(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation:

\((r^2)^2 = 6r^2 \sin \theta \cos \theta\)

\(r^4 = 6r^2 \sin \theta \cos \theta\)

\(r^2 = 3 \sin 2\theta\)

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_0^{\frac{1}{2}\pi} 3 \sin 2\theta \, d\theta\)

\(= \frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi}\)

= \(\frac{3}{2}\)

(d) To find the maximum distance from the initial line, set \(y = 3^{\frac{1}{2}} \sin^{\frac{1}{2}} 2\theta \sin \theta\) and solve \(\frac{dy}{d\theta} = 0\).

\(\sin^{\frac{1}{2}} 2\theta \cos \theta + \sin^{-\frac{1}{2}} 2\theta \cos 2\theta \sin \theta = 0\)

\(\sin 2\theta \cos \theta + \cos 2\theta \sin \theta = 0\)

\(\tan 2\theta = -\tan \theta\)

\(\frac{2 \tan \theta}{1 - \tan^2 \theta} = -\tan \theta\)

\(\theta = \frac{1}{3}\pi\)

Maximum distance is \(\frac{3^{\frac{3}{2}}}{2^{\frac{3}{2}}} = 1.40\).