Given:

Curves y = 4 – x² ...(1)

& y = x² ...(2)

Solving (1) & (2),we get

⇒ y = 4 – x²

⇒ x² = 4 – x²

⇒ 2x² = 4

⇒ x² = 2

⇒ x = ±

Substituting in y = x² ,we get

y = ()²

y = 2

The point of intersection of two curves are (,2) & (, – 2)

First curve y = 4 – x²

Differentiating above w.r.t x,

⇒ = 0 – 2x

⇒ m₁ = – 2x ...(3)

Second curve y = x²

Differentiating above w.r.t x,

⇒ = 2x

m₂ = 2x ...(4)

At (,2),we have,

m₁ = – 2x

⇒ – 2×

⇒ m₁ = – 2

At (,2),we have,

m₂ = – 2x

2 = 2

When m_{1 = –} 2 & m₂ = 2

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