Given:

Curves x² + 4y² = 8 ...(1)

& x² – 2y² = 2 ...(2)

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

from 2nd curve,

x² = 2 + 2y²

Substituting on x² + 4y² = 8,

⇒ 2 + 2y² + 4y² = 8

⇒ 6y² = 6

⇒ y² = 1

⇒ y = ±1

Substituting on y = ±1,we get,

⇒ x² = 2 + 2(±1)²

⇒ x² = 4

⇒ x = ±2

∴ The point of intersection of two curves (2,1) & ( – 2, – 1)

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

⇒ x² + 4y² = 8

⇒ 2x + 8y. = 0

⇒ 8y. = – 2x

...(3)

⇒ x² – 2y² = 2

⇒ 2x – 4y.0

⇒ x – 2y.0

⇒ 4yx

⇒ ...(4)

At (2,1) in equation(3),we get

m₁

At (2,1) in equation(4),we get

= 1

m₂ = 1

when m₁ & m₂ = 1

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}(3)

θ≅71.56