= Solution
The seven requested <parabolas> are $y=cx^2$ for $c=-3,-2,-1,0,1,2,3$; the case $c=0$ is the horizontal axis. Positive $c$ opens upward, negative $c$ downward, and larger $|c|$ gives a narrower <parabola>. All pass through the origin with a horizontal <tangent line>.
To find <orthogonal trajectories>, use the continuous family $y=cx^2$ containing these seven representative members. Its slope is $y'=2cx=2y/x$ away from $x=0$. Two finite slopes of perpendicular tangents multiply to $-1$, so the orthogonal slope is
$$
\frac{dy}{dx}=-\frac{x}{2y}.
$$
This <separable differential equation> gives $2y\,dy=-x\,dx$, hence
$$
\boxed{x^2+2y^2=C,\qquad C>0.}
$$
These <ellipses> are the regular real trajectories; $C=0$ is a single point, not a regular curve, and $C<0$ has no real points. One example is $x^2+2y^2=2$, with semiaxes $\sqrt2$ horizontally and $1$ vertically.
The implicit tangent relation $2x\,dx+4y\,dy=0$ also describes points with vertical tangents, where the slope quotient is unavailable. At intersections with the member $y=0$, the <ellipse> has a vertical <tangent line> and is perpendicular to that horizontal axis. The common origin of the <parabolas> is not on any regular trajectory with $C>0$. The sketch shows the requested seven members and the example <ellipse>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1-orthogonal-trajectories.png]
{title=Seven parabolic family members and an elliptical orthogonal trajectory}
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