The slope of normal at any point x , y , x > 0 , y > 0 on the curve y = y x is given by x 2 x y - x 2 y 2 - 1 . If the curve passes through the point 1 , 1 , then e · y e is equal to

Step-by-step solution

Given, Slope of normal = - d x d y = x 2 x y - x 2 y 2 - 1 x 2 y 2 d x + d x - x y d x = x 2 d y x 2 y 2 d x + d x = x 2 d y + x y d x x 2 y 2 d x + d x = x x d y + y d x x 2 y 2 d x + d x = x d x y d x x = d x y 1 + x 2 y 2 ln k x = tan - 1 x y         ⋯ i Curve passes though 1 , 1 So, In k = π 4 ⇒ k = e π 4 Now from equation i We get, π 4 + ln x = tan - 1 x y x y = tan π 4 + ℓ n x x y = 1 + tan ℓ n x 1 - tan ℓ n x         &#