Given:

Curves x² + y² = 2x ...(1)

& y² = x ...(2)

Solving (1) & (2),we get

Substituting y² = x in x² + y² = 2x

⇒ x² + x = 2x

⇒ x² – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 or (x – 1) = 0

⇒ x = 0 or x = 1

Substituting x = 0 or x = 1in y² = x ,we get,

when x = 0,

⇒ y² = 0

⇒ y = 0

when x = 1,

⇒ y² = 1

⇒ y = 1

The point of intersection of two curves are (0,0) & (1,1)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + y² = 2x

⇒ 2x + 2y. = 2

⇒ x + y. = 1

⇒ y. = 1 – x

...(3)

⇒ y² = x

⇒ 2y.1

⇒ ...(4)

At (1,1) in equation(3),we get

m₁ = 0

At (1,1) in equation(4),we get

m₂

when m₁ = 0 & m₂

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅26.56