The points on the curve $9y^2=x^3$ where normal to the curve makes equal intercepts with the axes is/are :

$\left(4, \frac{8}{3}\right),\left(4, \frac{-8}{3}\right)$

The correct answer is Option (1) → $\left(4, \frac{8}{3}\right),\left(4, \frac{-8}{3}\right)$

$9y^2=x^3$  ...(1)

differentiating wrt (x)

$18y\frac{dy}{dx}=3x^2$

so $\frac{dy}{dx}=\frac{x^2}{6y}$

slope of normal → $-\frac{dy}{dx}=(-\frac{x^2}{6y})^{-1}$

as it makes equal intersepts

$-\frac{x^2}{6y}=-1⇒6y=x^2$

or $y=\frac{x^2}{6}$

from (1) $9×\frac{x^4}{36}=x^3$

so $x=4$

from (1) $9y^2=4^3$

$y^2=\frac{64}{9}$

$y=±\frac{8}{3}$

Points $\left(4, ±\frac{8}{3}\right)$