Find shape operator of a parallel surface

Question

Let $S$ be a regular surface and $\mathbb{X}(u, v)$ a local parametrization of $S$. If $a$ is a small number, let's consider $$\tilde{\mathbb{X}}(u,v)=\mathbb{X}(u,v)+a\textbf{N}(\mathbb{X}(u,v))$$ where $\textbf{N}$ is the normal vector to the surface. Prove that $$\tilde{\mathcal{F}}(\tilde{\mathbb{X}}_u)=\mathcal{F}(\mathbb{X}_u) \quad \quad\tilde{\mathcal{F}}(\tilde{\mathbb{X}}_v)=\mathcal{F}(\mathbb{X}_v)$$ where $\mathcal{F}$ is the shape operator. I've already verified the following (which I think may be useful): $$\tilde{\mathbb{X}}_u = \mathbb{X}_u -a\mathcal{F}(\mathbb{X}_u) \quad \quad \tilde{\mathbb{X}}_v = \mathbb{X}_v-a\mathcal{F}(\mathbb{X}_v)$$ $$\tilde{\mathbb{X}}_u \times \tilde{\mathbb{X}}_v = J \: \mathbb{X}_u \times \mathbb{X}_v$$ where $J=1-2aH+a^2K$ and: $$H = \frac{1}{2}\cdot \: \mbox{trace}\: (\mbox{matrix of} \: \mathcal{F}_p) \quad ;\quad K=\mbox{det}\: (\mbox{matrix of} \: \mathcal{F}_p)$$ Can someone please help me?