The points of the curve $9{y^2} = {x^3}$, where the normal to the curve makes equal intercepts with the axes are

(A) $\left( {4,\; \pm \cfrac{8}{3}} \right)$

(B) $\left( {4,\;\cfrac{{ - 8}}{3}} \right)$

(C) $\left( {4,\; \pm \cfrac{3}{8}} \right)$

(D) $\left( { \pm 4,\;\cfrac{3}{8}} \right)$

Option A is correct

We have,$9{y^2} = {x^3}$ …(i)

Differentiating (i) w.r.t. $x$, we get $18y\cfrac{{dy}}{{dx}} = 3{x^2} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{{x^2}}}{{6y}}$

Let $P({x_1},{y_1})$be the point on (i).
Therefore, $9y_1^2 = x_1^3$ …(ii)

Now, slope of tangent at$({x_1},{y_1})$ is $\cfrac{{x_1^2}}{{6{y_1}}}$

Slope of the normal at $({x_1},{y_1})$ is $- \cfrac{{6{y_1}}}{{x_1^2}}$

It is given that the normal makes equal intercepts on the axes

Therefore, its slope $= \pm 1 \Rightarrow \cfrac{{6{y_1}}}{{x_1^2}} = \pm 1$
$\Rightarrow 6{y_1} = \pm x_1^2$ …(iii)

Now from (ii) and (iii), we get

$9{\left( {\cfrac{{x_1^2}}{6}} \right)^2} = x_1^3$

$\Rightarrow \cfrac{{x_1^4}}{4} = x_1^3 \Rightarrow {x_1} = 4$

Putting ${x_1} = 4$ in (iii), ${y_1} = \pm \cfrac{{x_1^2}}{6} = \pm \cfrac{{16}}{6} = \pm \cfrac{8}{3}$

Hence, the points are $\left( {4,\; \pm \cfrac{8}{3}} \right)$.