Question

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

Answer 1

Let the point be (h, k) Given curve is 9y2 = x3

Since the point (h, k) **** on the curve, so 9k2 = h3

⇒ 9(h2/6)2 = h3

⇒ 9h4 = 36h3

⇒ h = 4

when h = 4, then k = ± 8/3

Hence, the points are (4, 8/3) & (4, – 8/3)