Conormal bundle and lagrangian submanifold 
Let $Q_1^{n_1},Q_2^{n_2}$ be smooth manifolds, $\phi:Q_1\to Q_2$ a smooth map and:
  $$R_\phi:=\{(x, \xi, y,\eta)\mid y=\phi(x), \xi=(d\phi)^*\eta\}\subset T^*Q_1\times T^*Q_2$$
  $$\text{graph}(\phi)=\{(q,\phi(q))\mid q\in Q_1\}\subset Q_1\times Q_2$$
  verifiy that $R_\phi$ is a Lagrangian submanifold and describe the relation between $R_\phi$ and the conormal bundle $N^*\text{graph}(\phi)$.

I was able prove that $R_\phi$ is Lagrangian, but I don't know where the conormal bundle fits into the picture. Since $R_\phi$ is Lagrangian, it has dimension $n_1+n_2$, while $N^*\text{graph}(\phi)$ has dimension $n_2$. Is the conormal bundle embbeded in $R_\phi$, maybe? If so,  why is this interesting?
 A: I'd like to thank Jordan Payette for the crucial help. 
Locally, $R_\phi$ is made of points $(x,(d\phi)^*\eta,\phi(x),\eta)$, so that the coordinates $x,\eta$ are just enough to describe it. This means $R_\phi$ is embedded with dimension $n_1+n_2$, which is half the dimension of  $T^*Q_1\times T^*Q_2$. So to prove $R_\phi$ is Lagrangian, we only need to prove it is isotropic, i.e., $i^*\omega=0$, where $i:R_\phi\hookrightarrow T^*Q_1\times T^*Q_2$ with $ (x,\eta)\mapsto (x,(d\phi)^*\eta,\phi(x),\eta)$ is the inclusion.
If $u(x,\eta):=(d\phi)_x^*(\eta)$, notice that:
$$u_i=u(x,\eta)\left(\frac{\partial}{\partial x_i}\right)=\eta\left(d\phi\left(\frac{\partial}{\partial x_i}\right)\right)=\sum_{j=1}^{n_2}\eta_j\frac{\partial \phi_j}{\partial x_i}$$
$$\frac{\partial u_i}{\partial x_k}-\frac{\partial u_k}{\partial x_i}=\sum_{j=1}^{n_2}\eta_j\frac{\partial^2 \phi_j}{\partial x_k\partial x_i}-\eta_j\frac{\partial^2 \phi_j}{\partial x_i\partial x_k}=0$$
$$\frac{\partial u_i}{\partial \eta_k}=\frac{\partial \phi_k}{\partial x_i}$$
Now we have:
\begin{align*}
i^*\omega &=i^*\left(\sum_{i=1}^{n_1}dx_i\wedge d\xi_i-\sum_{j=1}^{n_2}dy_j\wedge d\eta_j\right)\\
&=\sum_{i=1}^{n_1}dx_i\wedge du_i-\sum_{j=1}^{n_2}d\phi_j\wedge d\eta_j\\
&=\sum_{i=1}^{n_1}dx_i\wedge\left(\sum_{k=1}^{n_1}\frac{\partial u_i}{\partial x_k}dx_k+\sum_{j=1}^{n_2}\frac{\partial u_i}{\partial\eta_j}d\eta_j\right)-\sum_{j=1}^{n_2}\left(\sum_{i=1}^{n_1}\frac{\partial\phi_j}{\partial x_i}dx_i\right)\wedge d\eta_j\\
&=\sum_{i=1}^{n_1}\sum_{k=1}^{n_1}\frac{\partial u_i}{\partial x_k}dx_i\wedge dx_k+\sum_{i=1}^{n_1}\sum_{j=1}^{n_2}\underbrace{\left(\frac{\partial u_i}{\partial\eta_j}-\frac{\partial\phi_j}{\partial x_i}\right)}_{=0}dx_i\wedge d\eta_j\\
&=\sum_{1\leq i<k\leq n_1}\underbrace{\left(\frac{\partial u_i}{\partial x_k}-\frac{\partial u_k}{\partial x_i}\right)}_{=0}dx_i\wedge dx_k=0
\end{align*}
Now we will prove $R_\phi$ is diffeomorphic to $N^*\Gamma_\phi$, where $\Gamma_\phi:=\text{graph}(\phi)$. If $i':\Gamma_\phi\hookrightarrow Q_1\times Q_2$ with $x\mapsto (x,\phi(x))$ is the inclusion, we have $di'\left(\frac{\partial}{\partial x_i}\right)=\frac{\partial}{\partial x_i}+d\phi\left(\frac{\partial}{\partial x_i}\right)$, which means $T_{(x,\phi(x))}\Gamma_\phi=\text{span}\left(\frac{\partial}{\partial x_i}+d\phi\left(\frac{\partial}{\partial x_i}\right)\right)$. Moreover, we have $T^*(Q_1\times Q_2)\simeq T^*Q_1\times T^*Q_2$ and $N^*\Gamma_\phi=\{(x,\xi)\in T^*(Q_1\times Q_2)\mid x\in \Gamma_\phi,\,\xi|_{T_x\Gamma_\phi}\equiv 0\}$. Therefore:
\begin{align*}
N^*\Gamma_\phi &\simeq\{(x,\xi,y,\eta)\mid (x,y)\in \Gamma_\phi,\,(\xi,\eta)|_{T_{(x,y)}\Gamma_\phi}\equiv 0\}\\
&=\left\{(x,\xi,y,\eta)\mid y=\phi(x),\,\xi\left(\frac{\partial}{\partial x_i}\right)+\eta\left(d\phi\left(\frac{\partial}{\partial x_i}\right)\right)=0\,\forall i\right\}\\
&=\left\{(x,\xi,y,\eta)\mid y=\phi(x),\,(\xi+(d\phi)^*\eta)\left(\frac{\partial}{\partial x_i}\right)=0\,\forall i\right\}\\
&=\{(x,\xi,y,\eta)\mid y=\phi(x),\,\xi=-(d\phi)^*\eta\}\subset T^*Q_1\times T^*Q_2
\end{align*}
Since $f:(x,\xi,y,\eta)\mapsto (x,\xi,y,-\eta)$ is a diffeomorphism (in fact a symplectomorphism) from $(T^*Q_1\times T^*Q_2,\omega_1\oplus\omega_2)$ to $(T^*Q_1\times T^*Q_2,\omega_1\oplus-\omega_2)$, we get a diffeomorphism:
$$\left.f\right|_{N^*\Gamma_\phi}:N^*\Gamma_\phi\to R_\phi$$
