Given:

The curve y² = 2x³ and The Slope of tangent is 3

y² = 2x³

Differentiating the above w.r.t x

⇒ 2y^{2 – 1} = 2x^{3 – 1}

⇒ y = 3x²

⇒

Since, The Slope of tangent is 3

= 3

⇒ = 1

⇒ x² = y

Substituting x² = y in y² = 2x³,

(x²)² = 2x³

x⁴ – 2x³ = 0

x³(x – 2) = 0

x³ = 0 or (x – 2) = 0

x = 0 or x = 2

If x = 0

⇒

⇒ , which is not possible.

So we take x = 2 and substitute it in y² = 2x³,we get

y² = 2(2)³

y² = 28

y² = 16

y = 4

Thus, the required point is (2,4)