Using a simple saddle point approximation one can derive the following:
$$ [A] \quad K_y(a) \sim \frac{1}{2} \exp{(-r + y\,s)}\sqrt{\frac{\pi}{2r}}\Big( 1 +
\text{erf}\big(\sqrt{\frac{r}{2}}\, s\big) \Big) \quad$$
where
$$ r=\sqrt{a^2+y^2} \text{ and } s=\text{arcsinh}(y/a) $$
I derived it from the known integral relation
$$ K_y(a) = \int_0^\infty \exp(-a\,\cosh(t))\cosh(y\,t) dt$$
Approximate $\cosh(yt) \sim 1/2 \exp{(yt)} .$ The integrand sans the 1/2 is $\exp(-a\,\cosh(t) + yt).$ Expand the argument of the exponential around its saddle point $s = \text{arcsinh}(y/a)$ and you get
$$ K_y(a) \sim \frac{1}{2} \exp{(-r + y\,s)} \int_0^\infty
\exp{(-\frac{r}{2} (t-s)^2)}\, dt $$
The integral evaluates to an error function, which is shown in [A]. Alternatively, if the argument of the error function is sufficiently large, say, >5, the the expression in the big parentheses can be approximated by 1.

Using Mathematica, for $y=x-1/2=10000,$ and $a=1/2$ I get about 6 significant figures agreement. For $a=11/2,$ I get about 5 sig figs. It is expected that this approximation will deteriorate as $a$ becomes comparable to $x$ but I haven't studied the limits. Comments seem to indicate that $x>>a$ so this formula may be sufficient. One can also get more terms by doing a complete expansion.

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