InterviewSolution
Saved Bookmarks
| 1. |
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`= constantB. `ny^(2)+x^(2)`= constantC. `n^(2)x+y^(n)`= constantD. `n^(2)x-y^(n)`= constant |
|
Answer» Correct Answer - B The equaiton of the given family of curves is `a^(n-1)y=x^(n)" …(i)"` `rArr" "(n-1)loga+logy=n logx` Differentiaing w.r.t. x, we get `(1)/(y)(dy)/(dx)=(n)/(x)" …(ii)"` This is the differential equation of the family of curves given in (i). The differential equation of the orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)by-(dx)/(dy)` in (ii). Replacing `(dy)/(dx)by-(dx)/(dy)` in (ii), we get `(1)/(y)xx-(dx)/(dy)=(n)/(x)rArr xdx +ny dy=0` On integrating, we get `(x^(2))/(2)+n(y^(2))/(2)=C rArr x^(2)+ny^(2)=2C` |
|