Question

The family of orthogonal trajectories of $xy=C_1$ is:
(where $C_1,C_2$ are arbitrary constants)

• A. $2x^2-y^2=C_2$
• B. $x^2-2y^2=C_2$
• C. $x^2-y^2=C_2$
• D. $x^2+y^2=C_2$

Answer 1

The correct option is C $x^2-y^2=C_2$

Differentiating the given expression w.r.t. $x,$ we get
$x\frac{dy}{dx}+y=0$
To get the orthogonal trajectory, replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$
$x\frac{dx}{dy}-y=0\Rightarrow xdx-ydy=0$
Integrating both sides, we get
$\Rightarrow x^2-y^2=C_2$
This is the family of the required orthogonal trajectories.