Orthogonal trajectories of family of the curve `x^(2/3)+y^(2/3)=a^((2/3))` , where `a` is any arbitrary constant, is

Options

A. `x^(2//3)-y^(2//3)=c`
B. `x^(4//3)-y^(4//3)=c`
C. `x^(4//3)+y^(4//3)=c`
D. `x^(1/3)-y^(1//3)=c`

Answer

B

Explanation

`x^(2//3)+y^(2//3)=a^(2//3)`
`rArr 2/3x^(-1//3)+2/3y^(-1//3)(dy)/(dx)=0`
Replacing `(dy)/(dx) = x^(-1//3)/y^(-1//3)`
`rArr intx^(1//3)dx=inty^(1//3)dy`
`rArr x^(4//3)-y^(4//3)=(4/3)c`