Answer
\[y=ke^{-\frac{2}{m}x}\]
![](https://gradesaver.s3.amazonaws.com/uploads/solution/4612b7ee-79a7-4e6d-bbc1-3005ad12127d/result_image/1595876916.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20250218%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20250218T120320Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=6e554cedc1b026bc622eedce5c3f3a7ae574b14fd39eb32751488e41cdd03525)
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 $y=ke^{-\frac{2}{m}x}$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/4612b7ee-79a7-4e6d-bbc1-3005ad12127d/steps_image/small_1595876916.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20250218%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20250218T120320Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=4300e9312ef9623367df80b87c838e4bf78141a9092d366ddf755313c3b673df)