(i) Differentiate both sides of \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) with respect to \(x\):

\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0\)

Rearrange to find \(\frac{dy}{dx}\):

\(\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\)

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)

(ii) At the intersection point \(P\), \(y = x\), so substitute into the curve equation:

\(\sqrt{x} + \sqrt{x} = \sqrt{a}\)

\(2\sqrt{x} = \sqrt{a}\)

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

\(x = \frac{1}{4}a\)

Thus, \(P\) is \(\left( \frac{1}{4}a, \frac{1}{4}a \right)\).

The slope of the tangent at \(P\) is \(-\frac{\sqrt{y}}{\sqrt{x}} = -1\).

The equation of the tangent line is:

\(y - \frac{1}{4}a = -1(x - \frac{1}{4}a)\)

\(y = -x + \frac{1}{2}a\)

Rearranging gives:

\(x + y = \frac{1}{2}a\)