The point on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes is

Given curve 9y2 = x3 ….(1)


Differentiate w.r.t. x,




Equation of normal:



it makes equal intercepts with the axes


slope of the normal = ±1


x2 = ±6y


Squaring both the sides,


x4 = ± 36y2


From (1),


x= 0, 4


and


But the line making equal intercept cannot pass through origin.


So, the required points are

1