$\text{ If the tangent at any point }P\left(4m^2,8m^3\right)\text{ of }{x}^3-y^2=0\text{ is also a normal to curve }$$x^3-y^2=0\text{ , then }m=$

 

$\begin{array}{l}{x}^3-y^2=0\dots \dots \dots \text{ (1) }\\ \text{ Differentiating with respect to }x\end{array}$

$\begin{array}{l}2y\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2\\ \therefore \text{ Slope of tangent at }P=\frac{\mathrm{d}y}{\mathrm{d}x}={\left(\frac{3x^2}{2y}\right)}_{\left(4m^2,8m^3\right)}=3m\end{array}$

$\begin{array}{l}\therefore \text{ Equation of tangent at }P\text{ is }y-8m^3=3m\left(x-4m^2\right)\\ \text{ Or }y=3mx-4m^3\dots \dots ..\text{ (2) }\end{array}$

$\begin{array}{l}\text{ Solving (1) and (2) }\\ \text{ (i.e) }{x}^3=y^2\text{ and }y=3mx-4m^3\\ {\left(3mx-4m^3\right)}^2=x^3\\ x^3-9m^2{x}^2+24m^{4}x-16m^6=0\end{array}$

By trial and error method, $x=m^2$ satisfies above equations.

After further factorization we get

$\begin{array}{l}{\left(x-4m^2\right)}^2(x-m{)}^2=0\\ \therefore x=4m^2,m^2\\ \text{ Put }x=m^2\text{ in }\left(2\right);\text{ then }y=-m^3\\ \therefore Q\left(m^2,-m^3\right)\end{array}$

$\begin{array}{l}\text{ Slope of tangent at }Q={\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}_{\left(m^2,-m^3\right)}=-\frac{3}{2}m\\ \text{ Slope of normal at }Q=\frac{2}{3m}\end{array}$

$\therefore \frac{2}{3m}=3m$

$\begin{array}{l}\text{ (Or) }9m^2=2\\ \therefore m=\pm \frac{\sqrt{2}}{3}\\ \text{ Therefore, the correct answer is (1). }\end{array}$