(i) To find the equation of \(BC\), we first find the derivative of \(y = \sqrt{1 + 4x}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \times 4\).

At \(B(0, 1)\), the gradient of the tangent is \(2\). Therefore, the gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y - 1 = -\frac{1}{2}x\).

(ii) To find the area of the shaded region, we first find the x-coordinate of \(A\) by setting \(y = 0\) in \(y = \sqrt{1 + 4x}\), giving \(x = -\frac{1}{4}\).

The integral of \(\sqrt{1 + 4x}\) is \(\int \sqrt{1 + 4x} \, dx = \frac{(1 + 4x)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{4}\).

Evaluating from \(-\frac{1}{4}\) to \(0\), the area under the curve is \(\frac{1}{6}\).

The area of \(\triangle BOC\) is \(\frac{1}{2} \times 2 \times 1 = 1\).

Thus, the shaded area is \(1 + \frac{1}{6} = \frac{7}{6}\).