Answer
The orthogonal trajectories of the family of curves
$$
y=ke^{x}
$$
are
$$
x=C-\frac{1}{2} y^{2}
$$
Work Step by Step
To find the orthogonal trajectories of the family of curves
$$
y=ke^{x}
$$
First, we differentiate both sides:
$$
\frac{d}{d x}(y)=\frac{d}{d x}\left(k e^{x}\right) \Rightarrow y^{\prime}=k e^{x}=y,
$$
An orthogonal trajectory is the negative reciprocal of the tangent line slope. So the orthogonal trajectories must have:
$$
y^{\prime}=-\frac{1}{y} \Rightarrow \frac{d y}{d x}=-\frac{1}{y}
$$
We write the equation in terms of differentials and integrate both sides:
$$
y d y=-d x \Rightarrow \int y d y=-\int d x
$$
$
\Rightarrow
$
$$
\frac{1}{2} y^{2}=-x+C \Rightarrow x=C-\frac{1}{2} y^{2},
$$
Which are parabolas with a horizontal axis and $ C $ is an arbitrary constant.
So the orthogonal trajectories of the family of curves
$$
y=ke^{x}
$$
are
$$
x=C-\frac{1}{2} y^{2}
$$
