(a) The equation of the curve is $9 y^2=x^3$,
Differentiating $18 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2 \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^2}{6 \mathrm{y}}$
Let $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be the point where normal is drawn. Slope of tangent $=\frac{x_1^2}{6 y_1}$
$\therefore$ Slope of normal $=-6 y_1 / x_1^2$
Normal make equal intercepts on the curve
$\therefore \quad$ its slope $=\pm 1 \Rightarrow \frac{-6 \mathrm{y}_1}{\mathrm{x}_1^2}=\pm 1$
or $6 \mathrm{y}_1=\pm \mathrm{x}_1^2 \quad \ldots(i)$
$\left(x_1, y_1\right)$ lies on the curve $9 y^2=x^3$
or $9 \mathrm{y}_1^2=\mathrm{x}_1{ }^3 \quad \ldots(ii)$
Taking + ve sign, eliminating $\mathrm{y}_1$, from (i) \& (ii)
$\begin{aligned}
&9 \times\left(\frac{x_1^2}{6}\right)^2=x_1^3 \text { or } 9 x_1{ }^4=36 x_1{ }^3 \Rightarrow x_1=4 \\
&y_1=\pm \frac{x_1^2}{6}=\pm \frac{16}{6}\left[x_1=4\right], y_1=\pm \frac{8}{3}
\end{aligned}$
The point $\mathrm{P}$ is $\left(4, \pm \frac{8}{3}\right)$