Find the angle to intersection of the following curves :

and x² + y² = ab

Given:

Curves + 1 ...(1)

& x² + y² = ab ...(2)

Second curve is x² + y² = ab

y² = ab – x²

Substituting this in equation (1),

+ 1

1

x²b² + a³b – a²x² = a²b²

x²b² – a²x² = a²b² – a³b

x²(b² – a²) = a²b(b – a)

x²

x²

x²

∴a² – b² = (a + b)(a – b)

x ...(3)

since , y² = ab – x²

y² = ab – ()

y²

y²

y = ± ...(4)

since ,curves are + 1 & x² + y² = ab

Differentiating above w.r.t x,

⇒ . = 0

⇒ . =

⇒

⇒

⇒ m₁ ...(5)

Second curve is x² + y² = ab

⇒ 2x + 2y.0

⇒ m₂ ...(6)

Substituting (3) in (4), above values for m₁ & m_2,we get,

At (, ) in equation(5),we get

⇒ m₁

At (, ) in equation(6),we get

m₂

when m₁ & m₂

tanθ

tanθ

tanθ

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}()