= Solution
The <Brownian reflection principle> gives, for $t>0$,
$$
\mathbb P(\tau_x\leq t)=\mathbb P\left(\sup_{s\leq t}X_s\geq x\right)=2\mathbb P(X_t\geq x)=2\left(1-\Phi\left(\frac x{\sqrt t}\right)\right).
$$
For a standard <normal random variable> $N'$, this is also $\mathbb P(|N'|\geq x/\sqrt t)=\mathbb P(x^2/(N')^2\leq t)$. Thus the <inverse-square Gaussian law of Brownian first passage> is
$$
\boxed{\tau_x\overset d=\frac{x^2}{(N')^2}.}
$$
In particular, $\tau_x$ is finite almost surely.
The <Brownian motions> $X,Y$ are independent, so conditional on $\tau_x=s$, the ordinate $Y_{\tau_x}$ has the same distribution as $\sqrt s\,N$, with $N$ a standard normal independent of the random time. By <Brownian scaling> and the hitting-time identity,
$$
Y_{\tau_x}\overset d=\frac{xN}{|N'|}\overset d=\frac{xN}{N'}.
$$
The last equality uses symmetry of $N$ and independence of $N,N'$: inserting the independent sign of $N'$ does not change the distribution of the numerator. Therefore
$$
\boxed{Y_{\tau_x}\overset d=xC,\qquad C\text{ standard Cauchy}.}
$$
For completeness, the ratio has density $\int_{\mathbb R}|v|\varphi(cv)\varphi(v)\,dv=1/(\pi(1+c^2))$, where $\varphi$ is the standard normal density. The scaled exit ordinate has density $x/(\pi(x^2+y^2))$.
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