(i) The slope of the normal is \(-\frac{1}{3}\), so the slope of the tangent is \(3\) (since \(m_1 m_2 = -1\)).

Set \(\frac{dy}{dx} = 3 = \frac{12}{\sqrt{4 \times 2 + a}}\).

\(3 = \frac{12}{\sqrt{8 + a}}\)

\(\sqrt{8 + a} = 4\)

\(8 + a = 16\)

\(a = 8\)

(ii) Integrate \(\frac{dy}{dx} = \frac{12}{\sqrt{4x + 8}}\) to find \(y\).

\(y = \int 12(4x + 8)^{-\frac{1}{2}} \, dx\)

\(y = 12 \times 2 \sqrt{4x + 8} + c\)

\(y = 24 \sqrt{4x + 8} + c\)

Using point \(P(2, 14)\):

\(14 = 24 \sqrt{16} + c\)

\(14 = 24 \times 4 + c\)

\(14 = 96 + c\)

\(c = -82\)

Thus, the equation of the curve is \(y = 24 \sqrt{4x + 8} - 82\).