Answer
$$
y^{2}=2Cx, \quad\quad \text{(parabolas)}
$$
the orthogonal trajectories of the given family are
$$
y^{2}+ 2x^{2}=K, \quad\quad \text{(ellipse)}
$$
where $K$ is an arbitrary constant.
Work Step by Step
$$
y^{2}=2Cx, \quad\quad \text{(parabolas)}
$$
First, solve the given equation by differentiating implicitly with respect to $x$ we obtain the differential equation:
$$
2y y^{\prime}=2C
$$
$\Rightarrow $
$$
\begin{aligned} y^{\prime} &=\frac{C}{y}\\
\\ &=\frac{1}{y}.\frac{y^{2}}{2x}\\
& =\frac{y}{2x}
\end{aligned}
$$
Slope of given family is
$$
y^{\prime}=\frac{y}{2x}
$$
Because represents the slope of the given family of curves at it follows that
the orthogonal family has the negative reciprocal slope so,
$$
y^{\prime}=\frac{-2x}{y}
$$
Now you can find the orthogonal family by separating variables and integrating.
$$
\int (y ) dy=\int (-2x)dx
$$
$\Rightarrow$
$$
\frac{1}{2}y^{2}=-x^{2}+K \Rightarrow y^{2}+ 2x^{2}=K, \quad\quad \text{(ellipse)}
$$
where $K$ is an arbitrary constant.
