If the point $P(a, b)$ lies on the curve $9y^2 = x^3$ such that the normal to the curve at$P$ makes equal intercepts with the axes then  the value of $(a + 3b)$  is

Normal makes equal intercept at $P\Rightarrow {\frac{dy}{dx}}_P =1$
Given $9y^2=x^3$
Let the point on the curve be $x=t^2$ and $y=\frac{t^3}{3}$
$\frac{dx}{dt}=2t ; \frac{dy}{dt}=t^2$
$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx} =\frac{t^2}{2t}=\frac{t}{2} \Rightarrow$slope of the normal $=-\frac{2}{t}$

normal makes equal intercept
hence $-\frac{2}{t}=-1 \Rightarrow t=2$

Hence $P=(4,\frac{8}{3}) \Rightarrow a+3b=4+3.\frac{8}{3}=4+8=12 Ans$