The equation of the family of curves which intersect the hyperbola xy=2 orthogonally is

Options

A. `y=(x^(3))/(6)+C`
B. `y=(x^(2))/(4)+C`
C. `y=(-x^(3))/(6)+C`
D. `y=(-x^(2))/(4)+C`

Answer

A

Explanation

We have,
`xy=2rArry=(2)/(x)rArr(dy)/(dx)=-(2)/(x^(2))`
Let y = f(x) be the required family of curves. Then,
`((dy)/(dx))_(C_(1))xx((dy)/(dx))_(C_(2))=-1`
`rArr" "(dy)/(dx)xx(-2)/(x^(3))=-1rArr(dy)/(dx)=(x^(2))/(2)rArry=(x^(3))/(6)+C`
This is the required family of curves.