Orthogonal trajectories of family of the curve `x^(2/3)+y^2/3=a^((2/3))` , where `a` is any arbitrary constant, is (a) `( b ) (c) (d) x^(( e ) (f) (g)2/( h )3( i ) (j) (k))( l )-( m ) y^(( n ) (o) (p)2/( q )3( r ) (s) (t))( u )=c (v)` (w) (b) `( x ) (y) (z) x^(( a a ) (bb) (cc)4/( d d )3( e e ) (ff) (gg))( h h )-( i i ) y^(( j j ) (kk) (ll)4/( m m )3( n n ) (oo) (pp))( q q )=c (rr)` (ss) (c) `( d ) (e) (f) x^(( g ) (h) (i)4/( j )3( k ) (l) (m))( n )+( o ) y^(( p ) (q) (r)4/( s )3( t ) (u) (v))( w )=c (x)` (y) (d) `( z ) (aa) (bb) x^(( c c ) (dd) (ee)1/( f f )3( g g ) (hh) (ii))( j j )-( k k ) y^(( l l ) (mm) (nn)1/( o o )3( p p ) (qq) (rr))( s s )=c (tt)` (uu)

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)=c`