To find the values of \(k\) and \(a\), we equate the equations of the curve and the line at the given \(x\)-coordinates.

1. For \(x = 0\):

\(4(0)^2 - k(0) + \frac{1}{2}k^2 = 0 - a\)

\(\frac{1}{2}k^2 = -a\) (Equation 1)

2. For \(x = \frac{3}{4}\):

\(4\left(\frac{3}{4}\right)^2 - k\left(\frac{3}{4}\right) + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(4\left(\frac{9}{16}\right) - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\) (Equation 2)

Substitute \(a = -\frac{1}{2}k^2\) from Equation 1 into Equation 2:

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} + \frac{1}{2}k^2\)

\(\frac{9}{4} - \frac{3}{4}k = \frac{3}{4}\)

\(\frac{9}{4} - \frac{3}{4} = \frac{3}{4}k\)

\(\frac{6}{4} = \frac{3}{4}k\)

\(k = 2\)

Substitute \(k = 2\) back into Equation 1:

\(\frac{1}{2}(2)^2 = -a\)

\(2 = -a\)

\(a = -2\)

Thus, \(k = 2\) and \(a = -2\).