Orthonormal frame on hyperbolic plane I'm having trouble comprehending a question from Do Carmo's Differential Forms and Applications. The question (in its entirety) is as follows:

(Exercise 5-2 in Do Carmo). Let $H^2$ be the upper half-plane, that is,
  $$
H^2=\{(x,y)\in\mathbb{R}^2;y>0\}.
$$
  Consider in $H^2$ the following inner product: If $(x,y)\in H^2$ and $u,v\in T_pH^2$, then
  \begin{equation}
\langle u,v \rangle_p=\frac{u\cdot v}{y^2}
\end{equation}
  where $u\cdot v$ is the canonical inner product of $\mathbb{R}^2$. Prove that this is a Riemannian metric in $H^2$ whose Gaussian curvature is $K\equiv -1$; with this Riemannian metric $H^2$ is called the hyperbolic plane. [Hint: Choose the orthonormal frame $e_1=\frac{a_1}{y}$, $e_2=\frac{a_2}{y}$, where $\{a_1,a_2\}$ is the canonical frame of $\mathbb{R}^2$.]

I'm not all that troubled with the question, except that I don't understand the hint to the point where I think there's a typo (though there probably isn't). 
I get how $e_1,e_2$ are orthogonal with respect to the inner product $\langle\cdot\,,\cdot\rangle$, but
$$
\langle e_1,e_1\rangle_p=\frac{e_1\cdot e_1}{y^2}=\frac{(a_1\cdot a_1)/y^2}{y^2}=\frac{1}{y^4}\neq 1.
$$
So $e_1,e_2$ can't be orthonormal with the inner product. It would work perfectly with $e_1=ya_1$ and $e_2=ya_2$, but then why does the hint say something different? Am I misunderstanding what he means by "orthonormal?" Any help is appreciated. Thank you in advance.

Since $\{ ye_i \}$ is ON so $\{\omega_1:= \frac{dx}{y},\ \omega_2:=\frac{dy}{y}\}$ is coframe  Then $$ \omega_{12}\omega_2=d\omega_1=-\frac{1}{y^2} dydx=\omega_1\omega_2 $$ $$ \omega_{21}\omega_1=d\omega_2= 0 $$
Hence $$ \omega_{12}=\omega_1 $$
And apply $ d\omega_{12}=-K\omega_1\omega_2$
