Find the points on the curve `9y^2=x^3` where normal to the curve makes equal intercepts with the axes.

Answer

Point is `(4,8 / 3)`

Explanation

$$
\begin{aligned}
&9 y^{2}=x^{3} \\
&y=\frac{x^{3 / 2}}{3} \\
&\frac{d y}{d x}=\frac{3}{2} \times \frac{x^{1 / 2}}{3}=\frac{1}{2} \sqrt{x} \\
&\text { slope of Normal }=\frac{-2}{\sqrt{x}}=-1 \\
&x=4 \\
&y=\sqrt{\frac{64}{9}}=\frac{8}{3}
...