Answer
\[\frac{x^2-y^2}{x^2y^2}=K\]
Work Step by Step
$x^4+y^4=c$ ____(1)
Differentiate (1) with respect to $x$
\[4x^3+4y^3\frac{dy}{dx}=0\]
$\Large\frac{dy}{dx}=-\Large\frac{x^3}{y^3}$ ___(2)
Replace $\Large\frac{dy}{dx}$ by $-\Large\frac{dx}{dy}$ in (2) [ For orthogonal trajectories]
\[-\frac{dx}{dy}=-\frac{x^3}{y^3}\]
Separating variables,
\[x^{-3}dx=y^{-3}dy\]
Integrating,
\[\int x^{-3}dx=\int y^{-3}dy+C_{1}\]
$C_{1}$ is constant of integration
\[\frac{x^{-2}}{-2}=\frac{y^{-2}}{-2}+C_{1}\]
\[\frac{1}{2y^2}-\frac{1}{2x^2}=C_{1}\]
$\Large\frac{x^2-y^2}{x^2y^2}$=$K$, where $K=2C_{1}$
Hence orthogonal trajectories of (1) is $\Large\frac{x^2-y^2}{x^2y^2}$=$K$
