Let $k$ be a field with $\operatorname{char}(k)\neq 2,3$ and $E: y^2=x^3+Ax+B$ an elliptic curve over $k$ , where $4A^3+27B^2\neq 0$ and let $P=(\alpha,\beta)$ be a point defined over $k$.

Show that if $\beta\neq 0$, then $x-\alpha$ is a uniformizer of $P$

So basically we want to show that the maximal ideal $(y-\beta,x-\alpha)$ is equal to $(x-\alpha)$ by somehow utilizing the curve $E$. Our teacher showed the following example during a lecture so I'd like to apply a similar strategy, but don't know how.

Example: Let $k$ be as above, $E: x^2+y^2=1$ and $P=(1,0)$, show that $(y)$ is a uniformizer. Then, using $E$:
$y^2=1-x^2=(1-x)(1+x)$. 

Now (1+x) is nonzero at p, so $\frac{1}{1+x}\in\mathcal{O}\_{E,p}$, thus $(1-x)=\frac{1}{1+x}\cdot y^2\in(y)$ and thus $(x-1,y)=(y)$ which proves our statement. 

However, I have no idea how to apply this to the complicated example since it won't factor in such a nice way.