2)=K(\alpha_1)\cong K[t]/(t^2+a_1t-a_2)$, a quadratic extension of $K$.
As the question mentions, by (i) we have $E_{\text{ns}}(K)\subset E_{\text{ns}}(L)\cong L^*$, where $E_{\text{ns}}(L)\rightarrow L^*$ is given by $(x,y)\mapsto \frac{y-\alpha_1 x}{y-\alpha_2 x}$. The goal is to prove that
$$E_{\text{ns}}(K)\cong \left\{t\in L^* \mid N_{L/K}(t)=1\right\}.$$
In one direction, if $x,y\in K$, then computing $N(y-\alpha_1 x)$ with respect to the basis $\left\{1,\alpha_1\right\}$ and $N(y-\alpha_2 x)$ with respect to basis $\left\{1,\alpha_2\right\}$, it follows that $N(y-\alpha_1 x)=N(y-\alpha_2 x)$, hence $$N\left(\frac{y-\alpha_1 x}{y-\alpha_2 x}\right)=1.$$
Hence we indeed have a map $E_{\text{ns}}(K)\rightarrow \left\{t\in L^* \mid N_{L/K}(t)=1\right\}.$ Since it is simply the restriction of the map $E_{\text{ns}}(L)\rightarrow L^*$, I suppose that injectivity follows from this. What remains is to prove surjectivity.
I've attempted to show that $$N\left(\frac{y-\alpha_1 x}{y-\alpha_2 x}\right)=1\iff x,y\in K,$$ but it quickly got messy (and I'm not sure if it leads to the correct result). Is my approach reasonable, or is there a nicer way to prove the existence of the bijection 
$$E_{\text{ns}}(K)\cong \left\{t\in L^* \mid N_{L/K}(t)=1\right\}?$$
 A: As far as I know, answering my own question is considered okay. Here is my very computational approach: there is likely still a much more elegant solution.
Suppose $(x,y)\in E_{\text{ns}}(L)$ such that $N(y-\alpha_1 x)=N(y-\alpha_2 x)$. The goal is to show that $(x,y)\in E_{\text{ns}}(K)$, or $x,y\in K$. 
Write $x=x_1+\alpha_1 x_2$ and $y=y_1+\alpha_1 y_2$. Using the fact that $\alpha_1+\alpha_2=-a_1$ and $\alpha_1\alpha_2=-a_2$ we write that
$$y-\alpha_1 x=y_1-a_2x_2+\alpha_1(y_2-x_1+a_1x_2)$$ and
$$y-\alpha_2 x=y_1+a_2x_2+a_1x_1+\alpha_1(y_2+x_1).$$
Therefore a norm computation shows that
$$N(y-\alpha_1 x)=\begin{vmatrix}y_1-a_2x_2 & a_2y_2-a_2x_1+a_1a_2x_2 \\ y_2-x_1+a_1x_2 & y_1-a_2x_2-a_1y_2+a_1x_1-a_1^2x_2\end{vmatrix}$$
and
$$N(y-\alpha_2 x)=\begin{vmatrix}y_1+a_2x_2+a_1x_1 & a_2y_2+a_2x_1 \\ y_2+x_1 & y_1+a_2x_2-a_1y_2\end{vmatrix}.$$
Now it follows that $N(y-\alpha_1 x)-N(y-\alpha_2 x) = (a_1^2+4a_2)(x_1y_2-x_2y_1)$. If $x_2=0$ then it must follow that $y_2=0$ and the other way around. So suppose that $x_2\neq 0$ and $y_2\neq 0$. Then $x_1x_2^{-1}=y_1y_2^{-1}$. 
So define $s=x_1x_2^{-1}$. Then we can write $x=x_2(s+\alpha_1)$ and $y=y_2(s+\alpha_1)$. Plugging $x,y$ into the curve equation $y^2+a_1xy-a_2x^2-x^3=0$ we obtain that
$$(s+\alpha_1)^2\left(y_2^2+a_1x_2y_2-a_2x_2^2-x_2^3(s+\alpha_1)=0\right).$$
Note that if $s=\alpha_1$, then $x=0$ which contradicts $x_2\neq 0$. Hence we can conclude that 
$$y_2^2+a_1x_2y_2-a_2x_2^2-x_2^3s-\alpha_1x_2^3=0.$$
But then $x_2^3=0$, again contradicting that $x_2\neq 0$. Therefore we conclude that $x_2=0$, from which follows that $y_2=0$. Thus $x=x_1\in K$ and $y=y_1\in K$ and we are done.
