Step 1: Calculate the first derivative, $\frac{dy{dx}$.}
First, find the derivatives with respect to the parameter $t$.
\[
\frac{dx}{dt} = 1 - \cos t, \quad \frac{dy}{dt} = \sin t.
\]
The first derivative is given by the chain rule:
\[
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t}{1-\cos t}.
\]
Using half-angle identities, this simplifies to $\frac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \cot(t/2)$.
Step 2: Calculate the second derivative, $\frac{d^2y{dx^2}$.}
We use the formula $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) / \frac{dx}{dt}$.
\[
\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(\cot(t/2)) = -\csc^2(t/2) \cdot \frac{1}{2}.
\]
\[
\frac{d^2y}{dx^2} = \frac{-\frac{1}{2}\csc^2(t/2)}{1-\cos t} = \frac{-\frac{1}{2}\csc^2(t/2)}{2\sin^2(t/2)} = -\frac{1}{4\sin^4(t/2)} = -\frac{1}{4}\csc^4(t/2).
\]
Step 3: Use the given condition to find the value of K.
We are given that at $t=K$, $\frac{d^2y}{dx^2} = -1$.
\[
-\frac{1}{4}\csc^4(K/2) = -1 \implies \csc^4(K/2) = 4.
\]
Taking the fourth root, $\csc(K/2) = \sqrt{2}$ (since cosecant of half angle for cycloid is positive).
\[
\sin(K/2) = \frac{1}{\sqrt{2}}.
\]
Since $K>0$, the smallest positive solution is $K/2 = \pi/4$, which gives $K = \pi/2$.
Step 4: Evaluate the required limit.
We need to find $\lim_{t \to K} \frac{y}{x} = \lim_{t \to \pi/2} \frac{1-\cos t}{t-\sin t}$.
Since the function is continuous at $t=\pi/2$, we can directly substitute the value.
\[
\frac{1-\cos(\pi/2)}{\pi/2 - \sin(\pi/2)} = \frac{1-0}{\pi/2 - 1} = \frac{1}{(\pi-2)/2}.
\]
\[
\boxed{\frac{2}{\pi-2}}.
\]