Answer
$y= Kx^{\frac{1}{m}}$
Work Step by Step
$y^2+mx^2= c$______(1)
Differentiating with respect to $x$
\[2y\frac{dy}{dx}+2mx=0\]
$\frac{dy}{dx}=\frac{-mx}{y}$ _____(2)
Replace $\frac{dy}{dx}$ by -$\frac{dx}{dy}$ in (2) [For orthogonal Trajectories]
\[-\frac{dx}{dy}=-\frac{mx}{y}\]
\[\frac{dx}{dy}=\frac{mx}{y}\]
\[\frac{dx}{x}=m\frac{dy}{y}\]
Integrating, $\int \frac{dx}{x}=m\int \frac{dy}{y} $
$lnk+lnx=mlny$, $lnk$ is constant of integration
$lnkx=lny^m$
$y=Kx^{\frac{1}{m}}$, where $K=k^{\frac{1}{m}}$
Hence family of orthogonal trajectories of (1) is $y=Kx^{\frac{1}{m}}$.
