Question

The orthogonal trajectories of the family of curves `y=Cx^(2)`, (C is an arbitrary constant), is
A. `x^(2)+2y^(2)=2C`
B. `2x^(2)+y^(2)=2C`
C. `x^(2)+y^(2)=2C`
D. `x^(2)-2y^(2)=2C`

Answer 1

Correct Answer - A
The equation of the given family of curves is
`y=Cx^(2)" …(i)"`
Differentiating (i) w.r.t. x, we get
`(dy)/(dx)=2Cx" …(ii)"`
Eliminating C between (i) and (ii), we obtain
`y=((1)/(2x)(dy)/(dx))x^(2)rArr2y=x(dy)/(dx)" ...(iii)"`
This is the differential equation of the family of curves given in (i).
The differential equaiton of the orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)` by `-(dx)/(dy)` in equation (iii).
Replacing `(dy)/(dx)` by `(-dx)/(dy)` in (iiii), we obtain
`2y=-x(dx)/(dy)rArr 2ydy=-xdx`
On integrating, we obtain
`y^(2)=-(x^(2))/(2)+CrArr x^(2)+2y^(2)=2C`
This is the required family of orthogonal trajectories.