Answer
\[y=ke^{-\frac{2}{m}x}\]
Work Step by Step
Given that, $y^2=mx+c$ _____(1)
Differentiate with respect to $x$
\[2y\frac{dy}{dx}=m\]
$\frac{dy}{dx}=\frac{m}{2y}$ ______(2)
Replace $\frac{dy}{dx}$ by -$\frac{dx}{dy}$ in (2) [ For orthogonal Trajectories]
\[-\frac{dx}{dy}=\frac{m}{2y}\]
\[-2\frac{dx}{m}=\frac{dy}{y}\]
Integrating, $-\frac{2}{m}\int dx=\int \frac{dy}{y}$
$lnk-\frac{2}{m}x=lny$, where $lnk$ is constant of integration
$\frac{-2}{m}x=ln\frac{y}{k}$
$y=ke^{-\frac{2}{m}x}$
Hence, family of orthogonal trajectories to (1) is $y=ke^{-\frac{2}{m}x}$
