Given:

Curves y = x² ...(1)

& x² + y² = 20 ...(2)

First curve y = x²

⇒ m₁ = 2x ...(3)

Second curve is x² + y² = 20

Differentiating above w.r.t x,

⇒ 2x + 2y. = 0

⇒ y. = – x

⇒ m₂ ...(4)

Substituting (1) in (2),we get

⇒ y + y² = 20

⇒ y² + y – 20 = 0

We will use factorization method to solve the above Quadratic equation

⇒ y² + 5y – 4y – 20 = 0

⇒ y(y + 5) – 4(y + 5) = 0

⇒ (y + 5)(y – 4) = 0

⇒ y = – 5 & y = 4

Substituting y = – 5 & y = 4 in (1) in (2),

y = x²

when y = – 5,

⇒ – 5 = x²

⇒ x

when y = 4,

⇒ 4 = x²

⇒ x = ±2

Substituting above values for m₁ & m_2,we get,

when x = 2,

m₁4

when x = 1,

m₁4

Values of m₁ is 4 & – 4

when y = 4 & x = 2

m₂

when y = 4 & x = – 2

m₂

Values of m₂ is &

when m₁ = ∞ & m₂ = 0

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅77.47