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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`A. `x^(n)+n^(2)y` = constB. `ny^(2)+x^(2)` = constC. `n^(2)x+y^(n)` = constD. `n^(2)x - y^(n)` = const |
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Answer» Correct Answer - B 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. |
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