Show that the following set of curves intersect orthogonally :

x² + 4y² = 8 and x² – 2y² = 4.

Given:

Curves x² + 4y² = 8 ...(1)

& x² – 2y² = 4 ...(2)

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

from 2nd curve,

x² = 4 + 2y²

Substituting on x² + 4y² = 8,

⇒ 4 + 2y² + 4y² = 8

⇒ 6y² = 4

⇒ y²

⇒ y = ±

Substituting on y = ±, we get,

⇒ x² = 4 + 2(±)²

⇒ x² = 4 + 2()

⇒ x² = 4 +

⇒ x²

⇒ x = ±

⇒ x = ±

∴ The point of intersection of two curves (,) & (,)

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

⇒ x² + 4y² = 8

⇒ 2x + 8y. = 0

⇒ 8y. = – 2x

...(3)

⇒ x² – 2y² = 4

⇒ 2x – 4y.0

⇒ x – 2y.0

⇒ 4yx

⇒ ...(4)

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

m₁

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

m₂ = 1

when m₁ & m₂

⇒ × = – 1

∴ Two curves x² + 4y² = 8 & x² – 2y² = 4 intersect orthogonally.