Question

Find the point on curve 9y² = x³ where normal to the curve makes equal intercepts with the axes.

Answer 1

9 y² = x³                               y = +-  [x^³/² ] /3
18 y dy/dx = 3 x²

slope of tangent = m = dy/dx = x²/6y = + 3 x²/(6x^³/²) = + √x / 2
Slope of normal to the curve at x,y:  +  2/√x

 When a straight line makes equal intercepts with the axes, the slope = 1 or -1.

So    x = 4. 
       y = + √x³ / 3  = + 8/3

so the points are:   (4, 8/3)  and  (4, -8/3)