Question

The orthogonal trajectories of the family of curves 2 = 2y - 1+ Ce"where is an arbitrary constant is given by the family of curves 2y=**** where x arbitrary constant º + where x ban arbitrary constant None of them + y = + where Karbary.com + is where is an arbitrary constant

Answer 1

In this problem first, we find the slope in the curve given and the apply the orthogonal trajectory slope concept to find the solpe of the orthogonal trajectory.

and then put the value of C in terms of orthogonal trajectory slope in the original equation to eliminate it and form a differential equation and finally calculate the required family of the curve.

comment if any query.

x² = 24-1+ cesy axax =ady-acezydy =ady -acety dsc dx. dx dx. y (1-ce24 x = 1-cetry yi Now, we consider thogonal trajectory with slope M so, ma= -1 Sm, = y) . m =-ce-dy put it in equation

x2 = 2y-(- ceny) x² =24+ x m₂ x² =2y + x dy Here Now, osc represent slope o outhogonal Sody t ау = 32 ato a S % dx I.= e e Banu? 23 x²y = ² x da tk. xey sch where Kis an arbitrally constant so option (a) is