This is going to involve a lot of TeX, so will take rather more time thanOriginally Posted bybraddy

I have available to do it in one go, so I will do this incrementally.

(PH you were wise to leave your post to be checked, you seem to have

gotten the wrong end of the stick, as we say over here).

You want to find a family of functions/curves which are orthogonal

at every point to the parabola from the given family passing through

the point.

Lets start by deriving the equations of the family of parabolas with vertex

at and axes parallel to the vertical axis. Well they

are:

,

where parameterises the family. Now if one of these

parabolas passess through the point , we have:

,

so:

and so the parabola through is:

We are interested in the slope of this parabola at as

we are interested in the slope of the curve in the orthoganal family through

the point, which as it is orthogonal to the parabola has a slope of minus the

reciprical of the slope of the parabola at this point.

For the parabola we have:

so:

.

Hence the slope of the member of the orthogonal family through is:

,

so the ODE satisfied by the orthogonal trajectory is:

.

Which is of variables seperable type, and I think you will find that the

solution is a circle with centre .

RonL