Question

Find the condition for the curves `x^2/"a"^2 - y^2/"b"^2` = 1; xy = c² to interest orthogonally.

Answer 1

Let the curves intersect at (x₁, y₁).

Therefore, `x^2/"a"^2 - y^2/"b"^2` = 1

⇒ `(2x)/"a"^2 - (2y)/"b"^2 "dy"/"dx"` = 0

⇒ `"dy"/"dx" = ("b"^2x)/("a"^2y)`

⇒ Slope of tangent at the point of intersection (m₁) = `("b"^2x_1)/("a"^2y_1)`

 Again xy = c²

⇒ `x "dy"/"dx" + y` = 0

⇒ `"dy"/"dx" = (-y)/x`

⇒ m₂ = `(-y)/x_1`

For orthoganality, m₁ × m₂ = – 1

⇒ `"b"^2/"a"^2` = 1 or a² – b² = 0.