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