Find the point on

Given:

The curve is y = x³

y = x³

Differentiating the above w.r.t x

⇒ = 3x^{2 – 1}

⇒ = 3x² ...(1)

Also given the The Slope of the tangent is equal to the x – coordinate,

= x ...(2)

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

i.e, 3x² = x

⇒ x(3x – 1) = 0

⇒ x = 0 or x =

Substituting x = 0 or x = this in y = x³,we get,

when x = 0

⇒ y = 0³

⇒ y = 0

when x =

⇒ y = )³

⇒ y =

Thus, the required point is (0,0) & (,)