The abscissa of the point on the curve $9y^2 = x^3$, the normal at which cuts off equal intercepts on the coordinate axes is

Differentiating $9y^2=x^3$ we have $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^2}{6y}$

Any point on the curve is of the form $\left(t^2,t^3/3\right)$ and so $\frac{\mathrm{d}y}{\mathrm{d}x}$ at this point is $t/2$ Thus an equation of normal is $y-t^3/3=(-2/t)\left(X-t^2\right)$  This will intersect coordinate axes at $\left(0,2t+t^3/3\right)$ and $\left(\frac{t^4}{6}+t^2,0\right)$

Hence we must have  $2t+\frac{t^3}{3}=t^2+\frac{t^4}{6}\Rightarrow 2+\frac{t^2}{3}=t+\frac{t^3}{6}$

Clearly $t = 2$ satisfies, the last equation. Hence the abscissa of the required point is 4. (For$t = 0,$ the normal meets both the axes only at origin.)