Definitions:
Given $V(I=\{F\})=\mathbf{\mathcal{X}}_F\subset \mathbb{P}^{2}(K)$ a smooth projective curve, and $P$ a point of the curve, $\mathcal{O}_P$ is the ring of rational funcions that are regular at $P$. If $Q_F$ is the field of fractions of $K[X,Y,Z]/V(F)$ I have this definition: $$\mathcal{O}_P:=\Big{\{}\;\frac{G}{H}\;\;\Big{|}\;\; (\frac{G}{H}\in Q_F)\;\land\;(G\;\text{y}\;H\;\text{homogeneus})\;\land\;(deg(G)=deg(H))\land\;(G(P)\neq 0)\Big{\}}$$
The claim:
The maximal ideal $\mathcal{M}_P=\{f\in\mathcal{O}_P\;|\; f(P)=0\}$ of $\mathcal{O}_P$ is a principal ideal.
The proof I don't fully understand
So this is the proof of book I am reading:
"We want to show that the maximal ideal $\mathcal{M}_{P}$ is a principal ideal, that is to say, generated by one element. Let $\mathbf{\mathcal{X}}$ be a smooth curve in $\mathbb{A}^{2}$ defined by the equation $F=0$, and let $P = (a, b)$ be a point on $\mathbf{\mathcal{X}}$. The maximal ideal $\mathcal{M}_P$ is generated by $(x-a)$ and $(y-b)$. Now $$F_{X}(P)(x - a) + F_Y (P)(y - b) \equiv 0\pmod {\mathcal{M}_P^2}$$
Hence the K-vector space $\mathcal{M}_P/\mathcal{M}_P^2$ has dimension 1, and therefore $\mathcal{M}_P$ has one generator."
I don't understand how the tangent being $0\pmod {\mathcal{M}_P^2}$ implies that $\mathcal{M}_P/\mathcal{M}_P^2$ has dimension 1, and how that translates to $\mathcal{M}_P$ having 1 generator.