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