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