# Thread: Help for family of curves.

1. ## Help for family of curves.

Hi please can I have some help with this problem:
It is about first order ODE.

Give the family of orthogonal trajectories of the family of parabolas with vertical axis and vertex at the point ( 4 , 2 ).

The solution is
(x-4)^2 +(y-2)^2=C

But I dont know how to find it.
Thank you
B

Hi please can I have some help with this problem:
It is about first order ODE.

Give the family of orthogonal trajectories of the family of parabolas with vertical axis and vertex at the point ( 4 , 2 ).

The solution is
(x-4)^2 +(y-2)^2=C

But I dont know how to find it.
Thank you
B
This is going to involve a lot of TeX, so will take rather more time than
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 $( 4,2)$ and axes parallel to the vertical axis. Well they
are:

$
(y-2)=\lambda (x-4)^2
$
,

where $\lambda$ parameterises the family. Now if one of these
parabolas passess through the point $(u,v)$, we have:

$
v-2= \lambda (u-4)^2
$
,

so:

$
\lambda=\frac{v-2}{(u-4)^2}
$

and so the parabola through $(u,v)$ is:

$
y-2= \frac{v-2}{(u-4)^2} (x-4)^2
$

We are interested in the slope of this parabola at $(u,v)$ 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:

$
\frac{dy}{dx}=\frac{v-2}{(u-4)^2}\ 2\ (x-4)
$

so:

$
\frac{dy}{dx}\left|_{{x=u}\atop{y=v}}=2\ \frac{v-2}{u-4}
$
.

Hence the slope of the member of the orthogonal family through $(u,v)$ is:

$
\frac{dy}{dx}\left|_{{x=u} \atop {y=v}}=-\frac{1}{2}\ \frac{u-4}{v-2}
$
,

so the ODE satisfied by the orthogonal trajectory is:

$
\frac{dy}{dx}=-\frac{1}{2}\ \frac{x-4}{y-2}
$
.

Which is of variables seperable type, and I think you will find that the
solution is a circle with centre $(4,2)$.

RonL