Substitute \(a = -\frac{7}{2}\) into the line equation, giving \(y = x + \frac{7}{2}\).

Set the curve equal to the line: \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}\).

Rearrange to form a quadratic equation: \(4x^2 - (k+1)x + \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

For the line to be tangent, the discriminant must be zero: \((k+1)^2 - 4 \times 4 \times \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

Simplify: \((k+1)^2 - 2k^2 + 14 = 0\).

Further simplify to: \(7k^2 - 2k - 57 = 0\).

Factor or use the quadratic formula to solve: \((k-3)(7k+19) = 0\).

Thus, \(k = 3\) or \(k = \frac{19}{7}\).