To find the point of tangency, equate the line and curve equations:

\(kx - k = -\frac{1}{2x}\)

Multiply through by \(2x\) to clear the fraction:

\(2kx^2 - 2kx = -1\)

Rearrange to form a quadratic equation:

\(2kx^2 - 2kx + 1 = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 2k\), \(b = -2k\), \(c = 1\).

\((-2k)^2 - 4(2k)(1) = 0\)

\(4k^2 - 8k = 0\)

Factor out \(4k\):

\(4k(k - 2) = 0\)

Since \(k\) is positive, \(k = 2\).

Substitute \(k = 2\) back into the line equation:

\(y = 2x - 2\)

Equate with the curve equation:

\(2x - 2 = -\frac{1}{2x}\)

Multiply through by \(2x\):

\(4x^2 - 4x + 1 = 0\)

\((2x - 1)^2 = 0\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into the line equation:

\(y = 2 \times \frac{1}{2} - 2 = -1\)

The point of tangency is \(\left( \frac{1}{2}, -1 \right)\).