The orthogonal trajectories of the family of curves an `a^(n-1)y = x^n` are given by (A) `x^n+n^2y=constant` (B) `ny^2+x^2=constant` (C) `n^2x+y^n=constant` (D) `y=x`

Options

A. `x^(n)+n^(2)y` = const
B. `ny^(2)+x^(2)` = const
C. `n^(2)x+y^(n)` = const
D. `n^(2)x - y^(n)` = const

Answer

B

Explanation

Differentiating, we have
`a^(n-1)(dy)/(dx) = nx^(n-1)`
`rArr" "a^(n-1)= n x^(n-1) (dx)/(dy)`
Putting this value in the given equation, we have
`nx^(n-1)(dx)/(dy) y = x^(n)`
Replacing `(dy)/(dx)` by `-(dx)/(dy)`, we have `ny = -x (dx)/(dy)`
`rArr" "ny dy + x dx = 0`
`rArr" "ny^(2) + x^(2) =` const.
Which is the required family of orthogonal trajectories.