Given:

The curve is 2a²y = x³ – 3ax²

Differentiating the above w.r.t x

⇒ 2a² = 3x^{3 – 1} – 32ax^{2 – 1}

⇒ 2a² = 3x² – 6ax

⇒ = ...(1)

= The Slope of the tangent = tan

Since, the tangent is parallel to x – axis

i.e,

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ 3x² – 6ax = 0

⇒ 3x(x – 2a) = 0

⇒ 3x = 0 or (x – 2a) = 0

⇒ x = 0 or x = 2a

Substituting x = 0 or x = 2a in 2a²y = x³ – 3ax²,

when x = 0

⇒ 2a²y = (0)³ – 3a(0)²

⇒ y = 0

when x = 2

⇒ 2a²y = (2a)³ – 3a(2a)²

⇒ 2a²y = 8a³ – 12a³

⇒ 2a²y = – 4a³

⇒ y = – 2a

Thus, the required point is (0,0) & (2a, – 2a)